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Fakaruddin Jagdish Ganguly
Asked: 2022-11-03T15:42:35+05:30 In:

For the reaction `N_(2)(g) + 3H_(2) rarr 2NH_(3)(g)`
`Delta H = – 95.4 kJ and Delta S = -198.3 JK^(-1)`
Calculate the temperature at which Gibbs energy change `(Delta G)` is equal to zero. Predict the nature of the reaction at this temperature and above it.

For the reaction `N_(2)(g) + 3H_(2) rarr 2NH_(3)(g)`
`Delta H = – 95.4 kJ and Delta S = -198.3 JK^(-1)`
Calculate the temperature at which Gibbs energy change `(Delta G)` is equal to zero. Predict the nature of the reaction at this temperature and above it.
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  1. `Delta G = Delta H – T Delta S`
    `Delta G = 0`
    `Delta H – T Delta S = 0`
    or `Delta H = T Delta S`
    `T = (Delta H)/(Delta S)=(-95.4 xx 1000 J)/(-198.3 JK^(-1))=481 K`
    At this temperature, the reaction would be in equilibrium and with the increase in temperature the opposing factor `T Delta S` would become more and hence, `Delta G` would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous at the temperature below 481 K.

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