Find out the value of equilibrium constant for the following reaction at 298 K.
`2NH_(3)(g)+CO_(2)(g)hArrNH_(2)CONH_(2)(aq)+H_(2)O(1)`
Standard Gibbs energy change, `Delta_(r)G^(Ө)` at the given temperature is `-13.6 kJ "mol"^(-1)`

Explanation

T= 298K
`DeltaG_(r)^(@)=-13.6 "kJ.mol"^(-1)`
`=-13600 "J mol"^(-1)`
`DeltaG_(r)^(@)=-2.303RT log K_(eq)`
`log K_(eq)=(13.6"KJmol"^(-1))/(2.303xx8.314xx10^(-3)JK^(-1)mol^(-1)xx298K)`
`logK_(eq)=2.38`
`K_(eq)= ` antilog (2.38) = 239.88.