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The equilibrium constant at 25^(@)C for ...

The equilibrium constant at `25^(@)C` for the process
`CO^(3+)(aq) + 6NH_(3)(aq) hArr [CO(NH_(3))_(6)]^(3)(aq)` is `2.5 xx 10^(6)`. Calculate the value of `Delta G^(@)` at `25^(@)C`. `(R = 8.314 JK^(-1) mol^(-1))`. In which direction is the reaction spontaneous under standard conditions ?

Text Solution

Verified by Experts

`Delta G^(@) = -2.303 RT log K`
`= -2.303 xx 3.134 xx 298 log (2.5 xx 10^(6))`
`= -5705.8[0.3980 + 60000] = -5705.8 xx 6.3980 = -36.505 k//mol`
The reaction is spontaneous in the forward direction under standard conditions.
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