To determine the enthalpy change (ΔH) for the reaction [tex]\( X \rightleftharpoons Y \)[/tex], we can use the van't Hoff equation, which relates the change in the equilibrium constant (K) with temperature to the enthalpy change. The necessary steps to solve the problem are as follows:
1. Given Data:
- Equilibrium constant at 40°C ([tex]\( K_{c,40°C} \)[/tex]) = 0.430
- Equilibrium constant at 90°C ([tex]\( K_{c,90°C} \)[/tex]) = 0.675
- Temperature in Kelvin:
[tex]\[ T_1 = 40 + 273.15 = 313.15 \text{ K} \][/tex]
[tex]\[ T_2 = 90 + 273.15 = 363.15 \text{ K} \][/tex]
2. Van't Hoff Equation:
The van't Hoff equation in its linear form is:
[tex]\[ \ln \left(\frac{K_2}{K_1}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]
3. Define Constants:
- [tex]\( R \)[/tex] (universal gas constant) = 8.314 J/(mol·K)
4. Calculate the natural logarithm of the ratio of K values:
[tex]\[ \ln \left(\frac{K_{c,90°C}}{K_{c,40°C}}\right) = \ln \left(\frac{0.675}{0.430}\right) \approx 0.4509274821849218 \][/tex]
5. Calculate the difference in inverse temperatures:
[tex]\[ \left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \left(\frac{1}{363.15} - \frac{1}{313.15}\right) \approx -0.00043967476466243334 \text{ K}^{-1} \][/tex]
6. Rearrange the van’t Hoff equation to solve for ΔH:
[tex]\[ \Delta H = -R \cdot \frac{\ln \left(\frac{K_{c,90°C}}{K_{c,40°C}}\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)} \][/tex]
Substituting the known values:
[tex]\[ \Delta H = -8.314 \, \text{J/(mol·K)} \cdot \frac{0.4509274821849218}{-0.00043967476466243334 \, \text{K}^{-1}} \approx 8526.78249515593 \text{ J/mol} \][/tex]
7. Convert from J/mol to kJ/mol:
[tex]\[ \Delta H (kJ/mol) = \frac{8526.78249515593 \text{ J/mol}}{1000} \approx 8.526782495155931 \text{ kJ/mol} \][/tex]
Thus, the enthalpy change (ΔH) for the reaction is approximately [tex]\( \boxed{8.53 \text{ kJ/mol}} \)[/tex].
