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Chapter 16: Problem 110

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 atm and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a 1.00-L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

Short Answer

Expert verified
In this reaction at the given conditions, we have the equilibrium constant \(K \approx 1.62 \times 10^{-9}\), \(\Delta G^{\circ} \approx -59.2\,\frac{\text{kJ}}{\text{mol}}\), and \(\Delta S^{\circ} \approx 149.5\,\frac{\text{J}}{\text{mol}\cdot\text{K}}\).

Step by step solution

01

Determine the initial moles of H2 and Br2

The initial pressures of H2 and Br2 are both 1.00 atm, which are equal. Given that the volume of the container is 1.00 L, we can use the ideal gas law (\(PV = nRT\)) to find the initial moles of the gases: For H2: $$n_{H_{2}} = \frac{P_{H_{2}}V}{RT} = \frac{1.00 \,\text{atm} \cdot 1.00\, \text{L}}{0.0821\, \frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot\text{K}} \cdot (25+273.15)\,\text{K}}\approx 0.0409\, \text{mol}$$ Similarly, for Br2: $$n_{Br_{2}} = \frac{P_{Br_{2}}V}{RT} = \frac{1.00 \,\text{atm} \cdot 1.00\, \text{L}}{0.0821\, \frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot\text{K}} \cdot (25+273.15)\,\text{K}}\approx 0.0409\, \text{mol}$$
02

Calculate the equilibrium concentrations of H2, Br2, and HBr

Given the number of H2 molecules at equilibrium (1.10 x 10^13), we can convert it to moles using Avogadro's number (\(6.022\times10^{23}\, \text{molecules/mol}\)): $$n_{H_{2\,eq}} = \frac{1.10 \times 10^{13}\, \text{molecules}}{6.022\times10^{23}\, \text{molecules/mol}} \approx 1.83\times10^{-11}\, \text{mol}$$ We can find the change in moles of H2 and Br2, and the moles of HBr formed: $$\Delta n_{H_{2}} = \Delta n_{Br_{2}} = 0.0409\, \text{mol} - 1.83\times10^{-11}\, \text{mol} \approx 0.0409\, \text{mol}$$ $$n_{HBr} = 2 \times (0.0409\, \text{mol} - \Delta n_{H_{2}}) = 2 \times 1.83\times10^{-11}\, \text{mol} = 3.66\times10^{-11}\, \text{mol}$$ Now we can calculate the concentrations at equilibrium using the volume of the flask (1.00 L): $$[\mathrm{H}_{2}]_{\text{eq}} \approx \frac{1.83\times10^{-11}\, \text{mol}}{1.00\, \text{L}} = 1.83\times10^{-11}\, \text{M}$$ $$[\mathrm{Br}_{2}]_{\text{eq}} \approx \frac{0.0409\, \text{mol}}{1.00\, \text{L}} = 0.0409\, \text{M}$$ $$[\mathrm{HBr}]_{\text{eq}} \approx \frac{3.66\times10^{-11}\, \text{mol}}{1.00\, \text{L}} = 3.66\times10^{-11}\, \text{M}$$
03

Find the equilibrium constant (K)

We can find the equilibrium constant using the equilibrium concentrations. $$K = \frac{[\mathrm{HBr}]_{\text{eq}}^{2}}{[\mathrm{H}_{2}]_{\text{eq}} \cdot [\mathrm{Br}_{2}]_{\text{eq}}} = \frac{(3.66\times10^{-11}\, \text{M})^2}{(1.83\times10^{-11}\, \text{M}) \cdot (0.0409\, \text{M})}\approx 1.62 \times 10^{-9}$$
04

Calculate \(\Delta G^{\circ}\) using \(\Delta H^{\circ}\) and the equilibrium constant

We can use the following relationship to find \(\Delta G^{\circ}\): $$\Delta G^{\circ} = -RT\ln{K} = -8.314\,\frac{\text{J}}{\text{mol}\cdot\text{K}}\cdot298.15\,\text{K}\ln{(1.62 \times 10^{-9})} \approx -59.2\,\frac{\text{kJ}}{\text{mol}}$$
05

Determine 螖S掳 using 螖G掳 and 螖H掳

Finally, we can calculate the standard entropy change by rearranging the Gibbs-Helmholtz equation: $$\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} = \frac{(-103.8\,\mathrm{kJ/mol}-(-59.2\,\mathrm{kJ/mol}))}{298.15\,\mathrm{K}} \approx 149.5\,\frac{\text{J}}{\text{mol}\cdot\text{K}}$$ Thus, for this reaction at given conditions, we have the equilibrium constant \(K \approx 1.62 \times 10^{-9}\), \(\Delta G^{\circ} \approx -59.2\,\frac{\text{kJ}}{\text{mol}}\), and \(\Delta S^{\circ} \approx 149.5\,\frac{\text{J}}{\text{mol}\cdot\text{K}}\).

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Key Concepts

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

Equilibrium Constant (K)
In chemical reactions, reaching equilibrium is like finding a balance. For any reversible reaction, the equilibrium constant, symbolized as \( K \), tells us the ratio of product concentrations to reactant concentrations when the reaction is at equilibrium. This constant is specific to a particular reaction at a given temperature.
Typically, for a reaction of the type \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is defined by the expression:
  • \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
"Products over reactants" is the basic rule for calculating \( K \). It's important because it helps us predict the direction the reaction will shift if disturbed. When \( K \) is large, products are favored. When \( K \) is small, like in our problem (\( 1.62 \times 10^{-9} \)), it indicates that reactants are favored.
Understanding how to calculate \( K \) is key to predicting the concentrations of substances at equilibrium, providing insight into the reaction's behavior.
Gibbs Free Energy (螖G掳)
Gibbs Free Energy, represented as \( \Delta G^{\circ} \), is a thermodynamic quantity that helps us understand whether a reaction is spontaneous. A negative \( \Delta G^{\circ} \) means a reaction can occur without needing additional energy 鈥� it's spontaneous. Conversely, a positive \( \Delta G^{\circ} \) means the reaction requires energy input.
The relationship between \( \Delta G^{\circ} \) and the equilibrium constant \( K \) is given by the expression:
  • \( \Delta G^{\circ} = -RT\ln{K} \)
Here, \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin. In our solved exercise, \( \Delta G^{\circ} \) was calculated to be \(-59.2 \ \mathrm{kJ/mol}\), showing that the reaction is spontaneous under the given conditions.
This connection between \( \Delta G^{\circ} \) and \( K \) allows chemists to determine the tendency of reactions to occur and adjust reaction conditions accordingly to make desired products.
Entropy Change (螖S掳)
Entropy Change, \( \Delta S^{\circ} \), tells us about the disorder or randomness in a system. When a reaction results in more disorder (positive \( \Delta S^{\circ} \)), the system becomes more favorable. If it decreases disorder (negative \( \Delta S^{\circ} \)), the system tends to be less favorable.
The change in standard entropy can be determined by using the equation:
  • \( \Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T} \)
For the reaction provided in the exercise, \( \Delta S^{\circ} \) was found to be \( 149.5 \ \mathrm{J/mol\cdot K} \). This positive value indicates an increase in the disorder during the reaction, which aligns with the spontaneous nature of the process described by \( \Delta G^{\circ} \).
Understanding entropy is crucial for predicting how changes in pressure, volume, temperature, or concentration impact the spontaneity and progression of chemical reactions. This concept helps chemists devise strategies for optimizing reaction conditions.

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Most popular questions from this chapter

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Is \(\Delta S_{\text {surr favorable or unfavorable for exothermic reactions? }}\) Endothermic reactions? Explain.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3}\) ). The elevated boiling point of the solution was \(355.4 \mathrm{K} .\) What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{kJ} / \mathrm{mol},\) the entropy of vaporization is \(95.95 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol},\) and the boiling-point elevation constant is \(2.5 \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}.\)

The equilibrium constant \(K\) for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{K}\) and a \(y\) -intercept of \(-14.51 .\) Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79.

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right).\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction.鈥�
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