Question

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 \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

Short answer

(a) The chemical equation for the equilibrium that corresponds to Kb is: \(CH_{3}NH_{2} (aq) + H_2O (l) \rightleftharpoons CH_{3}NH_{3}^{+} (aq) + OH^{-} (aq)\) (b) The standard Gibbs free energy change (ΔG°) for the equilibrium is approximately 10,851 J/mol. (c) The value of ΔG at equilibrium is 0. (d) The value of ΔG for given concentrations is approximately 10,905 J/mol.

Step-by-step solution

a) Chemical equation for methylamine equilibrium

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

b) Calculation of ΔG° using Kb

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

c) Value of ΔG at equilibrium

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

d) Calculation of ΔG for given concentrations

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

Key concepts

Base Dissociation Constant

In the context of chemical equilibrium, the base dissociation constant, commonly denoted as \(K_{b}\), is a measure of the ability of a base to accept protons. Specifically, it is used to describe the equilibrium position of a weak base reaction in water. For example, when methylamine \((\text{CH}_3\text{NH}_2)\) dissolves in water, it participates in an equilibrium reaction: \(\text{CH}_3\text{NH}_2 (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{CH}_3\text{NH}_3^{+} (aq) + \text{OH}^{-} (aq)\).
The \(K_{b}\) value provides insight into how far to the right this reaction proceeds under given conditions. A larger \(K_{b}\) indicates a stronger base with a greater tendency to accept protons, thus forming more \(\text{CH}_3\text{NH}_3^{+}\) and \(\text{OH}^{-}\). Conversely, a smaller \(K_{b}\) means the base is weaker. The significance of \(K_{b}\):

• Helps in calculating the pH of a basic solution by understanding the degree of proton acceptance by the base.
• Essential in predicting the position of the equilibrium, which further assists in reaction forecasting.

Gibbs Free Energy

Gibbs Free Energy, expressed as \( \Delta G \), is a critical concept in understanding chemical reactions and equilibrium. It represents the maximum reversible work a system can do at constant temperature and pressure. A reaction proceeds spontaneously when \( \Delta G \) is negative. For our equilibrium context, \( \Delta G^\circ \) (standard Gibbs free energy change) is calculated using the relation between \( \Delta G^\circ \) and \( K_{b} \): \[ \Delta G^\circ = -RT \ln(K_{b}) \] where \( R \) is the ideal gas constant and \( T \) is the temperature in Kelvin.
The standard Gibbs free energy change informs us about the spontaneity of the reaction under standard conditions (1 atm pressure and 298.15 K).Key Points about Gibbs Free Energy:

• \( \Delta G = 0 \) at equilibrium, indicating no net change.
• A negative \( \Delta G^\circ \) suggests the equilibrium lies to the right, while a positive value indicates the equilibrium lies to the left.
• Understanding \( \Delta G \) helps chemists control reaction conditions for desired yields.

Reaction Quotient

The Reaction Quotient, denoted as \( Q \), serves as a useful tool for predicting the direction of a reaction and its relation to equilibrium. It is calculated using the same expression for the equilibrium constant but with the concentrations at any point in time: \[ Q = \frac{[\text{CH}_3\text{NH}_3^{+}][\text{OH}^{-}]}{[\text{CH}_3\text{NH}_2]} \] This differs from \( K_{b} \), which uses concentrations at equilibrium. By comparing \( Q \) with \( K_{b} \), we can determine the reaction's progress:

• If \( Q = K_{b} \), the system is at equilibrium.
• If \( Q < K_{b} \), the reaction will proceed forward (to the right) to achieve equilibrium.
• If \( Q > K_{b} \), the reaction will reverse (to the left) to reach equilibrium.

If calculations show that \( Q \) aligns with the conditions given, using the concentrations helps in understanding how close the system is to equilibrium, thus making adjustments possible to optimize chemical processes.