Question

Which of following lines correctly show the temperature dependence of equilibrium constant $K$, for an exothermic reaction ?

• A. A and B
• B. B and C
• C. C and D
• D. A and D

Answer 1

Answer: (A)

Equilibrium constant $K=\left(\frac{A_f}{A_b}\right)e^{-\frac{\Delta H^{\circ }}{RT}}$

ln $K=$ ln $\left(\frac{A_f}{A_b}\right)-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T}\right)$

$y=C+mx$

Comparing with equation of straight line,

Slope $=\frac{-\Delta H^{\circ }}{R}$

Since, reaction is exothermic, $\Delta H^{\circ }=-ve$, therefore, slope $=+ve$