6.17   For the reaction at 298 K,
            2A + B \rightarrow C
          \Delta H = 400 kJ mol^ {-1 }\: \: and \: \: \Delta S = 0.2 kJ K ^{-1} mol^{-1}
At what temperature will the reaction become spontaneous considering \Delta H and \Delta S to be constant over the temperature range.

Answers (1)
M manish

From the equation,

\Delta G = \Delta H-T\Delta S
Suppose the reaction is at equilibrium, So the change in temperature is given as;

T=\frac{\Delta H -\Delta G}{\Delta S}
      = \frac{400}{0.2} =2000K  (\Delta G at equilibrium is zero)
    To reaction should be spontaneous, \Delta G should be neagtive. So, that for the given reaction T should be greater than 2000 K

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