Question

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

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

Answer 1

The correct option is D

A and B

A correct answer is an option(d) And B

The explanation for the incorrect option

$$\begin{gathered} ∆\mathrm{G}=∆\mathrm{G}°+\mathrm{RT}\ln \mathrm{q}\mathrm{A}+\mathrm{B}\underset{}{\rightleftharpoons \mathrm{C}+\mathrm{D}} \\ \mathrm{at}\mathrm{equillibrium},∆\mathrm{g}=0{\mathrm{k}}_{\mathrm{c}}=\left(\mathrm{C}\right)+\left(\mathrm{D}\right)÷\left(\mathrm{A}\right)+\left(\mathrm{B}\right) \\ 0=∆\mathrm{G}°+\mathrm{RT}\ln \mathrm{K}{\mathrm{q}}_{\mathrm{c}=}\left(\mathrm{C}\right)+\left(\mathrm{D}\right)÷\left(\mathrm{A}\right)+\left(\mathrm{B}\right) \\ ∆\mathrm{G}°=-\mathrm{RT}\ln \mathrm{K}\to \mathrm{equation}1 \\ ∆\mathrm{G}°=∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}°\to \mathrm{equation}1 \\ -\mathrm{RT}\ln \mathrm{K}=∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}° \\ \ln \mathrm{K}=\raisebox{1ex}{$-∆\mathrm{H}°$}\!\left/ \!\raisebox{-1ex}{$\mathrm{R}$}\right.\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.\right)+\raisebox{1ex}{$∆\mathrm{S}°$}\!\left/ \!\raisebox{-1ex}{$\mathrm{R}$}\right. \\ \mathrm{y}=\mathrm{mx}+\mathrm{c} \end{gathered}$$

For endothermic reaction,$∆\mathrm{H}°=\mathrm{positive},\mathrm{slope}=\mathrm{negative}$

The explanation for the correct option

For endothermic reaction,$∆\mathrm{H}°=\mathrm{negative},\mathrm{slope}=\mathrm{positive}$

Conclusion:

Hence Option (d) A and B is correct.