AP Chemistry : Thermodynamics

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

Example Question #6 : Entropy

\displaystyle NO_2               \displaystyle \Delta S_f=240J

\displaystyle O_{2(g)}               \displaystyle \Delta S_f=205J

\displaystyle N_{2(g)}               \displaystyle \Delta S_f=191.5J

What is the change in entropy for the following reaction?

\displaystyle N_{2(g)}+O_{2(g)}\rightarrow NO_2

Possible Answers:

\displaystyle -348\frac{J}{mol}

\displaystyle -121.5\frac{J}{mol}

\displaystyle -553\frac{J}{mol}

\displaystyle 83.5\frac{J}{mol}

\displaystyle -156.5\frac{J}{mol}

Correct answer:

\displaystyle -121.5\frac{J}{mol}

Explanation:

The change in entropy is calculated by:

\displaystyle \Delta S_{products}-\Delta S_{reactants}

When \displaystyle \Delta S cannot be measured, it can be calculated from known entropies of formation.

\displaystyle NO_2               \displaystyle \Delta S_f=240J

\displaystyle O_{2(g)}               \displaystyle \Delta S_f=205J

\displaystyle N_{2(g)}               \displaystyle \Delta S_f=191.5J

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

\displaystyle N_{2(g)}+2O_{2(g)}\rightarrow 2NO_2

\displaystyle \Delta S_{rxn}=2(\Delta S_{NO_2})-\Delta S_{N_{2(g)}}-2(\Delta S_{O_{2(g)}})

\displaystyle \Delta S_{rxn}=2(240J)-191.5J-2(205J)

\displaystyle \Delta S_{rxn}=-121.5\frac{J}{mol}

Example Question #72 : Thermochemistry And Kinetics

\displaystyle Mg                     \displaystyle \Delta S_f=32.7J

\displaystyle O_2                       \displaystyle \Delta S_f=205J

\displaystyle MgO                 \displaystyle \Delta S_f=26.9J

What is the change in entropy for the following reaction?

\displaystyle 2Mg +O_2\rightarrow 2MgO

Possible Answers:

\displaystyle -816.6\frac{J}{mol}

\displaystyle 199.2\frac{J}{mol}

\displaystyle -216.6\frac{J}{mol}

\displaystyle -433.5\frac{J}{mol}

\displaystyle -73.9\frac{J}{mol}

Correct answer:

\displaystyle -216.6\frac{J}{mol}

Explanation:

The change in entropy is calculated by:

\displaystyle \Delta S_{products}-\Delta S_{reactants}

When \displaystyle \Delta S cannot be measured, it can be calculated from known entropies of formation.

\displaystyle Mg                     \displaystyle \Delta S_f=32.7J

\displaystyle O_2                       \displaystyle \Delta S_f=205J

\displaystyle MgO                 \displaystyle \Delta S_f=26.9J

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

\displaystyle 2Mg +O_2\rightarrow 2MgO

\displaystyle \Delta S_{rxn}=2(26.9J)-2(32.7J)-205J

\displaystyle \Delta S_{rxn}=-216.6\frac{J}{mol}

Example Question #1 : Entropy

\displaystyle C_4H_{10}                             \displaystyle \Delta S_f=310J

\displaystyle O_2                                    \displaystyle \Delta S_f=205J

\displaystyle CO_2                                \displaystyle \Delta S_f=213.6J

\displaystyle H_2O                                \displaystyle \Delta S_f=69.9J

What is the change in entropy for the following equation?

\displaystyle C_4H_{10}+O_2\rightarrow CO_2+H_2O

Possible Answers:

\displaystyle -17.9\frac{J}{mol}

\displaystyle -877.2\frac{J}{mol}

\displaystyle -21.8\frac{J}{mol}

\displaystyle -362.8\frac{J}{mol}

\displaystyle -231.5\frac{J}{mol}

Correct answer:

\displaystyle -877.2\frac{J}{mol}

Explanation:

The change in entropy is calculated by:

\displaystyle \Delta S_{products}-\Delta S_{reactants}

When \displaystyle \Delta S cannot be measured, it can be calculated from known entropies of formation.

\displaystyle C_4H_{10}                             \displaystyle \Delta S_f=310J

\displaystyle O_2                                    \displaystyle \Delta S_f=205J

\displaystyle CO_2                                \displaystyle \Delta S_f=213.6J

\displaystyle H_2O                                \displaystyle \Delta S_f=69.9J

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

\displaystyle 2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

\displaystyle \Delta S_{rxn}=8(\Delta S_{CO_2})+10(\Delta S_{H_2O})-2(\Delta S_{C_2H_{10}})-13(\Delta S_{O_2})

\displaystyle \Delta S_{rxn}=8(213.6J)+10(69.9J)-2(310J)-13(205J)

\displaystyle \Delta S_{rxn}=-877.2\frac{J}{mol}

Example Question #1 : Entropy

\displaystyle A                    \displaystyle \Delta S_f=-10J

\displaystyle B                    \displaystyle \Delta S_f=+15J

\displaystyle C                    \displaystyle \Delta S_f=-75J

\displaystyle D                    \displaystyle \Delta S_f=-45J

What is the entropy for the following reaction?

\displaystyle A+B\rightarrow C+D

Possible Answers:

\displaystyle -5\frac{J}{mol}

\displaystyle -25\frac{J}{mol}

\displaystyle -35\frac{J}{mol}

\displaystyle -125\frac{J}{mol}

\displaystyle -145\frac{J}{mol}

Correct answer:

\displaystyle -125\frac{J}{mol}

Explanation:

The change in entropy is calculated by:

\displaystyle \Delta S_{products}-\Delta S_{reactants}

When \displaystyle \Delta S cannot be measured, it can be calculated from known entropies of formation.

\displaystyle A                    \displaystyle \Delta S_f=-10J

\displaystyle B                    \displaystyle \Delta S_f=+15J

\displaystyle C                    \displaystyle \Delta S_f=-75J

\displaystyle D                    \displaystyle \Delta S_f=-45J

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

\displaystyle A+B\rightarrow C+D

\displaystyle \Delta S_{rxn}=\Delta S_{C}+\Delta S_{D}-\Delta S_{A}-\Delta S_{B}

\displaystyle \Delta S_{rxn}=-75J+(-45J)-(-10J)-15J

\displaystyle \Delta S_{rxn}=-125\frac{J}{mol}

Example Question #1 : Entropy

Which reaction will result in a negative entropy?

Possible Answers:

Correct answer:

Explanation:

A reaction will have a positive change in entropy when the disorder of the system is increasing. This can happen when a liquid transitions to a gas, or more gas molecules are formed in the reaction.

Since two molecules of gaseous nitrogen dioxide can combine to form one molecule of dinitrogen tetroxide, the amount of gaseous molecules is halved. This increases order, which decreases the entropy of the system.

Example Question #41 : Thermodynamics

A reaction will reach equilibrium when the system's entropy __________.

Possible Answers:

entropy does not influence equilibrium at all

is at a minimum

is at a maximum

is zero

Correct answer:

is at a maximum

Explanation:

Equilibrium is the basis of entropy, which can essentially be thought of as the distribution of energy in the system. For a reaction at a given temperature, the system's entropy will be at a maximum when it reaches equilbrium because this is the ideal energy state, and the forward and reverse reactions are occurring at a steady rate.

Example Question #72 : Thermochemistry And Kinetics

Consider the following reaction:

Below is a list of the standard entropy of formation for the compounds in the reaction:

\displaystyle \Delta S_{N_{2}} = 191.49\frac{J}{molK}

\displaystyle \Delta S_{H_{2}} = 130.59\frac{J}{molK}

\displaystyle \Delta S_{NH_{3}} = 192.5\frac{J}{molK}

What is the entropy of the reaction when one mole of ammonia is formed?

Possible Answers:

\displaystyle -65\frac{J}{molK}

\displaystyle 968\frac{J}{molK}

\displaystyle -99\frac{J}{molK}

\displaystyle -198\frac{J}{molK}

Correct answer:

\displaystyle -99\frac{J}{molK}

Explanation:

Since we have the entropies of formation, we can find the entropy of the reaction using the following equation:

\displaystyle \Delta S_{reaction} = \Delta S_{products} - \Delta S_{reactants}

We will need to use the coefficients from the balanced equation to calculate the entropy.

\displaystyle \Delta S_{reaction}=2(\Delta S_{NH_3})-\Delta S_{N_2}-3(\Delta S_{H_2})

\displaystyle \Delta S_{reaction}=2(192.5\frac{J}{molK})-(191.49\frac{J}{molK})-3(130.59\frac{J}{molK})

\displaystyle \Delta S_{reaction}=-197.77\frac{J}{molK}

Since this is the entropy for the reaction when two moles of ammonia are formed, we must divide this entropy by two in order to solve for the entropy of one mole.

\displaystyle \frac{197.77\frac{J}{molK}}{2}=-99\frac{J}{molK}

Example Question #1 : Gibbs Free Energy And Spontaneity

Suppose that a rxn has ∆H = -28 kJ and ∆S= -60 J/K. At what temperature will it
change from spontaneous to non-spontaneous?

Possible Answers:

500 K

467 K

50

46.7 K

Correct answer:

467 K

Explanation:

Approximately 467 K. ∆G=∆H-T∆S and a rxn proceeds spontaneously when ∆G < 0
and is non-spontaneous when ∆G > 0. So if we set ∆G=0 and solve the equation for T, we will see that the crossover from spontaneous to non-spontaneous occurs when T=467K.

Example Question #1 : Gibbs Free Energy And Spontaneity

If the reaction quotient (Q) is greater than the equilibrium constant (K), what is true about the Gibbs free energy?

Possible Answers:

It is equal to zero.

It is greater than zero.

More information is needed to determine the gibbs free energy.

It is less than zero.

Correct answer:

It is greater than zero.

Explanation:

If Q is greater than K, the reaction has exceeded the equilibrium state. It will proceed nonspontaneously (since equilibrium has already been reached), and this must mean that the ΔG (gibbs free energy) must be positive, or greater than zero. 

Example Question #2 : Gibbs Free Energy And Spontaneity

The entropy and enthalpy of a reaction are both negative. Is the reaction spontaneous?

Possible Answers:

The reaction will be spontaneous if and only if the magnitude of the entropy is greater than the magnitude of the enthalpy times the temperature

The reaction will be spontaneous if and only if the magnitude of the enthalpy is greater than the magnitude of the entropy

The reaction is not spontaneous

The reaction will be spontaneous if and only if the magnitude of the enthalpy is greater than the magnitude of the entropy times the temperature

The reaction is spontaneous

Correct answer:

The reaction will be spontaneous if and only if the magnitude of the enthalpy is greater than the magnitude of the entropy times the temperature

Explanation:

A reaction is spontaneous if the Gibb's Free Energy of the reaction is negative.

\displaystyle \Delta G=\Delta H-T\Delta S

If \displaystyle \Delta H, the enthalpy, and \displaystyle \Delta S, the entropy, are both negative, then the reaction will be spontaneous if and only if the magnitude of the enthalpy is greater than the magnitude of the entropy times the temperature.

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