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Chapter 17: Problem 109

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ 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 the reaction at 25°C with the given concentrations of OH⁻ and H⁺ is approximately 74370.3 J/mol.

Step by step solution

01

Identify the given values

We are given that: 1. ΔG° = 79.9 kJ/mol 2. T = 25°C = 298.15 K (converted to Kelvin) 3. [OH⁻] = 0.15 M 4. [H⁺] = 0.71 M
02

Calculate the reaction quotient (Q)

For the reaction: H2O(l) ⮌ H⁺(aq) + OH⁻(aq), the reaction quotient Q is given by: Q = [H⁺][OH⁻], Now, substituting the given values of [H⁺] and [OH⁻], Q = (0.71 M)(0.15 M) = 0.1065
03

Calculate ΔG using ΔG°, RT, and ln(Q)

We have the equation: ΔG = ΔG° + RT ln(Q), We have already found the values for ΔG° (= 79.9 kJ/mol) and Q (= 0.1065). Let's convert ΔG° from kJ/mol to J/mol, which is 79.9 × 1000 = 79,900 J/mol. The gas constant R = 8.314 J/(mol·K). Now we can calculate ΔG: ΔG = 79900 J/mol + (8.314 J/(mol·K))(298.15 K)ln(0.1065) ΔG ≈ 79900 J/mol + (8.314 J/(mol·K))(298.15 K)(-2.2398) ΔG ≈ 79900 J/mol - 5529.7 J/mol ΔG ≈ 74370.3 J/mol So, the value of ΔG for the reaction at 25°C with the given concentrations of OH⁻ and H⁺ is approximately 74370.3 J/mol.

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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.

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 $\mathrm{J} / \mathrm{K} \cdot$ mol. a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

Consider a weak acid, HX. If a \(0.10-M\) solution of \(\mathrm{HX}\) has a pH of 5.83 at \(25^{\circ} \mathrm{C},\) what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C} ?\)

Consider the following reaction at \(35^{\circ} \mathrm{C} :\) $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \Delta G^{\circ}=20 . \mathrm{kJ} $$ If 2.0 atm of NOCl are reacted in a rigid container at $35^{\circ} \mathrm{C},$ calculate the equilibrium partial pressure of NO.

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is (are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K\) is greater than 1.00 .
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