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Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\). (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.

Short Answer

Expert verified
(a) To calculate the standard Gibbs free energy change (螖G鈦), first find the standard enthalpy change (螖H鈦) and standard entropy change (螖S鈦) using the data in Appendix C: 螖H鈦(reaction) = 危 螖H鈦(products) - 危 螖H鈦(reactants) and 螖S鈦(reaction) = 危 螖S鈦(products) - 危 螖S鈦(reactants). Then, use the equation 螖G鈦 = 螖H鈦 - T螖S鈦 to find the standard Gibbs free energy change. (b) To calculate 螖G at 298 K given the partial pressures, first find the reaction quotient (Q) using the formula Q = (P(N鈧侽鈧))/(P(NO鈧)虏) with the given partial pressures of NO鈧 and N鈧侽鈧 as 0.40 atm and 1.60 atm, respectively. Next, use the equation 螖G = 螖G鈦 + RT ln(Q) to find 螖G, where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K).

Step by step solution

01

Find the standard enthalpy change and standard entropy change

Look up the standard enthalpy change (螖H鈦) and the standard entropy change (螖S鈦) for the reaction in Appendix C. We can find the values for the individual species and use the equation 螖H鈦(reaction) = 危 螖H鈦(products) - 危 螖H鈦(reactants) and 螖S鈦(reaction) = 危 螖S鈦(products) - 危 螖S鈦(reactants).
02

Calculate the standard Gibbs free energy change

Use the equation 螖G鈦 = 螖H鈦 - T螖S鈦 to find the standard Gibbs free energy change. (b) Calculate 螖G at 298K given the partial pressures.
03

Find the reaction quotient (Q)

The reaction quotient Q is defined as the ratio of the product of partial pressures of products to the product of partial pressures of reactants. In this case, Q = (P(N鈧侽鈧))/(P(NO鈧)虏). Given the partial pressures of NO鈧 and N鈧侽鈧 as 0.40 atm and 1.60 atm respectively, calculate the value of Q.
04

Calculate 螖G using the reaction quotient

Use the equation 螖G = 螖G鈦 + RT ln(Q) to find 螖G. R is the gas constant (8.314 J/mol K), T is the temperature (298 K), and Q is the reaction quotient calculated in step 3. Following these steps above will give you the answers for 螖G鈦 and 螖G at 298 K for the given reaction and partial pressures.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Enthalpy Change
In a chemical reaction, the standard enthalpy change, \( \Delta H^{\circ} \), is a way to express the heat absorbed or released under standard conditions (1 atm pressure and 298 K). It is calculated using the equation:
  • \( \Delta H^{\circ}(\text{reaction}) = \Sigma \Delta H^{\circ}(\text{products}) - \Sigma \Delta H^{\circ}(\text{reactants}) \)
Standard enthalpy values are usually provided for individual substances in tables like Appendix C of textbooks.
By using these values, you can determine if the reaction releases heat (exothermic, \( \Delta H^{\circ} < 0 \)) or absorbs heat (endothermic, \( \Delta H^{\circ} > 0 \)). This helps in predicting how energy changes during the formation of products from reactants.
Remember to ensure your units match when making calculations, as enthalpy is often given in kilojoules per mole (kJ/mol). Understanding \( \Delta H^{\circ} \) is crucial for assessing the energy profile of chemical reactions.
Standard Entropy Change
Entropy, represented by \( \Delta S^{\circ} \), measures the degree of disorder or randomness in a system. The standard entropy change of a reaction indicates how the entropy of the system changes when reactants convert to products under standard conditions.
  • \( \Delta S^{\circ}(\text{reaction}) = \Sigma \Delta S^{\circ}(\text{products}) - \Sigma \Delta S^{\circ}(\text{reactants}) \)
A positive \( \Delta S^{\circ} \) indicates an increase in disorder, implying more randomness in the products compared to reactants. Conversely, a negative \( \Delta S^{\circ} \) suggests that the reaction results in a more ordered system.
Changes in entropy are a crucial component when calculating Gibbs Free Energy, as they provide insight into the feasibility of a reaction proceeding under certain conditions. Thermodynamics tells us that reactions with increasing entropy are generally more favored, from an order-disorder perspective.
Reaction Quotient
The reaction quotient, \( Q \), is a snapshot of a reaction's position at any given point in time, distinct from the equilibrium state. It is calculated using the partial pressures or concentrations of the reactants and products at a particular moment:
  • For the given reaction, \( Q = \frac{P_{\text{products}}}{P_{\text{reactants}}^{\text{coefficients}}} \)
In our specific case:
  • \( Q = \frac{P(\mathrm{N}_{2} \mathrm{O}_{4})}{P(\mathrm{NO}_{2})^{2}} \)
Here, \( Q \) helps us compare the current reaction concentration to the equilibrium state, aiding in predicting which direction the reaction will shift to reach equilibrium.
If \( Q < K \), the reaction will move towards the products. If \( Q > K \), the reaction will favor the formation of reactants. Thus, \( Q \) plays a pivotal role in understanding real-time reaction dynamics and is integral in computing non-standard Gibbs Free Energy changes.
Partial Pressures
Partial pressures are a critical concept in understanding gas mixtures. Each gas in a mixture exerts pressure independently of other gases, known as its partial pressure. For a reaction involving gases, like the one in our exercise, understanding partial pressures is key to calculating reactions under non-standard conditions.
The total pressure is the sum of the partial pressures of all gases present, based on Dalton's Law:
  • Total Pressure = sum of all individual partial pressures
In terms of chemical reactions, the partial pressures of reactants and products help determine the reaction quotient \( Q \).
For instance, in the reaction \(2 \mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}\), if \(P(\mathrm{NO}_{2})\) is 0.40 atm and \(P(\mathrm{N}_{2} \mathrm{O}_{4})\) is 1.60 atm, these values are plugged into the expression for \( Q \) to understand how far the reaction is from equilibrium. Partial pressures thus serve as a vital tool in thermodynamic calculations for gaseous systems, providing valuable insight into the state of chemical processes.

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Most popular questions from this chapter

For the majority of the compounds listed in Appendix \(\mathrm{C},\) the value of \(\Delta G_{f}^{\circ}\) is more positive (or less negative) than the value of \(\Delta H_{f}^{\circ} .\) (a) Explain this observation, using \(\mathrm{NH}_{3}(g), \mathrm{CCl}_{4}(l)\), and \(\mathrm{KNO}_{3}(s)\) as examples. (b) An exception to this observation is \(\mathrm{CO}(g)\). Explain the trend in the \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) values for this molecule.

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C,\) calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500{ }^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S) ?(\mathbf{c})\) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7 . (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

For each of the following pairs, indicate which substance possesses the larger standard entropy: (a) \(1 \mathrm{~mol}\) of \(\mathrm{P}_{4}(g)\) at \(300^{\circ} \mathrm{C}, 0.01 \mathrm{~atm},\) or \(1 \mathrm{~mol}\) of \(\mathrm{As}_{4}(g)\) at \(300^{\circ} \mathrm{C}, 0.01 \mathrm{~atm} ;\) (b) \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm},\) or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm} ;\) (c) \(0.5 \mathrm{~mol}\) of \(\mathrm{N}_{2}(g)\) at \(298 \mathrm{~K}, 20-\mathrm{L}\) volume, or \(0.5 \mathrm{~mol} \mathrm{CH}_{4}(g)\) at \(298 \mathrm{~K}, 20-\mathrm{L}\) volume; (d) \(100 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) at \(30^{\circ} \mathrm{C}\) or \(100 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) at \(30^{\circ} \mathrm{C}\)

Isomers are molecules that have the same chemical formula but different arrangements of atoms, as shown here for two isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12} .\) (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. [Section 19.4\(]\)

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated under atmospheric pressure. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

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