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

Question

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

cdquestions Admin · Accepted Answer

From the expression, $∆G = ∆H – T∆S$ Assuming the reaction at equilibrium,  $∆T$  for the reaction would be: $T = (△H-△G)\frac {1}{△S}$ $T= \frac {△H}{△S}$          ( $∆G = 0$  at equilibrium) $T=\frac { 400\ kJ mol^{-1}}{0.2\ kJK^{-1} mol^{-1}}$ $T = 2000\ K$ For the reaction to be spontaneous,  $∆G$  must be negative. Hence, for the given reaction to be spontaneous,  $T$  should be greater than  $2000\ K$ .

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

Solution and Explanation

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Gibbs Free Energy

The Conditions of Equilibrium

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