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Chapter 19: Problem 83

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a}\). (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when $\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\(, \)\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} \mathrm{M},\( and \)\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?$

Short Answer

Expert verified
(a) The chemical equation for the equilibrium is: \(HNO_2 \rightleftharpoons H^+ + NO_2^-\) \(K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}\) (b) The ΔG° can be calculated using the equation: \(\Delta G^ο = -RT \ln{K_a}\) (c) At equilibrium, ΔG is 0. (d) To determine ΔG with given concentrations, use the equation: \(\Delta G = \Delta G^ο + RT \ln{Q}\) where Q is: \(Q = \frac{(5.0 \times 10^{-2})(6.0 \times 10^{-4})}{0.20}\) Calculate Q, and then use this value, ΔG°, R, and T to determine the ΔG value.

Step by step solution

01

Writing the Chemical Equation and Ka Expression

Write down the equation for the ionization of nitrous acid in water and the corresponding equilibrium constant (Ka): \(HNO_2 \rightleftharpoons H^+ + NO_2^-\) The equilibrium constant expression, Ka, for this reaction is: \(K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}\)
02

Calculating ΔG° for the Dissociation of Nitrous Acid

Use the provided \(K_a\) value and the relationship between \(K_a\) and ΔG°, given by the equation: \(\Delta G^ο = -RT \ln{K_a}\) where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K since it is 25°C). Once we have the ΔG° value, it can be used to find the ΔG at the given concentrations in the later parts of the problem. Note: Be sure to convert the temperature from Celsius to Kelvin (T = 25°C + 273.15 = 298.15 K).
03

Finding ΔG at Equilibrium

At equilibrium, the Gibbs free energy change (ΔG) is zero: \(\Delta G = 0\) This is because the concentrations of products and reactants are balanced at equilibrium.
04

Determining ΔG with Given Concentrations

Now, we need to find ΔG when the solution has specific concentrations of H⁺, NO₂⁻, and HNO₂. Use the provided concentrations and the reaction quotient Q, which is related to ΔG and ΔG° by the following equation: \(\Delta G = \Delta G^ο + RT \ln{Q}\) The reaction quotient Q for this reaction is: \(Q = \frac{[H^+][NO_2^-]}{[HNO_2]}\) Plug in the given concentrations of the species: \([H^+] = 5.0 \times 10^{-2} M\), \([NO_2^-] = 6.0 \times 10^{-4} M\), and \([HNO_2] = 0.20 M\): \(Q = \frac{(5.0 \times 10^{-2})(6.0 \times 10^{-4})}{0.20}\) Calculate Q, and then use this value, ΔG° from step 2, R, and T to determine the ΔG value for these given concentrations.

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Most popular questions from this chapter

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta H\) zero for the process?

The potassium-ion concentration in blood plasma is about $5.0 \times 10^{-3} \mathrm{M}$, whereas the concentration in muscle-cell fluid is much greater \((0.15 \mathrm{M})\). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to \(\mathrm{K}^{+}\). (a) What is \(\Delta G\) for the transfer of \(1 \mathrm{~mol}\) of \(\mathrm{K}^{+}\) from blood plasma to the cellular fluid at body temperature \(37^{\circ} \mathrm{C} ?\) (b) What is the minimum amount of work that must be used to transfer this \(\mathrm{K}^{+} ?\)

As shown here, one type of computer keyboard cleaner contains liquefied 1,1 -difluoroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{~F}_{2}\right),\) which is a gas at atmospheric pressure. When the nozzle is squeezed, the 1,1 -difluoroethane vaporizes out of the nozzle at high pressure, blowing dust out of objects. (a) Based on your experience, is the vaporization a spontaneous process at room temperature? (b) Defining the 1,1 -difluoroethane as the system, do you expect \(q_{\text {sys }}\) for the process to be positive or negative? (c) Predict whether \(\Delta S\) is positive or negative for this process. (d) Given your answers to \((a),(b),\) and \((c),\) do you think the operation of this product depends more on enthalpy or entropy? [Sections 19.1 and 19.2\(]\)

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A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and $\Delta S^{\circ}=\( \)+100.0 \mathrm{~J} / \mathrm{K} .$ (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at $298 \mathrm{~K} .(\mathbf{d})\( Is the reaction spontaneous at \)298 \mathrm{~K}$ under standard conditions?
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