Question

The \(K_{d}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25{ }^{\circ} \mathrm{C}\) is given in \(\mathrm{Ap}-\) pendix 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 \(\left[\mathrm{CH}_{2} \mathrm{NH}_{2}\right]=0.098 \mathrm{M}\) ?

Short answer

The chemical equilibrium equation for \(K_b\) is: \(CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-\). Using the given \(K_d\), we can find the value of \(K_b\) and then calculate ∆G° using the relationship \(\Delta G^{\circ} = -RT \ln{K_b}\). At equilibrium, ∆G is equal to 0. For the given concentrations, we calculate Q and then find ∆G using the equation \(\Delta G = \Delta G^{\circ} + RT \ln{Q}\).

Step-by-step solution

(a) Chemical equation for the equilibrium that corresponds to \(K_b\)

Methylamine acts as a weak base when it reacts with water. The chemical equation for this equilibrium can be written as: \(CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-\) Since we are given \(K_d\), we need to find \(K_b\). We know that: \(K_w = K_a \times K_b\) Where the \(K_w\) is the ion-product constant for water, and \(K_a\) is the acidity constant for the conjugate acid \(CH_3NH_3^+\). We can determine \(K_a\) using the given \(K_d\) value for methylamine (\(K_d = K_a\)). We can find the \(K_b\) value from \(K_a\) as follows: \(K_b = \frac{K_w}{K_a}\)

(b) Calculate ∆G° for the equilibrium

Use the relationship between ∆G° and \(K_b\) for the reaction: \(\Delta G^{\circ} = -RT \ln{K_b}\) Where R is the gas constant, T is the temperature in Kelvin, and \(K_b\) is the equilibrium constant for the base reaction. Since the temperature is given as \(25^\circ C\), we need to convert it to Kelvin by adding 273.15, so the temperature in Kelvin is 298.15 K.

(c) Value of ∆G at equilibrium

At equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (\(K_d\)), meaning that: \(\Delta G = \Delta G^{\circ} + RT \ln{Q} = \Delta G^{\circ} + RT \ln{K_d}\) Since ∆G° includes the term \(RT \ln{K_d}\), ∆G is equal to 0 at equilibrium.

(d) Calculate the value of ∆G with given concentrations

To find the value of ∆G with the given concentrations, we first need to find the reaction quotient (Q) value by substituting the concentrations into the expression: \(Q = \frac{\left[CH_3NH_3^+\right] \left[OH^-\right]}{\left[CH_3NH_2\right]\left[H^+\right]}\) Plug the given concentrations into the Q expression and solve for Q. Then, calculate ∆G using the relationship: \(\Delta G = \Delta G^{\circ} + RT \ln{Q}\) Finally, substitute the calculated values of ∆G° and Q into the equation to find the value of ∆G for the given concentrations.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for studying reactions and their behaviors. Chemical equilibrium occurs when a reaction and its reverse reaction proceed at equal rates, leading to stable concentrations of the reactants and products over time. This state doesn't mean that the reaction has stopped, but rather that it has reached a dynamic balance, with reactants turning into products and vice versa at the same rate.

For the equilibrium involving methylamine, \(CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-\), the equilibrium constant \(K_b\) expresses the ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their stoichiometric coefficients. When the system is at equilibrium, the ratio remains constant. It's important to note that the value of \(K_b\) itself doesn't indicate the extent of the reaction but rather the position of equilibrium. High values suggest the equilibrium lies to the right (product favored), while low values indicate an equilibrium position to the left (reactant favored).

Gibbs Free Energy

Gibbs free energy (\(\Delta G\)) is an essential concept in thermodynamics that indicates the spontaneity of a chemical reaction. When \(\Delta G\) is negative, a reaction is spontaneous, meaning it can occur without additional energy input. Conversely, a positive \(\Delta G\) indicates a nonspontaneous reaction, requiring external energy to proceed.

The relationship between \(\Delta G^\circ\), the standard change in free energy, and the equilibrium constant is given by the equation \(\Delta G^\circ = -RT \ln{K_b}\). As we calculated in the example, by knowing \(K_b\), we can determine the \(\Delta G^\circ\) for the reaction at standard conditions (1 bar pressure and concentration of 1M). At equilibrium, \(\Delta G\) is always zero because the system has no driving force to shift in either direction. Outside of equilibrium, the value of \(\Delta G\) helps predict the direction the reaction will proceed to achieve balance.

Reaction Quotient

The reaction quotient (Q) is like a snapshot of a reaction that hasn't yet reached equilibrium. It's a ratio that is identical in form to the equilibrium constant, but instead of using the equilibrium concentrations, it uses the current concentrations of reactants and products at any given point in time.

To illustrate, with the methylamine reaction, we calculate Q using \(Q = \frac{\left[CH_3NH_3^+\right] \left[OH^-\right]}{\left[CH_3NH_2\right]\left[H^+\right]}\). This formula allows us to determine the direction the reaction will proceed to reach equilibrium. If Q is less than the equilibrium constant \(K_b\), the forward reaction is favored, and the system will move towards equilibrium by forming more products. If Q is greater, the reverse reaction is favored, and the reactants will increase as the system returns to equilibrium. At equilibrium, Q equals \(K_b\), and no net change occurs because the forward and reverse reactions are balanced.

Acid-Base Equilibrium

Acid-base equilibrium is a specific type of chemical equilibrium that involves the transfer of protons (H+) between reactants. In water, weak bases like methylamine partially accept protons to form their conjugate acids, as depicted in our example equation.

The equilibrium constant for a base (\(K_b\)) describes its strength; the larger the \(K_b\), the stronger the base. Conversely, the equilibrium constant for an acid (\(K_a\)) quantifies the extent to which an acid donates protons to form its conjugate base.

• Strong acids completely dissociate in solution, resulting in a large \(K_a\).
• Weak acids only partially dissociate, leading to a smaller \(K_a\).

Understanding acid-base equilibrium is essential in many disciplines, including medicinal chemistry, environmental science, and biochemical pathways. It's particularly vital when discussing the pH of a solution, which is profoundly influenced by the presence of acids and bases and their respective equilibrium constants.