The maximum work (in kJ , mol^{-1}) that can | vedclass

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(A) The combustion reaction is: $CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}$Step $1$: Calculate the change in entropy $(\Delta S)$:$\De

Vedclass · Accepted Answer

$257$

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(A) The combustion reaction is: $CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}$Step $1$: Calculate the change in entropy $(\Delta S)$:$\Delta S = S_{CO_2} - (S_{CO} + \frac{1}{2} S_{O_2})$$\Delta S = 213.7 - (197.7 + \frac{1}{2} 	imes 205.1) \, J \, K^{-1} \, mol^{-1}$$\Delta S = 213.7 - (197.7 + 102.55) = -86.55 \, J \, K^{-1} \, mol^{-1} = -0.08655 \, kJ \, K^{-1} \, mol^{-1}$Step $2$: Calculate the change in Gibbs free energy $(\Delta G)$:$\Delta G = \Delta H - T \Delta S$$\Delta G = -283.0 - (298 	imes -0.08655)$$\Delta G = -283.0 + 25.79 = -257.21 \, kJ \, mol^{-1}$Step $3$: Maximum work $(w_{max})$ is equal to $-\Delta G$:$w_{max} = -(-257.21) \approx 257 \, kJ \, mol^{-1}$

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