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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)` |
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Answer» We know log, `K=(-Delta_(r)G^(Ө))/(2.303RT)` `=((-13.6xx10^(3)J"mol"^(-1)))/(2.303(8.314JK^(-1)"mol"^(-1))(298K))` `=2.38` Hence K = antilog `2.38 = 2.4xx10^(2)` |
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