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Answer :
We start with the relation between the standard Gibbs free energy change and the equilibrium constant given by
[tex]$$
\Delta G^\circ = -RT \ln K,
$$[/tex]
where
- [tex]$\Delta G^\circ$[/tex] is the standard Gibbs free energy change,
- [tex]$R$[/tex] is the universal gas constant, and
- [tex]$T$[/tex] is the temperature in Kelvin.
Given that [tex]$\Delta G^\circ = -97.7\;\text{kJ}$[/tex], we first convert this to Joules:
[tex]$$
\Delta G^\circ = -97.7 \times 1000 = -97700\;\text{J}.
$$[/tex]
The temperature is given as [tex]$10.0^\circ\text{C}$[/tex], which we convert to Kelvin:
[tex]$$
T = 10.0 + 273.15 = 283.15\;\text{K}.
$$[/tex]
The gas constant is
[tex]$$
R = 8.314\;\text{J/(mol$\cdot$K)}.
$$[/tex]
Rearranging the equation for [tex]$\ln K$[/tex], we have
[tex]$$
\ln K = -\frac{\Delta G^\circ}{RT}.
$$[/tex]
Substitute the known values:
[tex]$$
\ln K = -\frac{-97700}{8.314 \times 283.15}.
$$[/tex]
Notice that the negatives cancel:
[tex]$$
\ln K = \frac{97700}{8.314 \times 283.15}.
$$[/tex]
Calculating the product in the denominator:
[tex]$$
8.314 \times 283.15 \approx 2354.11,
$$[/tex]
we obtain
[tex]$$
\ln K \approx \frac{97700}{2354.11} \approx 41.50.
$$[/tex]
To find [tex]$K$[/tex], we exponentiate both sides:
[tex]$$
K = e^{41.50}.
$$[/tex]
Evaluating the exponential function gives
[tex]$$
K \approx 1.06 \times 10^{18}.
$$[/tex]
Thus, to 2 significant digits, the equilibrium constant is
[tex]$$
K \approx 1.06 \times 10^{18}.
$$[/tex]
[tex]$$
\Delta G^\circ = -RT \ln K,
$$[/tex]
where
- [tex]$\Delta G^\circ$[/tex] is the standard Gibbs free energy change,
- [tex]$R$[/tex] is the universal gas constant, and
- [tex]$T$[/tex] is the temperature in Kelvin.
Given that [tex]$\Delta G^\circ = -97.7\;\text{kJ}$[/tex], we first convert this to Joules:
[tex]$$
\Delta G^\circ = -97.7 \times 1000 = -97700\;\text{J}.
$$[/tex]
The temperature is given as [tex]$10.0^\circ\text{C}$[/tex], which we convert to Kelvin:
[tex]$$
T = 10.0 + 273.15 = 283.15\;\text{K}.
$$[/tex]
The gas constant is
[tex]$$
R = 8.314\;\text{J/(mol$\cdot$K)}.
$$[/tex]
Rearranging the equation for [tex]$\ln K$[/tex], we have
[tex]$$
\ln K = -\frac{\Delta G^\circ}{RT}.
$$[/tex]
Substitute the known values:
[tex]$$
\ln K = -\frac{-97700}{8.314 \times 283.15}.
$$[/tex]
Notice that the negatives cancel:
[tex]$$
\ln K = \frac{97700}{8.314 \times 283.15}.
$$[/tex]
Calculating the product in the denominator:
[tex]$$
8.314 \times 283.15 \approx 2354.11,
$$[/tex]
we obtain
[tex]$$
\ln K \approx \frac{97700}{2354.11} \approx 41.50.
$$[/tex]
To find [tex]$K$[/tex], we exponentiate both sides:
[tex]$$
K = e^{41.50}.
$$[/tex]
Evaluating the exponential function gives
[tex]$$
K \approx 1.06 \times 10^{18}.
$$[/tex]
Thus, to 2 significant digits, the equilibrium constant is
[tex]$$
K \approx 1.06 \times 10^{18}.
$$[/tex]
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