JEE MAIN - Chemistry (2018 (Offline) - No. 22)
Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an
exothermic reaction?
_en_22_1.png)
A and D
A and B
B and C
C and D
Explanation
We know from thermodynamics,
$$\Delta $$Go = $$\Delta $$Ho $$-$$ T$$\Delta $$So
$$ \Rightarrow \,\,\,\, - RT\ln K = \Delta {H^o} - T\Delta {S^o}$$
$$ \Rightarrow \,\,\,\,\ln K = - {{\Delta {H^o}} \over R} \times {1 \over T} + {{\Delta {S^o}} \over R}.......\left( 1 \right)$$
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.
_en_22_2.png)
$$\Delta $$Go = $$\Delta $$Ho $$-$$ T$$\Delta $$So
$$ \Rightarrow \,\,\,\, - RT\ln K = \Delta {H^o} - T\Delta {S^o}$$
$$ \Rightarrow \,\,\,\,\ln K = - {{\Delta {H^o}} \over R} \times {1 \over T} + {{\Delta {S^o}} \over R}.......\left( 1 \right)$$
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.
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