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GRE Subject Test: Chemistry : Physical Chemistry

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Example Question #1 : Reaction Rate And Rate Law

$A + 2B + C \rightarrow D$

You want to find the order of each reactant by manipulating the reactant concentrations in multiple trials. The table below shows the effect of altered reactant concentrations on initial reaction rate.

$\begin{matrix} \text{Trial} & \text{Initial [A]} & \text{Initial [B]} & \text{Initial [C]} & \text{Rate}\ (\frac{M}{s})\\ 1&0.1 &0.3 &0.2 &0.002 \\ 2& 0.1& 0.9& 0.2& 0.006\\ 3& 0.2& 0.3& 0.2& 0.002\\ 4& 0.1& 0.3& 0.4 & 0.008 \end{matrix}$

Using the above trials, write the rate law for the reaction.

Possible Answers:

$Rate = k[A][B][C]^{2}$

$Rate = k[B][C]^{2}$

Rate=k[A][B][C]

$Rate = k[A][B]^{2}[C]$

$Rate = k[B]^{2}[C]^{4}$

Correct answer:

$Rate = k[B][C]^{2}$

Explanation:

Keep in mind that the chemical equation and its coefficients have NOTHING to do with the rate law. The order of each reactant must be determined by experiment.

To find the order of each reactant, compare the initial reaction rates of two trials in which only one of the three reactants' concentrations is altered. For example, trials 1 and 4 keep A and B equal, but C is doubled. When C is doubled, we see that the initial reaction rate is quadrupled. As a result, we determine that the order of reactant C is 2. When the reactant is altered, but the initial reaction rate is kept constant, as seen in trials 1 and 3 with respect to A, the order of that reactant is 0. Finally, when the reactant is multiplied by the same factor that the initial reaction rate is multiplied, as seen in trials 1 and 2 with respect to B, the order of the reactant is 1.

Putting the data together: A is zeroth order, B is first order, and C is second order. Our rate law can thus be written Rate=k[B][C]^2 .

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Example Question #1101 : Mcat Physical Sciences

A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:

$NH_{4}^{+}(aq) + NO_{2}^{-}(aq) \rightarrow N_{2}(g) +2H_{2}O(l)$

In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.

$\begin{matrix} Trial & [NH_4^+] & [NO_2^-] &Rate \\ 1& 0.480M &0.120M &0.018\frac{M}{s} \\ 2& 0.240M & 0.120M& 0.009\frac{M}{s}\\ 3& 0.240M& 0.360M & 0.027\frac{M}{s} \end{matrix}$

Which of the following most closely approximates the rate law for this reaction?

Possible Answers:

Rate=k[NH_4^+][NO_2^-]^2

Rate=k[NH_4^+]^2[NO_2^-]^2

Rate=k[NH_4^+]

Rate=k[NH_4^+]^2[NO_2^-]

Rate=k[NH_4^+][NO_2^-]

Correct answer:

Rate=k[NH_4^+][NO_2^-]

Explanation:

The reaction table in the passage indicates that the reaction rate varies in a 1-to-1 fashion as you vary the each reactant, while holding the other constant.

The rate law is written as Rate=k[NH_4^+]^x[NO_2^-]^y .

Compare trials 1 and 2 to see that doubling the ammonium concentration doubles the rate. The reaction is first order for ammonium: x=1 .

Compare trials 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate: y=1 .

The final rate law is Rate=k[NH_4^+][NO_2^-] .

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Example Question #1 : Kinetics And Kinematics

For the reaction, $2X\ +\ Y\rightarrow\ X_{2}Y$ , the initial rate of the reaction was determined and the values are tabulated below.

Screen shot 2015 11 27 at 5.48.47 am

What is the overall order for the rate law of this reaction?

Possible Answers:

Correct answer:

Explanation:

For the reaction given:

$rate\ of\ reaction=k[X]^{a}[Y]^{b}$

We need to find the reaction order, therefore we have to find the values of a and b in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of Y was doubled in run 2 while keeping the concentration of X constant. We can observe that the reaction rate was quadrupled. Let's compare by ratios:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[X]_{2}^{a}[Y]_{2}^{b}}{[X]_{1}^{a}[Y]_{1}^{b}} \right )$

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[Y]_{2}^{b}}{[Y]_{1}^{b}} \right )$

Plugging the values into the equation gives:

$\frac{3.0\frac{M}{s}}{0.75\frac{M}{s}}=\frac{(0.01M)^{b}}{(0.005M)^{b}}$

Simplifying the above equation gives:

$4.0=2.0^{b}$

Therefore, b=2

We can also compare the results of run number 2 and run number 3. In that case, the concentration of Y was kept constant while the concentration of X was doubled. We can see that the reaction rate doubled. Therefore the reaction is first order with respect to X and second order with respect to Y. Therefore, the reaction is third order overall:

$rate\ of\ reaction=k[X][Y]^{2}$

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Example Question #1 : Reaction Rates And Reaction Order

For the reaction, $A\ +\ B\rightarrow\ C\ +\ D$ , the initial rate of the reaction was determined and the values are tabulated below.

Screen shot 2015 11 27 at 5.58.58 am

Determine the overall order for the rate law of this reaction.

Possible Answers:

Correct answer:

Explanation:

For the reaction given:

$rate\ of\ reaction=k[A]^{x}[B]^{y}$

We need to find the reaction order, therefore we have to find the values of x and y in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of A was doubled in run 2 while keeping the concentration of B constant. We can observe that the reaction rate was doubled. Let's compare by ratios:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[A]_{2}^{x}[B]_{2}^{y}}{[A]_{1}^{x}[B]_{1}^{y}} \right )$

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

$\frac{rate_{2}}{rate_{1}}=\left ( \frac{[A]_{2}^{x}}{[A]_{1}^{x}} \right )$

Plugging the values into the equation gives:

$\frac{5.0\times 10^{-5}}{2.0\times 10^{-5}}=\frac{(4.0\times 10^{-3})^{x}}{(2.0\times 10^{-3})^{x}}$

Simplifying the above equation gives:

$2.0=2.0^{x}$

Therefore, x=1

We can also compare the results of run number 2 and run number 3. In that case, the concentration of A was kept constant while the concentration of B was decreased by half. We can see that the reaction rate decreased by half. Therefore the reaction is first order with respect to A and first order with respect to B. Therefore, the reaction is second order overall:

rate=k[A][B]

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Example Question #1141 : Mcat Physical Sciences

For any given chemical reaction, one can draw an energy diagram. Energy diagrams depict the energy levels of the different steps in a reaction, while also indicating the net change in energy and giving clues to relative reaction rate.

Below, a reaction diagram is shown for a reaction that a scientist is studying in a lab. A student began the reaction the evening before, but the scientist is unsure as to the type of the reaction. He cannot find the student’s notes, except for the reaction diagram below.

Untitled

Upon further review, the scientist realizes that the reaction in question involved formation of a carbocation that quickly reacted again to form stable products. At which point would we most likely find this carbocation in the above diagram?

Possible Answers:

Correct answer:

Explanation:

Point 3 is where you would expect to find a relatively stable intermediate. An intermediate is more stable than a transition state, but not as stable as the original reactants and final products. Stability is inversely proportional to energy, thus we are looking for the point that is between the highest and lowest energies in the reaction. By this logic, point 1 is the reactants, 2 and 4 are transition states, 3 is a stable intermediate, and 5 is the products.

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Example Question #31 : Physical Chemistry

What kinetic equation describes the rate of an E1 process?

Possible Answers:

None of these

$k[\text{R-Br}]$

$k[\text{R-X}][\text{Base}]$

$k[\text{R-Br}]^2$

$k[\text{Base}]$

Correct answer:

$k[\text{R-Br}]$

Explanation:

Most organic reactions are carried out in multiple steps. The rate equation can be derived from the process that occurs in the slowest step of the mechanism. An E1 reaction's slow step is when a leaving group separates from the hydrocarbon and a carbocation is formed. The only reactant in this step is one molar equivalent of the hydrocarbon. Thus, the rate equation only depends on that substance. The molar equivalent determines the exponent of each reactant in the equation.

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Example Question #8 : General Thermodynamics

Use the following values for water as needed.

$\Delta H^o_{fus}=334\frac{J}{g}$

$\Delta H^o_{vap}=2260\frac{J}{g}$

$c=4.186\frac{J}{g^oC}$

$\rho = 1\frac{g}{mL}$

If burning wood releases $\small 20kJ$ of heat energy per gram of wood consumed, what mass of wood must be consumed to heat $\small 1L$ of water from $\small 20^oC$ to $\small 100^oC$ , and then to convert it to water vapor?

Possible Answers:

17g

335g

134g

130g

2595g

Correct answer:

130g

Explanation:

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from $\small 20^oC$ to $\small 100^oC$ (use $\Delta Q = mc\Delta T$ )

2. Boiling the water at a constant temperature of $\small 100^oC$ (use $\Delta Q = m\Delta H^o_{vap}$ )

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

$m = \rho*V= 1\frac{g}{mL}*1000mL=1000g$

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

$\Delta Q = (1000g)(4.186\frac{J}{g^oC})(100^oC - 20^oC)=334,880J$

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

$\Delta Q =(1000g) (2260\frac{J}{g})= 2,260,000J$

Sum the energies for step 1 and step 2.

$334,880 J + 2,260,000 J = 2,594,880J\approx 2,595kJ$

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

$2595kJ*(\frac{1 g}{20kJ})=129.7g\approx 130g$

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Example Question #1 : General Thermodynamics

What was the final temperature of water if a 19 g sample of water absorbs 810.5 J of heat energy and heats up from $25^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Possible Answers:

$18^{o}C$

$44^{o}C$

$25^{o}C$

35^oC

Correct answer:

35^oC

Explanation:

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1^oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$810.5J=19g\times4.179\frac{J}{g^{o}C}\times(T_{f}-25^{o}C)$

$810.5J=79.40\frac{J}{^{o}C}\times(T_{f}-25^{o}C)$

$\frac{810.5J}{79.40\frac{J}{^{o}C}}=T_{f}-25^{o}C$

$10.21^{o}C=T_{f}-25^{o}C$

$T_{f}=35^{o}C$

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Example Question #1 : General Thermodynamics

An 81.9 g metal at $54.2^{o}C$ was put into 40.7g of water at $25^{o}C$ . The final temp of the water and metal was $29.3^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ .

Possible Answers:

$6.9\frac{J}{g^{o}C}$

$0.358\frac{J}{g^{o}C}$

$2.4\frac{J}{g^{o}C}$

$4.8\frac{J}{g^{o}C}$

Correct answer:

$0.358\frac{J}{g^{o}C}$

Explanation:

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(81.9g\cdot C\cdot (29.3^{o}C-54.2^oC)=40.7g\cdot4.18\frac{J}{g^o}C(29.3^{o}C-25^{o}C)$

$-(-2039g^{o}C\cdot C)=731J$

$2039g^{o}C\cdot\ C=731J$

$C=\frac{731J}{2039g^{o}C}=0.358\frac{J}{g^{o}C}$

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Example Question #1 : Heat And Temperature

Calculate the final temperature when a 44.0g sample of metal (specific heat of the metal= $0.52 \frac{J}{g^{o}C}$ ) at $45.0^{o}C$ is placed into 95.0g of water at $23.0^{o}C$ . (specific heat of water is $4.18\frac{J}{g^{o}C}$ )

Possible Answers:

$132^{o}C$

$24^{o}C$

$76^{o}C$

$17^{o}C$

Correct answer:

$24^{o}C$

Explanation:

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(44.0g\cdot 0.52\frac{J}{g^{o}C}\cdot (T_{f}-45.0^oC)=95.0g\cdot4.18\frac{J}{g^o}C(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}(T_{f}-45.0^{o}C))=397\frac{J}{^{o}C}(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-22.9\frac{J}{^{o}C}\cdot 45.0^{o}C)=397\frac{J}{^{o}C}\cdot T_{f}-397\frac{J}{^{o}C}\cdot 23.0^{o}C$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-1031J)=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}+1031J=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}-397\frac{J}{^{o}C}\cdot T_{f}=-1031J-9131J$

$-419\frac{J}{^{o}C}\cdot T_{f}=-10162J$

$T_{f}=\frac{-10162^{o}C}{-419.9}=24^{o}C$

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GRE Subject Test: Chemistry : Physical Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #1 : Reaction Rate And Rate Law

Example Question #1101 : Mcat Physical Sciences

Example Question #1 : Kinetics And Kinematics

Example Question #1 : Reaction Rates And Reaction Order

Example Question #1141 : Mcat Physical Sciences

Example Question #31 : Physical Chemistry

Example Question #8 : General Thermodynamics

Example Question #1 : General Thermodynamics

Example Question #1 : General Thermodynamics

Example Question #1 : Heat And Temperature

All GRE Subject Test: Chemistry Resources

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