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 \(K_{b r}\) 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 \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }\right]=\left[\mathrm{H}^{+}\right]=1.5 \times 10^{-8} \mathrm{M}\) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]=5.5 \times 10^{-4} \mathrm{M}\), and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.120 \mathrm{M} ?\)

Short answer

The chemical equation for the equilibrium that corresponds to \(K_b\) is \(CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^−\). The standard Gibbs free energy change (ΔG°) for this equilibrium is approximately 21.9 kJ/mol. At equilibrium, the Gibbs free energy change (ΔG) is 0 J/mol. When given specific concentrations of CH3NH2, CH3NH3+, and H+, the value of ΔG is approximately -32.2 kJ/mol.

Step-by-step solution

Write the chemical equation for the equilibrium

Methylamine (CH3NH2) is a weak base. When it reacts with water, it generates the conjugate acid, CH3NH3+ and hydroxide ions (OH-). The chemical equation for the equilibrium is: \[CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^−\]

Calculate ΔG° for the equilibrium

First, we need to find the value of the equilibrium constant, Kb, for methylamine from Appendix D. \(K_b = 4.38 \times 10^{-4}\). We can then use the relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (Kb) to calculate ΔG°: \[\Delta G^\circ = -RT\ln K_b\] Where R is the universal gas constant (\(8.314\,J\,mol^{-1}\,K^{-1}\)) and T is the temperature in Kelvin (assuming the temperature is 25°C, so T = 25 + 273.15 = 298.15 K). Plug in the values and solve for ΔG°: \(\Delta G ^\circ = -(8.314\,J\,mol^{-1}\,K^{-1})(298.15\,K)\ln(4.38 \times 10^{-4})\) \(\Delta G ^\circ \approx 21.9\,kJ\,mol^{-1}\)

Find the value of ΔG at equilibrium

At equilibrium, the Gibbs free energy change (ΔG) is equal to zero: \[\Delta G = 0\,J\,mol^{-1}\]

Determine the value of ΔG when given specific concentrations

We are given the concentrations of CH3NH2, CH3NH3+, and H+: \[[CH_3NH_2] = 0.120\,M\] \[[CH_3NH_3^+] = 5.5 \times 10^{-4}\,M\] \[[H^+] = 1.5 \times 10^{-8}\,M\] We can use the reaction quotient (Q) for the given concentrations: \[Q=\frac{[CH_3NH_3^+][OH^{-}]}{[CH_3NH_2]}\] Since \([OH^{-}] = \frac{[H^+]}{K_w}\), where \(K_w\) is the ion product of water (\(1\times10^{-14}\)), we can plug that into the Q expression to obtain: \(Q=\frac{[CH_3NH_3^+][(1.5 \times 10^{-8}\,M)/10^{-14}]}{[CH_3NH_2]}\) \(Q=\frac{(5.5 \times 10^{-4}\,M)(1.5 \times 10^{-8}\,M)}{0.120\,M}\) Now, we can find the Gibbs free energy ΔG using the relationship between ΔG, ΔG°, and Q: \[\Delta G = \Delta G^\circ + RT\ln Q\] Plug in the values and solve for ΔG: \(\Delta G = (21.9\,kJ\,mol^{-1}) +(8.314\,J\,mol^{-1}\,K^{-1})(298.15\,K)\ln Q\) \(\Delta G \approx -32.2\, kJ\,mol^{-1}\)

Key concepts

Equilibrium Constant

The equilibrium constant, symbolized as 'K', is a critical concept in chemical equilibrium that quantifies the ratio of the concentrations of products to reactants at equilibrium. Each equilibrium reaction has a unique constant, denoted as Kc when expressed in terms of concentrations or Kp when in terms of partial pressures. For the equilibrium involving a weak base like methylamine (CH3NH2), the specific constant is represented as Kb for the base dissociation constant.
For the reaction of methylamine with water, the Kb value is the focal point in determining how the base dissociates:
\[CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-\]The higher the Kb value, the stronger the weak base, implying a greater degree of ionization in water. When solving problems involving weak bases, Kb provides a straightforward measure of the base's ability to accept a proton from water to form hydroxide ions. By mastering the use of the equilibrium constant, students can predict the direction and extent of chemical reactions under equilibrium conditions.

Gibbs Free Energy

Gibbs free energy, denoted by the symbol ΔG, is an essential thermodynamic quantity that helps predict the direction of chemical reactions. It represents the maximum amount of reversible work a system can perform at constant temperature and pressure. At equilibrium, ΔG is zero, signifying that the reaction can proceed in both the forward and reverse directions equally. If ΔG is negative, the reaction tends to occur spontaneously in the forward direction, whereas a positive ΔG indicates non-spontaneity.
In the context of chemical equilibrium, the standard Gibbs free energy change (ΔG°) can be calculated from the equilibrium constant using the equation:
\[\begin{array}{r}\forall\text{For a given temperature}, T, \text{ and gas constant }, R:\text{ }\text{ }\text{ }\Delta G^\circ = -RT\ln K\end{array}\]In our exercise dealing with methylamine, we use Kb as our equilibrium constant to calculate ΔG°. This relationship between ΔG° and Kb is crucial because it allows the prediction of spontaneity for a reaction at standard conditions. Understanding this connection helps students in determining whether a reaction is thermodynamically favorable or not.

Reaction Quotient

The reaction quotient, Q, plays a pivotal role in determining the direction in which a reaction at non-equilibrium conditions is likely to proceed. It is calculated similarly to the equilibrium constant, K, but with the initial concentrations instead of the equilibrium concentrations. The formula for Q is generally the same as K, reflecting the stoichiometry of the reaction, but without implying the equilibrium state.
For the equilibrium involving methylamine, the reaction quotient Q would be:
\[Q=\frac{[CH_3NH_3^+][OH^{-}]}{[CH_3NH_2]}\]When comparing Q to the equilibrium constant (Kb for bases), there are three possible scenarios:

• If Q = Kb, the system is at equilibrium.
• If Q < Kb, there's a tendency for the reaction to proceed forward to reach equilibrium.
• If Q > Kb, the reaction will shift backwards to reach equilibrium.

By calculating Q and analyzing how it relates to Kb, students can infer the direction of the reaction's shift to achieve equilibrium. This understanding is instrumental in predicting the outcomes of reactions when the initial conditions are altered.

Weak Base Behavior

Exploring weak base behavior involves understanding how weak bases, like methylamine, partially dissociate in water. Unlike strong bases that dissociate completely, weak bases do so to a limited extent, governed by the base dissociation constant Kb. The equilibrium set up by a weak base dictates its property of raising the pH of a solution by less than a strong base would.
In aqueous solutions, weak bases react with water to form a conjugate acid and hydroxide ions, slightly increasing the solution's pH. The value of Kb, as seen in the exercise, determines the strength of the weak base. Larger Kb values signify a greater tendency for the base to accept a proton, corresponding to stronger weak base behavior.
Through understanding weak base behavior, students can better grasp various biological and chemical processes, such as buffering and drug interactions, where the pH level and the strength of a base play critical roles.