Interpreting Graphs and Tables - AP Chemistry
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If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time, 
= initial concentration of reactant at time 
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of 
. In this case, 
represents the 
-axis, 
represents the 
-axis, 
represents the 
-intercept, and 
represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
If a solution of water has a pressure of 200atm and a temperature of 474 degrees Celsius, what changes could be made to bring the sample to its critical point?
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
The critical point for water is at 218atm and 374 degrees Celcius, but on this graph it can be found at the end of the line between the gas and liquid portions of the water phase graph (you can see it end with a dot). This means we need to increase our pressure and decrease our temperature in order to reach the critical point from our initial conditions.
Compare your answer with the correct one above
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A chemistry student is trying to identify the reaction order for a reaction based on her interpretation of the above graph. The student identifies the reaction as a first-order reaction. Is this student correct in their conclusion? Why?
A first-order reaction can be expressed in different ways. One way is the following:
where:
= concentration of reactant at a given time,
= initial concentration of reactant at time
= rate constant of the reaction
= amount of time that has passed since the start of the reaction
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of . In this case, represents the -axis, represents the -axis, represents the -intercept, and represents the slope. Thus, the graph in the question stem is indeed indicative of a first-order reaction.
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
A first-order reaction can be expressed in different ways. One way is the following:
where:
By taking the natural logarithm of both sides, we can obtain an alternative form of the equation, expressed as:
This second equation is useful for graphing, because it fits the general equation of
In the case of a zeroth-order reaction, the equation is:
Therefore, for a zeroth-order reaction, a straight line with a negative slope will only result from a graph of the concentration of a reactant versus time.
And lastly, in the case of a second-order reaction, the equation is:
Hence, a graph of the inverse concentration of a reactant versus time will yield a straight line with a positive slope, and this is indicative of a second-order reaction.
Compare your answer with the correct one above