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Chapter 17: Problem 70

A \(0.143 M\) solution of the weak base \(\mathrm{RNH}_{2}\) has \(\mathrm{pH} 9.11\) at \(25^{\circ} \mathrm{C}\). What is \(\Delta G^{\circ}\) for the dissociation of the weak base in water? $$ \mathrm{RNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{RNH}_{3}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Short Answer

Expert verified
Answer: The standard Gibbs free energy (ΔG°) for the dissociation of the weak base RNH2 in water is approximately 61.0 kJ/mol.

Step by step solution

01

Determine the concentration of OH- ions

Since we are given the pH of the solution (9.11), we can determine the concentration of OH- ions by using the relationship between pH and pOH (pH + pOH = 14 at 25°C). pOH = 14 - pH pOH = 14 - 9.11 pOH = 4.89 Now, we can find the concentration of OH- ions using pOH = -log10[OH-]: [OH-] = 10^(-pOH) = 10^(-4.89) ≈ 1.29 × 10^(-5) M
02

Determine K_b (base dissociation constant) for RNH2

We can write the expression for K_b using the dissociation formula provided in the exercise: K_b = [RNH3+][OH-] / [RNH2] We know that the initial concentration of RNH2 ([RNH2]_initial) is 0.143 M and [OH-] is 1.29 × 10^(-5) M at equilibrium. Assuming x M of RNH2 dissociates at equilibrium: [RNH3+]_eq = [OH-]_eq = 1.29 × 10^(-5) + x, as equal amounts of RNH3+ and OH- are formed [RNH2]_eq = 0.143 - x, the initial concentration minus the amount that dissociated K_b = [(1.29 × 10^(-5) + x)(1.29 × 10^(-5) + x)] / (0.143 - x) Since x is very small compared to 1.29 × 10^(-5) and 0.143, we can approximate and simplify the expression as follows: K_b ≈ (1.29 × 10^(-5))^2 / 0.143 K_b ≈ 1.14 × 10^(-11)
03

Calculate ΔG° using the relationship between ΔG° and K_b

The relationship between ΔG° and K is given by: ΔG° = -RT ln(K) Where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298 K in this case), and K is the equilibrium constant (K_b in this case). ΔG° = - (8.314 J/(mol·K)) × (298 K) × ln(1.14 × 10^(-11)) ΔG° ≈ 61.0 kJ/mol The standard Gibbs free energy for the dissociation of the weak base RNH2 in water is approximately 61.0 kJ/mol.

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Most popular questions from this chapter

Oxygen can be made in the laboratory by reacting sodium peroxide and water. $$ \begin{gathered} 2 \mathrm{Na}_{2} \mathrm{O}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{NaOH}(s)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=-109.0 \mathrm{~kJ} ; \Delta G^{\circ}=-148.4 \mathrm{~kJ} \text { at } 25^{\circ} \mathrm{C} \end{gathered} $$ (a) Calculate \(\Delta S^{\circ}\). Is the sign reasonable? (b) Calculate \(S^{\circ}\) for \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)\). (c) Calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)\).

Consider the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ At \(500^{\circ} \mathrm{C}\), a flask initially has all three gases, each at a partial pressure of \(0.200\) atm. When equilibrium is established, the partial pressure of HI is determined to be \(0.48\) atm. What is \(\Delta G^{\circ}\) for the reaction at \(500^{\circ} \mathrm{C}\) ?

\(K_{a}\) for acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is \(1.754 \times 10^{-5} .\) At \(50^{\circ} \mathrm{C}, K_{\mathrm{a}}\) is \(1.633 \times 10^{-5}\). Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not affected by a change in temperature, calculate \(\Delta S^{\circ}\) for the ionization of acetic acid.

Discuss the effect of temperature change on the spontaneity of the following reactions at 1 atm. (a) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s) \longrightarrow 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) $$ \Delta H^{\circ}=+851.5 \mathrm{~kJ} ; \Delta S^{\circ}=+38.5 \mathrm{~J} / \mathrm{K} $$ (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)\) $$ \Delta H^{\circ}=-50.6 \mathrm{~kJ} ; \Delta S^{\circ}=0.3315 \mathrm{~kJ} / \mathrm{K} $$ (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\) $$ \Delta H^{\circ}=98.9 \mathrm{~kJ} ; \Delta S^{\circ}=+0.0939 \mathrm{~kJ} / \mathrm{K} $$

Given the following standard free energies at \(25^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{SO}_{2}(g)+3 \mathrm{CO}(g) \longrightarrow \operatorname{COS}(g)+2 \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-246.5 \mathrm{~kJ} \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) & \Delta G^{\circ}=-28.5 \mathrm{~kJ} \end{array} $$ find \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction. $$ \mathrm{SO}_{2}(g)+\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{COS}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$
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