The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix \(D\). (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 $\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}$ and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

Methylamine (CH3NH2) is a weak base, and it accepts a proton from water. The chemical equation for the equilibrium corresponding to Kb is: \(CH_{3}NH_{2} (aq) + H_2O (l) \rightleftharpoons CH_{3}NH_{3}^{+} (aq) + OH^{-} (aq)\)

To calculate the standard Gibbs free energy change (ΔG°) from the base dissociation constant Kb, we can use the following equation: \(ΔG° = -RT \ln K_{b}\) where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Kb is the base dissociation constant. Given, the temperature is 25°C, which is equal to 298.15 K. The Kb for methylamine is 4.38 × 10^(-4) (from Appendix D). So, \(ΔG° = -(8.314\,J/(mol\cdot K))\cdot(298.15\,K)\cdot\ln(4.38\times10^{-4})\) Calculating the above expression, we get: ΔG° ≈ 10,851 J/mol

At equilibrium, the reaction has no Gibbs free energy change. The system is at its lowest energy when it is in equilibrium. Therefore, ΔG = 0 at equilibrium

To calculate the Gibbs free energy change (ΔG) for the given concentrations of H+ (C1), CH3NH3+ (C2), and CH3NH2 (C3), we can use the equation: \(ΔG = ΔG^\circ + RT \ln(Q)\) where Q is the reaction quotient calculated as: \(Q = \dfrac {[CH_3NH_3^{+}] [OH^{-}]}{[CH_3NH_2]}\) We notice that the reaction quotient has the term [OH-] which we are not given the concentration of directly. But, we can figure it out from the concentration of H+ given, through the water ionization constant (Kw): \[K_w = [H^+][OH^-]\] Given, [H+] = 6.7 × 10^(-9) M and Kw at 25°C = 1 × 10^(-14), hence: \([OH^-] =\dfrac{1\times10^{-14}}{6.7\times10^{-9}}\) Now, we can calculate Q using the given concentrations and the calculated [OH-]: \(Q=\dfrac{(2.4\times10^{-3})(1.48\times10^{-6})}{(0.098)}\) Finally, we can calculate ΔG using the calculated ΔG°, temperature, R-value, and Q: \(ΔG = (10,851\, J/mol) + (8.314\,J/(mol\cdot K))(298.15\,K)\cdot\ln\left(\dfrac{(2.4\times10^{-3})(1.48\times10^{-6})}{(0.098)}\right)\) Calculating this expression, we get: ΔG ≈ 10,905 J/mol