`DeltaH^(Theta)` and `DeltaS^(Theta)` for the reaction:
`Br_(2)(l) +CI_(2)(g) hArr 2BrCI(g)`
at `298K` are `29.3 kJ mol^(-1)` and `104.1 J K^(-1) mol^(-1)`, respectively. Calculate the equilibrium constant for the reaction.

Answer

`2.003`

Explanation

For the reaction:
`Br_(2)(l) + CI_(2)(g) to 2BrCI(G)`
` DeltaH^(@) = 29.3 kJ "mol"^(-1), DeltaS^(@) = 104.1 J K^(-1) "mol"^(-1)`
`DeltaG^(@) = DeltaH^(@) - TDeltaS^(@)`
` = 29.3 xx 10^(3) - 298 xx 104.1`
`= 29300 - 31021.8`
` = -1721.8 J "mol"^(-1)`
Now `log K = -(DeltaG^(@))/(2.303 RT)`
` = -(-1721.8)/(2.303 xx 8.314 xx 298)`
` = 0.3018`
or `K = 2.003`.