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Chapter 16: Problem 102

The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from \(300.0 \mathrm{K}\) to \(350.0 \mathrm{K}\). Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

Short Answer

Expert verified
The standard change in enthalpy for this reaction, \(\Delta H^{\circ}\), can be calculated using the van't Hoff equation: \[\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Given \(K_2/K_1 = 6.67\), \(T_1 = 300.0\,\mathrm{K}\), and \(T_2 = 350.0\,\mathrm{K}\), we can rearrange the equation and find that: \[\Delta H^{\circ} \approx -3.27 \times 10^{3}\:\mathrm{J\, mol^{-1}}\]

Step by step solution

01

Write down the van't Hoff equation.

The van't Hoff equation is given by: \[\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Here, \(K_1\) and \(K_2\) are the initial and final equilibrium constants, \(T_1\) and \(T_2\) are the initial and final temperatures, \(R\) is the gas constant, and \(\Delta H^{\circ}\) is the standard change in enthalpy.
02

Plug in given values.

We have the following information from the exercise: - \(K_2/K_1 = 6.67\) - \(T_1 = 300.0\,\mathrm{K}\) - \(T_2 = 350.0\,\mathrm{K}\) - \(R = 8.314\,\mathrm{J\, mol^{-1}K^{-1}}\) Now let's plug these values into the van't Hoff equation: \[\ln \left(6.67\right) = \frac{-\Delta H^{\circ}}{8.314} \left(\frac{1}{350.0} - \frac{1}{300.0}\right)\]
03

Rearrange the equation and solve for \(\Delta H^{\circ}\).

We want to isolate \(\Delta H^{\circ}\) to find its value, so let's rearrange the equation: \[\Delta H^{\circ} = - 8.314 \times \frac{\ln\left(6.67\right)}{\frac{1}{350.0} - \frac{1}{300.0}}\] Now we can calculate the value for \(\Delta H^{\circ}\): \[\Delta H^{\circ} = - 8.314 \times \frac{1.897}{\frac{1}{350.0} - \frac{1}{300.0}} = -8.314 \times \frac{1.897}{\frac{50}{105000}}\] \[\Delta H^{\circ} \approx -3.266 \times 10^{3}\,\mathrm{J\, mol^{-1}}\]
04

Write down the result.

Therefore, the standard change in enthalpy for this reaction is approximately: \[\Delta H^{\circ} \approx -3.27 \times 10^{3}\:\mathrm{J\, mol^{-1}}\]

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Most popular questions from this chapter

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.”

Ethanethiol \((\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\) also called ethyl mercaptan) is commonly added to natural gas to provide the "rotten egg" smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is \(27.5 \mathrm{kJ} / \mathrm{mol} .\) What is the entropy of vaporization for this substance?

Consider the following reaction at \(800 . \mathrm{K}:\) $$\mathrm{N}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{atm}, P_{\mathrm{NF}_{3}}=0.48\) atm. Calculate \(\Delta G^{\circ}\) for the reaction at \(800 .\) K.

Calculate \(\Delta S_{\text {sur }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 atm. $$\text { a. } \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$$$$\begin{aligned} &\Delta H^{\circ}=-2221 \mathrm{kJ}\\\ &\text { b. } 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=112 \mathrm{kJ} \end{aligned}$$

Consider the following reaction at \(298 \mathrm{K}:\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix \(4,\) determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?
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