Entropy - AP Chemistry

Card 0 of 156

Question

Which of the following does not represent an increase in entropy?

Answer

Of phases, from greatest to least entropy is gas>liquid>solid. Dissolving a salt into solution is an increase in entropy because the salt has greater space to move. The only decrease in entropy is condensation of water on glass (gas to liquid)

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Question

Which of the following reactions represents a negative change in entropy?

Answer

Entropy for phases are Gas>Liquid>Solid. A salt dissolving into solution has a positive entropy change (the solution becomes more chaotic). Therefore all of the listed reactions are a positive change in entropy, making E the correct answer.

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Question

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

Answer

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.

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Question

Which of the following scenarios describes a process with a negative entropy change?

Answer

Entropy can be thought of as the tendency for a system to become disordered, or random. Osmosis, melting ice, and a clean room becoming dirty are all examples of entropy at work. Each of these processes will occur spontaneously without the input of outside energy or work.

If you make a skyscraper, you are incorporating a series of compounds and building products into a single, distinct formation. Think of it this way: what are the odds that mother nature would spontaneously build a building? This process is very ordered, which means that entropy of the universe has decreased.

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Question

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

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

$\Delta S_f=210.7J$

What is the change in entropy for the following reaction?

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

Answer

The change in entropy is calculated by:

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

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

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

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

$\Delta S_f=210.7J$

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

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

$\Delta S_{rxn}=2(\Delta S_{NO})-\Delta S_{N_{2(g)}}-\Delta S_{O_{2(g)}}$

$\Delta S_{rxn}=2(210.7J)-191.5J-205J$

$\Delta S_{rxn}=24.9\frac{J}{mol}$

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Question

$NO_{(g)}$ $\Delta S_f=210.7J$

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

$NO_{2}$ $\Delta S_f=240J$

What is the change in entropy for the following reaction?

$NO_{(g)}+O_{2(g)} \rightarrow NO_2$

Answer

The change in entropy is calculated by:

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

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

$NO_{(g)}$ $\Delta S_f=210.7J$

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

$NO_{2}$ $\Delta S_f=240J$

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

$2NO_{(g)}+O_{2(g)}\rightarrow 2NO_2$

$\Delta S_{rxn}=2(\Delta S_{NO_2})-2(\Delta S_{NO_{(g)}})-\Delta S_{O_{2(g)}}$

$\Delta S_{rxn}=2(240J)-2(210.7J)-205J$

$\Delta S_{rxn}=-146.4\frac{J}{mol}$

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Question

$\Delta S_f=240J$

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

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

What is the change in entropy for the following reaction?

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

Answer

The change in entropy is calculated by:

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

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

$\Delta S_f=240J$

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

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

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

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

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

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

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

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Question

$\Delta S_f=32.7J$

$\Delta S_f=205J$

$\Delta S_f=26.9J$

What is the change in entropy for the following reaction?

$2Mg +O_2\rightarrow 2MgO$

Answer

The change in entropy is calculated by:

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

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

$\Delta S_f=32.7J$

$\Delta S_f=205J$

$\Delta S_f=26.9J$

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

$2Mg +O_2\rightarrow 2MgO$

$\Delta S_{rxn}=2(\Delta S_{MgO})-2(\Delta S_{Mg})-\Delta S_O_2{}$

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

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

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Question

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

$\Delta S_f=205J$

$\Delta S_f=213.6J$

$\Delta S_f=69.9J$

What is the change in entropy for the following equation?

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

Answer

The change in entropy is calculated by:

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

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

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

$\Delta S_f=205J$

$\Delta S_f=213.6J$

$\Delta S_f=69.9J$

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

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

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

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

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

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Question

$\Delta S_f=-10J$

$\Delta S_f=+15J$

$\Delta S_f=-75J$

$\Delta S_f=-45J$

What is the entropy for the following reaction?

$A+B\rightarrow C+D$

Answer

The change in entropy is calculated by:

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

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

$\Delta S_f=-10J$

$\Delta S_f=+15J$

$\Delta S_f=-75J$

$\Delta S_f=-45J$

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

$A+B\rightarrow C+D$

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

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

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

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Question

Consider the following reaction:

$N_{2}_{(g)} + 3H_{2}_{(g)} \rightarrow 2NH_{3}_{(g)}$

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

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

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

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

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

Answer

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

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

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

$N_{2}_{(g)} + 3H_{2}_{(g)} \rightarrow 2NH_{3}_{(g)}$

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

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

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

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.

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

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Question

Which reaction will result in a negative entropy?

Answer

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.

$2NO_{2}_{(g)} \rightarrow N_{2}O_{4}_{(g)}$

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.

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Question

Which of the following does not represent an increase in entropy?

Answer

Of phases, from greatest to least entropy is gas>liquid>solid. Dissolving a salt into solution is an increase in entropy because the salt has greater space to move. The only decrease in entropy is condensation of water on glass (gas to liquid)

Compare your answer with the correct one above

Question

Which of the following reactions represents a negative change in entropy?

Answer

Entropy for phases are Gas>Liquid>Solid. A salt dissolving into solution has a positive entropy change (the solution becomes more chaotic). Therefore all of the listed reactions are a positive change in entropy, making E the correct answer.

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

Which of the following scenarios describes a process with a negative entropy change?

Answer

Entropy can be thought of as the tendency for a system to become disordered, or random. Osmosis, melting ice, and a clean room becoming dirty are all examples of entropy at work. Each of these processes will occur spontaneously without the input of outside energy or work.

If you make a skyscraper, you are incorporating a series of compounds and building products into a single, distinct formation. Think of it this way: what are the odds that mother nature would spontaneously build a building? This process is very ordered, which means that entropy of the universe has decreased.

Compare your answer with the correct one above

Question

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

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

$\Delta S_f=210.7J$

What is the change in entropy for the following reaction?

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

Answer

The change in entropy is calculated by:

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

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

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

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

$\Delta S_f=210.7J$

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

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

$\Delta S_{rxn}=2(\Delta S_{NO})-\Delta S_{N_{2(g)}}-\Delta S_{O_{2(g)}}$

$\Delta S_{rxn}=2(210.7J)-191.5J-205J$

$\Delta S_{rxn}=24.9\frac{J}{mol}$

Compare your answer with the correct one above

Question

$NO_{(g)}$ $\Delta S_f=210.7J$

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

$NO_{2}$ $\Delta S_f=240J$

What is the change in entropy for the following reaction?

$NO_{(g)}+O_{2(g)} \rightarrow NO_2$

Answer

The change in entropy is calculated by:

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

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

$NO_{(g)}$ $\Delta S_f=210.7J$

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

$NO_{2}$ $\Delta S_f=240J$

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

$2NO_{(g)}+O_{2(g)}\rightarrow 2NO_2$

$\Delta S_{rxn}=2(\Delta S_{NO_2})-2(\Delta S_{NO_{(g)}})-\Delta S_{O_{2(g)}}$

$\Delta S_{rxn}=2(240J)-2(210.7J)-205J$

$\Delta S_{rxn}=-146.4\frac{J}{mol}$

Compare your answer with the correct one above

Question

$\Delta S_f=240J$

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

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

What is the change in entropy for the following reaction?

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

Answer

The change in entropy is calculated by:

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

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

$\Delta S_f=240J$

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

$\Delta S_f=32.7J$

$\Delta S_f=205J$

$\Delta S_f=26.9J$

What is the change in entropy for the following reaction?

$2Mg +O_2\rightarrow 2MgO$

Answer

The change in entropy is calculated by:

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

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

$\Delta S_f=32.7J$

$\Delta S_f=205J$

$\Delta S_f=26.9J$

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

$2Mg +O_2\rightarrow 2MgO$

$\Delta S_{rxn}=2(\Delta S_{MgO})-2(\Delta S_{Mg})-\Delta S_O_2{}$

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

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

Compare your answer with the correct one above

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