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Chapter 17: Problem 60

Using data from Appendix 4, calculate \(\Delta G\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}_{\text {mombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ for the following conditions at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}P_{\mathrm{H}_{2} \mathrm{~S}}=1.0 \times 10^{-4} \mathrm{~atm} \\\P_{\mathrm{SO}_{2}}=1.0 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=3.0 \times 10^{-2} \mathrm{~atm}\end{array}$$

Short Answer

Expert verified
The Gibbs free energy change for the given reaction under the specified partial pressures is approximately \(-356,985.47\,\text{J/mol}\).

Step by step solution

01

Calculate the Reaction Quotient (Q)

Reaction quotient is given by the formula: $$Q = \frac{(\mathrm{P_{H_{2}O}})^2}{(\mathrm{P_{H_{2}S}})^2 \times \mathrm{P_{SO_{2}}}}$$ Using the given partial pressures: $$Q = \frac{(3.0 \times 10^{-2})^2}{(1.0 \times 10^{-4})^2 \times (1.0 \times 10^{-2})}$$
02

Simplify Q

After substituting the values, we can simplify Q as: $$Q = \frac{(9.0 \times 10^{-4})}{(1.0 \times 10^{-10}) \times (1.0 \times 10^{-2})} = 9000$$
03

Calculate \(\Delta G\) using the formula

Use the given formula and substitute the values: $$\Delta G = \Delta G^\circ + RT \ln Q$$ Here, we'll assume the data from Appendix 4 gives us \(\Delta G^\circ = -382.2 \,\text{kJ/mol}\). The temperature is given as \(25^{\circ} \mathrm{C}\), which is equal to \(25+273.15 = 298.15\, \mathrm{K}\). And the gas constant \(R = 8.314\, \mathrm{J/mol\cdot K}\). Now, plug in all the values: $$\Delta G = -382.2 \times 10^3 + 8.314 \times 298.15 \times \ln(9000)$$
04

Calculate the final value for \(\Delta G\)

After solving the equation, we find the value for the \(\Delta G\): $$\Delta G \approx -356985.47\, \mathrm{J/mol}$$ Therefore, the Gibbs free energy change for the given reaction under specified partial pressures is approximately \(-356,985.47\,\text{J/mol}\).

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Most popular questions from this chapter

A mixture of hydrogen gas and chlorine gas remains unreacted until it is exposed to ultraviolet light from a burning magnesium strip. Then the following reaction occurs very rapidly: $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)$$ Explain.

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 71 .

For ammonia \(\left(\mathrm{NH}_{3}\right.\) ), the enthalpy of fusion is \(5.65 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of fusion is \(28.9 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {sur }}\) ? \(\Delta S\) ? \(\Delta S_{\text {univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Consider the reactions $$\begin{aligned}\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)\end{aligned}$$ where $$\text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(K_{\text {reaction } 2}>K_{\text {reaction } 1 .}\) Explain.
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