First slide

Gibb's Helmholtz equation

Question

For the reaction 2A →B; $∆$ H = +45 kJ/mole at 27°C. The entropy values of A and B at 27°C are $90 J K^{- 1} {mole}^{- 1} and 300 J K^{- 1} {mole}^{- 1} .$ Free energy change for the reaction is and the nature of reaction is _.

Easy

Solution

2A →B; $∆$ H = +45 kJ/mole
$\begin{array}{l} ΔS = [+ 300 - 2 (90)] J K^{- 1} {mole}^{- 1} \\ = 120 J K^{- 1} {mole}^{- 1} \\ = 0 .12 kJ K^{- 1} {mole}^{- 1} \\ ΔG = ΔH - TΔS \\ ΔG = + 45 {kJmol}^{- 1} - (300 K) (0 .12 kJ K^{- 1} {mole}^{- 1}) \\ = + 9 kJ / mole \end{array}$

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