For the reaction,
`A(g)+B(g) rarr C(g)+D(g), DeltaH^(@)` and `DeltaS^(@)`
are respectively,
`-29.8 kJ mol^(-1)` and `-0.100 kJ K^(-1) mol^(-1)` at 298 K
The equilibrium constant for the reaction at 298 K is :

Options

A. `1.0xx10^(-10)`
B. 10
C. 1
D. `1.0xx10^(10)`

Answer

C

Explanation

Given `DeltaH^(@)=-29.8kJmol^(-1)`
`DeltaS^(@)=-1.00kJK^(-1)`
From the equation
`DeltaG=DeltaH^(@)-TDeltaS^(@)=-29.8-(298xx-0.100)`
`=-29.8+29.8=0`
Now, `DeltaG^(@)=-2.303RTlogK_(eq)`
`0=-2.303RTlogK_(eq)`
`:.K_(eq)=1`