To find the enthalpy change (ΔH°) for the reaction, we can use the Van't Hoff equation in its linear form:
[tex]\[
\ln\left(\frac{K2}{K1}\right) = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T2} - \frac{1}{T1}\right)
\][/tex]
Given:
- [tex]\( K1 = 8.27 \)[/tex]
- [tex]\( K2 = 6.52 \times 10^{-2} \)[/tex]
- [tex]\( T1 = 25^{\circ} C \)[/tex]
- [tex]\( T2 = 72^{\circ} C \)[/tex]
- [tex]\( R = 8.314 \, \text{J/(mol*K)} \)[/tex] (the universal gas constant)
Convert the temperatures from Celsius to Kelvin:
- [tex]\( T1 = 25 + 273.15 = 298.15 \, K \)[/tex]
- [tex]\( T2 = 72 + 273.15 = 345.15 \, K \)[/tex]
The natural logarithm of the ratio [tex]\( \frac{K2}{K1} \)[/tex] is:
[tex]\[
\ln\left(\frac{K2}{K1}\right) = \ln\left(\frac{6.52 \times 10^{-2}}{8.27}\right) = -4.84293031908513
\][/tex]
Calculate the difference in the inverse of the temperatures:
[tex]\[
\frac{1}{T2} - \frac{1}{T1} = \frac{1}{345.15} - \frac{1}{298.15} = -0.0004567254017962771
\][/tex]
Now, substitute these values into the Van't Hoff equation to solve for ΔH°:
[tex]\[
-4.84293031908513 = -\frac{\Delta H^{\circ}}{8.314} \times -0.0004567254017962771
\][/tex]
Isolate ΔH°:
[tex]\[
\Delta H^{\circ} = \frac{-4.84293031908513}{-0.0004567254017962771} \times 8.314 = 88158.27303346187 \, \text{J/mol}
\][/tex]
Therefore, the enthalpy change (ΔH°) for the reaction is:
[tex]\[
\Delta H^{\circ} = 88.16 \, \text{kJ/mol} \quad (\text{since } 1 \, \text{kJ} = 1000 \, \text{J})
\][/tex]
