In the In K vs. `(1)/(T)` plot of a chemical process having `triangle S^(0) gt 0` and `triangle H^(@) lt 0` the slope is proportional to (where k is equilibrium constant )

Options

A. `-|triangleH^(@)|`
B. `|triangleH^(@)|`
C. `triangleS^(@)`
D. `-triangleS^(@)|`

Answer

B

Explanation

log k =log A-`(Ea)/(2.303R).(1)/(T)`
and `triangle G =triangleH-T triangleS=2.303 Rt log k`
on ploting the graph between `(1)/(T)` and log k using straight line equation

`y=mx+c`
`log k prop (1)/(T)`
The slope is proportion to `|triangleH^(@)|`