Answer :
We start with the reaction
[tex]$$
HNO_2(aq) + H_2 O(l) \rightleftharpoons NO_2^{-}(aq) + H_3 O^{+}(aq),
$$[/tex]
with the equilibrium constant
[tex]$$
K = 7.2 \times 10^{-4} \quad \text{(at 298 K)}.
$$[/tex]
Under the given conditions, the concentrations are:
- [tex]$[HNO_2] = 1.0\, M$[/tex],
- [tex]$[NO_2^-] = 1.0 \times 10^{-5}\, M$[/tex],
- [tex]$[H_3O^+] = 1.0 \times 10^{-5}\, M$[/tex].
Since water is a pure liquid, its concentration is not included in the reaction quotient. The reaction quotient [tex]$Q$[/tex] is defined as
[tex]$$
Q = \frac{[NO_2^-][H_3O^+]}{[HNO_2]}.
$$[/tex]
Substitute the given values:
[tex]$$
Q = \frac{(1.0 \times 10^{-5})(1.0 \times 10^{-5})}{1.0} = 1.0 \times 10^{-10}.
$$[/tex]
The standard free energy change [tex]$\Delta G^\circ$[/tex] is related to the equilibrium constant by the equation
[tex]$$
\Delta G^\circ = -RT\ln K,
$$[/tex]
where
- [tex]$R = 8.314 \, \text{J/(mol·K)}$[/tex] is the universal gas constant, and
- [tex]$T = 298\, K$[/tex] is the temperature.
Similarly, the free energy change under non-standard conditions is given by
[tex]$$
\Delta G = \Delta G^\circ + RT\ln Q.
$$[/tex]
This can be rearranged to:
[tex]$$
\Delta G = -RT\ln K + RT\ln Q = RT\ln\left(\frac{Q}{K}\right).
$$[/tex]
Now, calculate the natural logarithms:
1. For the equilibrium constant:
[tex]$$
\ln K = \ln(7.2 \times 10^{-4}) \approx -7.2363.
$$[/tex]
2. For the reaction quotient:
[tex]$$
\ln Q = \ln(1.0 \times 10^{-10}) \approx -23.0259.
$$[/tex]
Substitute these into the expression for [tex]$\Delta G$[/tex]:
[tex]$$
\Delta G = RT \, (\ln Q - \ln K).
$$[/tex]
The difference in the logarithms is:
[tex]$$
\ln Q - \ln K = (-23.0259) - (-7.2363) = -15.7896.
$$[/tex]
Now, compute [tex]$RT$[/tex]:
[tex]$$
RT = (8.314\, \text{J/(mol·K)}) \times (298\, K) \approx 2478.7 \, \text{J/mol}.
$$[/tex]
Multiply to get [tex]$\Delta G$[/tex]:
[tex]$$
\Delta G = 2478.7\, \text{J/mol} \times (-15.7896) \approx -39120 \, \text{J/mol}.
$$[/tex]
Convert joules to kilojoules:
[tex]$$
\Delta G \approx -39.1\, \text{kJ/mol}.
$$[/tex]
Thus, the change in free energy under the given conditions is approximately
[tex]$$
\boxed{-39.1\, \text{kJ/mol}}.
$$[/tex]
This corresponds to answer option D.
[tex]$$
HNO_2(aq) + H_2 O(l) \rightleftharpoons NO_2^{-}(aq) + H_3 O^{+}(aq),
$$[/tex]
with the equilibrium constant
[tex]$$
K = 7.2 \times 10^{-4} \quad \text{(at 298 K)}.
$$[/tex]
Under the given conditions, the concentrations are:
- [tex]$[HNO_2] = 1.0\, M$[/tex],
- [tex]$[NO_2^-] = 1.0 \times 10^{-5}\, M$[/tex],
- [tex]$[H_3O^+] = 1.0 \times 10^{-5}\, M$[/tex].
Since water is a pure liquid, its concentration is not included in the reaction quotient. The reaction quotient [tex]$Q$[/tex] is defined as
[tex]$$
Q = \frac{[NO_2^-][H_3O^+]}{[HNO_2]}.
$$[/tex]
Substitute the given values:
[tex]$$
Q = \frac{(1.0 \times 10^{-5})(1.0 \times 10^{-5})}{1.0} = 1.0 \times 10^{-10}.
$$[/tex]
The standard free energy change [tex]$\Delta G^\circ$[/tex] is related to the equilibrium constant by the equation
[tex]$$
\Delta G^\circ = -RT\ln K,
$$[/tex]
where
- [tex]$R = 8.314 \, \text{J/(mol·K)}$[/tex] is the universal gas constant, and
- [tex]$T = 298\, K$[/tex] is the temperature.
Similarly, the free energy change under non-standard conditions is given by
[tex]$$
\Delta G = \Delta G^\circ + RT\ln Q.
$$[/tex]
This can be rearranged to:
[tex]$$
\Delta G = -RT\ln K + RT\ln Q = RT\ln\left(\frac{Q}{K}\right).
$$[/tex]
Now, calculate the natural logarithms:
1. For the equilibrium constant:
[tex]$$
\ln K = \ln(7.2 \times 10^{-4}) \approx -7.2363.
$$[/tex]
2. For the reaction quotient:
[tex]$$
\ln Q = \ln(1.0 \times 10^{-10}) \approx -23.0259.
$$[/tex]
Substitute these into the expression for [tex]$\Delta G$[/tex]:
[tex]$$
\Delta G = RT \, (\ln Q - \ln K).
$$[/tex]
The difference in the logarithms is:
[tex]$$
\ln Q - \ln K = (-23.0259) - (-7.2363) = -15.7896.
$$[/tex]
Now, compute [tex]$RT$[/tex]:
[tex]$$
RT = (8.314\, \text{J/(mol·K)}) \times (298\, K) \approx 2478.7 \, \text{J/mol}.
$$[/tex]
Multiply to get [tex]$\Delta G$[/tex]:
[tex]$$
\Delta G = 2478.7\, \text{J/mol} \times (-15.7896) \approx -39120 \, \text{J/mol}.
$$[/tex]
Convert joules to kilojoules:
[tex]$$
\Delta G \approx -39.1\, \text{kJ/mol}.
$$[/tex]
Thus, the change in free energy under the given conditions is approximately
[tex]$$
\boxed{-39.1\, \text{kJ/mol}}.
$$[/tex]
This corresponds to answer option D.
