QJEEM18Ch0948: JEE Main 2018 Chemistry - Class 11 JEEM18Ch0948 Solution: Which of the following lines correctly show the temperature…

We know from thermodynamics,

$\Delta$ G^o = $\Delta$ H^o $-$ T $\Delta$ S^o

$\Rightarrow \,\,\,\, - RT\ln K = \Delta {H^o} - T\Delta {S^o}$

$\Rightarrow \,\,\,\,\ln K = - {{\Delta {H^o}} \over R} \times {1 \over T} + {{\Delta {S^o}} \over R}.......\left( 1 \right)$

We know, equation of straight line is

$y = mx + c\,\,\,.........\,\,\,\left( 2 \right)$

by comparing (1) and (2) we get,

slope (m) = $-$ ${{\Delta {H^o}} \over R}$ and intercept (c) = ${{\Delta {S^o}} \over R}$

For exothermic reaction $\Delta {H^o} = - Ve$

$\Rightarrow \,\,\,\, - \Delta {H^o} = + ve$

$\Rightarrow \,\,\,\,{{ - \Delta {H^o}} \over R} = + ve$

$\therefore\,\,\,$ Slope is positive, so, graph A and B is possible.

Now if entropy $\left( {\Delta {S^o}} \right)$ is positive then intercept $\left( {{{\Delta {S^o}} \over R}} \right)$ is positive. It is represented by graph A.

If entropy $\left( {\Delta {S^o}} \right)$ is negative then intercept $\left( {{{\Delta {S^o}} \over R}} \right)$ is negative. It is represented by graph B.