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Chapter 19: Problem 84

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

Expert verified
(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 (螖G掳) for the equilibrium is approximately 10,851 J/mol. (c) The value of 螖G at equilibrium is 0. (d) The value of 螖G for given concentrations is approximately 10,905 J/mol.

Step by step solution

01

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: \(CH_{3}NH_{2} (aq) + H_2O (l) \rightleftharpoons CH_{3}NH_{3}^{+} (aq) + OH^{-} (aq)\)
02

b) Calculation of 螖G掳 using Kb

To calculate the standard Gibbs free energy change (螖G掳) from the base dissociation constant Kb, we can use the following equation: \(螖G掳 = -RT \ln K_{b}\) where R is the gas constant (8.314 J/mol路K), T is the temperature in Kelvin, and Kb is the base dissociation constant. Given, the temperature is 25掳C, which is equal to 298.15 K. The Kb for methylamine is 4.38 脳 10^(-4) (from Appendix D). So, \(螖G掳 = -(8.314\,J/(mol\cdot K))\cdot(298.15\,K)\cdot\ln(4.38\times10^{-4})\) Calculating the above expression, we get: 螖G掳 鈮� 10,851 J/mol
03

c) Value of 螖G 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, 螖G = 0 at equilibrium
04

d) Calculation of 螖G for given concentrations

To calculate the Gibbs free energy change (螖G) for the given concentrations of H+ (C1), CH3NH3+ (C2), and CH3NH2 (C3), we can use the equation: \(螖G = 螖G^\circ + RT \ln(Q)\) 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 脳 10^(-9) M and Kw at 25掳C = 1 脳 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 螖G using the calculated 螖G掳, temperature, R-value, and Q: \(螖G = (10,851\, J/mol) + (8.314\,J/(mol\cdot K))(298.15\,K)\cdot\ln\left(\dfrac{(2.4\times10^{-3})(1.48\times10^{-6})}{(0.098)}\right)\) Calculating this expression, we get: 螖G 鈮� 10,905 J/mol

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Carbon disulfide \(\left(C S_{2}\right)\) is a toxic, highly flammable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(I)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) \begin{tabular}{lcc} \hline & \(\Delta H_{i}(\mathrm{k} / \mathrm{mol})\) & \(\Delta G_{i}^{\prime}(\mathrm{kJ} / \mathrm{mol})\) \\ \hline\(C S_{2}(l)\) & 89.7 & 65.3 \\ \(C S_{2}(g)\) & 117.4 & 67.2 \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? \((\mathbf{b})\) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) burns in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C,\) calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part \((c) .\) Is the reaction exothermic? Is it spontaneous at \(298 \mathrm{~K} ?\) (e) Use the data in the table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(I) .\) Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(l)\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(101.3 \mathrm{kPa}\) ?

For each of the following processes, indicate whether the signs of \(\Delta S\) and \(\Delta H\) are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of \(\mathrm{Co}(s)\) is lowered from \(60^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C} .\) (c) Ethyl alcohol evaporates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}(g)\) or \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\), (b) \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) or \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)

A standard air conditioner involves a \(r\) frigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{~F}_{2}\). An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of \(q\) for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at \(31^{\circ} \mathrm{C}\). Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\). Is this process spontaneous of nonspontaneous?

The element sodium (Na) melts at \(97.8^{\circ} \mathrm{C},\) and its molar enthalpy of fusion is \(\Delta H_{\text {fus }}=2.60 \mathrm{~kJ} / \mathrm{mol}\). (a) When molten sodium solidifies to \(\mathrm{Na}(\mathrm{s})\), is \(\Delta S\) positive or negative? (b) Calculate the value of \(\Delta S\) when \(50.0 \mathrm{~g}\) of \(\mathrm{Na}(l)\) solidifies at \(97.8^{\circ} \mathrm{C}\).
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