To solve this problem, we can utilize the van't Hoff equation, which relates the change in the equilibrium constant of a reaction to the change in temperature. The equation is given by:
[tex]\[
\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\][/tex]
Where:
- [tex]\(K_1\)[/tex] and [tex]\(K_2\)[/tex] are the initial and final equilibrium constants, respectively.
- [tex]\(T_1\)[/tex] and [tex]\(T_2\)[/tex] are the initial and final temperatures in Kelvin, respectively.
- [tex]\(\Delta H^\circ\)[/tex] is the standard enthalpy change.
- [tex]\(R\)[/tex] is the gas constant, [tex]\(8.314 \, \text{J/mol·K}\)[/tex].
Given the data in the problem:
[tex]\[
K_1 = 8.27, \quad K_2 = 0.0652, \quad T_1 = 25^\circ\text{C} = 298.15\, \text{K}, \quad T_2 = 72^\circ\text{C} = 345.15\, \text{K}
\][/tex]
1. Convert the temperatures to Kelvin (if necessary):
[tex]\[
T_1 = 25 + 273.15 = 298.15 \, \text{K}
\][/tex]
[tex]\[
T_2 = 72 + 273.15 = 345.15 \, \text{K}
\][/tex]
2. Calculate the natural logarithm of the ratio [tex]\( \frac{K_2}{K_1} \)[/tex]:
[tex]\[
\ln\left(\frac{K_2}{K_1}\right) = \ln\left(\frac{0.0652}{8.27}\right) = -4.84293031908513
\][/tex]
3. Calculate the difference in reciprocal temperatures:
[tex]\[
\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \left(\frac{1}{345.15} - \frac{1}{298.15}\right) = -0.0004567254017962771 \, \text{K}^{-1}
\][/tex]
4. Rearrange the van't Hoff equation to solve for [tex]\(\Delta H^\circ\)[/tex]:
[tex]\[
\Delta H^\circ = -R \cdot \frac{\ln \left( \frac{K_2}{K_1} \right)}{\left( \frac{1}{T_2} - \frac{1}{T_1} \right)}
\][/tex]
5. Substitute the known values into the equation:
[tex]\[
\Delta H^\circ = -8.314 \, \text{J/mol·K} \cdot \frac{-4.84293031908513}{-0.0004567254017962771 \, \text{K}^{-1}} = -88158.27303346187 \, \text{J/mol}
\][/tex]
Converting Joules to kilojoules since enthalpy is often expressed in these units:
[tex]\[
\Delta H^\circ = -88.158 \, \text{kJ/mol}
\][/tex]
Therefore, the standard enthalpy change, [tex]\(\Delta H^\circ\)[/tex], for the reaction is approximately [tex]\(-88.16 \, \text{kJ/mol}\)[/tex].
