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Chapter 16: Problem 99

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C} .\) Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M\)

Short Answer

Expert verified
The value of ΔG for this reaction at 25°C with the given concentrations of OH⁻ and H⁺ ions is 47.7 kJ/mol.

Step by step solution

01

Convert temperature to Kelvin

To convert the temperature from Celsius to Kelvin, we will use the formula: T(K) = T(°C) + 273.15 For T = 25°C, T(K) = 25 + 273.15 = 298.15 K
02

Calculate the Reaction Quotient, Q

The Reaction Quotient for this reaction is given by: Q = \(\frac{[H^{+}][OH^{-}]}{[H_{2}O]}\) Since we are given the concentrations of H⁺ and OH⁻ ions, and given that the concentration of liquid water does not change, we can simply use the concentrations of H⁺ and OH⁻ ions in the expression for Q: Q = \(\frac{(0.71)(0.15)}{1}\) = 0.1065
03

Calculate ΔG

Now we have all the values needed to calculate ΔG using the equation: ΔG = ΔG° + RT ln Q Where ΔG° = 79.9 kJ/mol, R = 8.314 J/(mol K), T = 298.15 K, and Q = 0.1065. ΔG = \(79.9 \times 10^3 + (8.314)(298.15) \ln (0.1065)\) ΔG ≈ 79.9 kJ/mol - 32.2 kJ/mol = 47.7 kJ/mol The value of ΔG for this reaction at 25°C with the given concentrations of OH⁻ and H⁺ ions is 47.7 kJ/mol.

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Most popular questions from this chapter

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?

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 \(\Delta S^{\circ} .\)

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of ±1 in all quantities): a. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{atm}\) b. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=200 \mathrm{atm}, P_{\mathrm{H}_{2}}=600 \mathrm{atm}\) \(P_{\mathrm{NH}_{3}}=200 \mathrm{atm}\)

Using data from Appendix \(4,\) calculate \(\Delta H^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the production of ozone from oxygen: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$ At \(30 \mathrm{km}\) above the surface of the earth, the temperature is about \(230 . . \mathrm{K}\) and the partial pressure of oxygen is about \(1.0 \times 10^{-3}\) atm. Estimate the partial pressure of ozone in equilibrium with oxygen at \(30 \mathrm{km}\) above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.
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