InterviewSolution
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InterviewSolution
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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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Answer» `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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