The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix \(\mathrm{D}\) . (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b}\) . (b) By using the value of \(K_{b},\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} M,\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

To start, let's write down the base dissociation reaction of methylamine (CH3NH2) when it reacts with water (H2O). Since methylamine is a weak base, it will accept a proton (H+) from water, forming its conjugate acid (CH3NH3+) and hydroxide ions (OH-). \( CH_{3}NH_{2} + H_{2}O \rightleftharpoons CH_{3}NH_{3}^{+} + OH^{-}\)

The equilibrium constant for the base dissociation reaction, Kb, is related to the standard Gibbs free energy change (ΔG°) by the following equation: \( ΔG° = -RT \ln{K_b}\) , where \(R\) is the gas constant (= 8.314 J/(mol K)) and \(T\) is the temperature in Kelvin. We are given that the temperature is 25°C, which is equal to 298.15 K. The value of Kb for methylamine can be found in Appendix D. Kb = 4.4 x 10^(-4). Now, we plug in these values into the formula to calculate ΔG°.

At equilibrium, the value of ΔG equals 0. This means the system is in a state of minimum energy. Hence, ΔG = 0

The relationship between ΔG, ΔG°, and the reaction quotient (Q) is given by the formula: \( ΔG = ΔG° + RT \ln{Q}\) We are given the concentrations of H+ ([H+]), CH3NH3+ ([CH3NH3+]), and CH3NH2 ([CH3NH2]). However, we should note that the base dissociation reaction involves OH- ions instead of H+ ions. Therefore, we need to use the relation between the concentrations of H+ ions and OH- ions: \(K_w = [H^+][OH^-]\), where Kw is the ion product of water (1.0 x 10^(-14) at 25°C). Now, let's calculate the concentration of OH- ions: \([OH^{-}] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{6.7 \times 10^{-9}}\) Next, we determine the reaction quotient, Q: \(Q = \frac{[CH_{3}NH_{3}^{+}][OH^{-}]}{[CH_{3}NH_{2}]}\) Plug in the given concentrations of the species involved in the reaction and the calculated value of [OH-] into the equation. Finally, we can calculate ΔG using the expression: \( ΔG = ΔG° + RT \ln{Q}\) Plug in the calculated value of ΔG°, and the reaction quotient, Q, to determine the final value of ΔG. Now you have successfully calculated the ΔG value for the given concentrations of H+, CH3NH3+, and CH3NH2.