The schematic plot of In Keq versus inverse of temperature for a reaction is shown in the figure. The reaction must be

# The schematic plot of In ${K}_{\mathrm{eq}}$ versus inverse of temperature for a reaction is shown in the figure. The reaction must be

1. A

one with negligible enthalpy change

2. B

highly spontaneous at ordinary temperature

3. C

exothermic

4. D

endothermic

Register to Get Free Mock Test and Study Material

+91

Live ClassesRecorded ClassesTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

The variation of standard equilibrium constant with temperature is given by

$\frac{\mathrm{dln}{K}_{\mathrm{eq}}°}{\mathrm{dT}}=\frac{\Delta H°}{R{T}^{2}}$ which on integeration gives $\mathrm{ln}{K}_{\mathrm{eq}}°=-\frac{\Delta H°}{RT}+$ constant

The slope of the graph between $\mathrm{ln}{K}_{\mathrm{eq}}°$ and $1/T$ shown in the figure is positive. The slope as given by the above equation is $-\Delta H/R$ This means $∆\mathrm{H}$ is negative, hence, the reaction is exothermic.