Question

For a given exothermic reaction, K_p and K'_P are the equilibrium constants at temperatures T₁ and T₂, respectively. Assuming that heat of reaction is constant in temperature range between T₁ and T₂, it is readily observed that:

• A. K_p<K'_p
• B. K_p=K'_p
• C. \(Kp=\frac {1}{K'p}\)
• D. K_p>K'_p

Answer 1

The Correct Option is D

 To solve this question, we need to understand the relationship between temperature and equilibrium constants for an exothermic reaction. For an exothermic reaction, the equilibrium constant \(K_p\) decreases with an increase in temperature due to Le Chatelier's principle.

The relationship between the equilibrium constants at two different temperatures for a given reaction is given by the Van't Hoff equation:

\(\ln \frac{K'_p}{K_p} = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\)

Where:

• \(\Delta H\) is the change in enthalpy (negative for exothermic reactions).
• \(R\) is the universal gas constant.
• \(T_1\) and \(T_2\) are the temperatures (with \(T_2 > T_1\)).

For an exothermic reaction, \(\Delta H\) is negative, so the expression on the right-hand side of the Van't Hoff equation is positive if \(T_2 > T_1\). This implies:

\(\ln \frac{K'_p}{K_p}\) is positive, meaning \(\frac{K'_p}{K_p} > 1\) or \(K'_p > K_p\)

Therefore, for an exothermic reaction:

If the temperature increases from \(T_1\) to \(T_2\), the equilibrium constant \(K_p\) at \(T_1\) is greater than \(K'_p\) at \(T_2\).

Conclusion: The correct answer is \(K_p > K'_p\), which means that the equilibrium constant decreases with an increase in temperature for an exothermic reaction, aligning with the concept that equilibrium shifts to the left (towards reactants) to absorb extra heat.