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

A
one with negligible enthalpy change
B
highly spontaneous at ordinary temperature
C
exothermic
D
endothermic

Solution:

The variation of standard equilibrium constant with temperature is given by

$\frac{dln K_{eq} °}{dT} = \frac{Δ H °}{R T^{2}}$ which on integeration gives $\ln K_{eq} ° = - \frac{Δ H °}{R T} +$ constant

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

The schematic plot of In $K_{eq}$ versus inverse of temperature for a reaction is shown in the figure. The reaction must be

Solution:

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