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Chapter 17: Problem 110

The equilibrium constant for a certain reaction decreases from \(8.84\) to \(3.25 \times 10^{-2}\) when the temperature increases from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\). Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise \(79 .\) )

Short Answer

Expert verified
The temperature where K = 1.00 for this reaction is approximately 403.48 K. The value of ΔS° for this reaction is approximately 30.40 J/(mol K).

Step by step solution

01

Write down the given information

We are given: - Initial equilibrium constant, K1 = 8.84 - Final equilibrium constant, K2 = 3.25 × 10⁻² - Initial temperature, T1 = 25°C (convert to Kelvin: 25 + 273.15 = 298.15 K) - Final temperature, T2 = 75°C (convert to Kelvin: 75 + 273.15 = 348.15 K) We want to find the temperature where K = 1.00 and the value of ΔS°.
02

Write down the Van 't Hoff equation

The Van 't Hoff equation is: \( \ln \frac{K_2}{K_1} = - \frac{ \Delta H ^{\circ}}{R} (\frac{1}{T_2} - \frac{1}{T_1}) \), where R is the gas constant, approximately 8.314 J/(mol K).
03

Solve for ΔH°

We will first solve for ΔH° with the given information: 1. \( \ln \frac{K_2}{K_1} = \ln \frac{3.25 \times 10^{-2}}{8.84} \) 2. \( \ln \frac{3.25 \times 10^{-2}}{8.84}=- \frac{\Delta H^{\circ}}{8.314}(\frac{1}{348.15}-\frac{1}{298.15}) \) 3. Solve for ΔH°: \( \Delta H° \approx -9.56 \times 10^3 \, \mathrm{J/mol} \) Now we have the value of ΔH°.
04

Find the temperature where K=1.00

We will now find the temperature where K=1.00. Let's call it T3. 1. Replace all the known values in the Van 't Hoff equation: \( \ln \frac{1}{8.84} = - \frac{-9.56 \times 10^3}{8.314}(\frac{1}{T_3}-\frac{1}{298.15}) \) 2. Solve for T3: \( T_3 \approx 403.48 \, \mathrm{K } \) Thus, the temperature where K = 1.00 for this reaction is approximately 403.48 K.
05

Estimate the value of ΔS°

Now, we can estimate the value of ΔS° using the relationship between ΔH°, ΔS°, and the equilibrium constant: 1. ΔG° = ΔH° - TΔS°, and ΔG° = -RT ln K 2. At T1 (298.15 K): -RT ln K1 = ΔH° - TΔS° 3. Solve for ΔS°: \( \Delta S ^{\circ} \approx 30.40 \, \mathrm{J/(mol \cdot K)} \) So, the value of ΔS° for this reaction is approximately 30.40 J/(mol K).

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Most popular questions from this chapter

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\). a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process \(\mathrm{S}_{\text {rhombic }}(s) \longrightarrow \mathrm{S}_{\text {monoclinic }}(s)\) b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at STP) or 1 mole of \(\mathrm{H}_{2}\) (at \(\left.100^{\circ} \mathrm{C}, 0.5 \mathrm{~atm}\right)\) b. 1 mole of \(\mathrm{N}_{2}\) (at STP) or 1 mole of \(\mathrm{N}_{2}\) (at \(\left.100 \mathrm{~K}, 2.0 \mathrm{~atm}\right)\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(\left.0^{\circ} \mathrm{C}\right)\) or 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) (at \(\left.20^{\circ} \mathrm{C}\right)\)

In the text, the equation $$ \Delta G=\Delta G^{\circ}+R T \ln (Q) $$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of \(\mathrm{mol} / \mathrm{L}\) for the quantities in \(Q\), specifically for aqueous reactions. With this in mind, consider the reaction $$ \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) $$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C}\). a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{~F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} M\) Based on the calculated \(\Delta G\) values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

As \(\mathrm{O}_{2}(l)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{~J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?
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