Using the data given below estimate the value of the Gibbs energy change of reaction (kJ mol^-1) at 310 K, for the following reaction:
CH₃CH₂OH (l) + O₂ (g) → CH₃CO₂H (l) + H₂O (l)
$\begin{array}{|c|c|c|c|c|}\hline & \begin{array}{c}\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \\(\mathrm{l}) \end{array} & \mathrm{O}_{2}(\mathrm{~g}) & \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{I}) & \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\\hline \Delta_{\mathrm{f}} H_{298}^{\circ} / \mathrm{kJ} \mathrm{mol}^{-1} & -277.6 & 0 & -484.5 & -285.8 \\\hline S_{298}^{\circ} / \mathrm{JK}^{-1} \mathrm{~mol}^{-1} & 159.9 & 205.1 & 158.0 & 69.9 \\\hline C_{p} / \mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1} & 111.5 & 29.4 & 123.1 & 75.3 \\\hline\end{array}$

Question

Using the data given below estimate the value of the Gibbs energy change of reaction (kJ mol^-1) at 310 K, for the following reaction:
CH₃CH₂OH (l) + O₂ (g) → CH₃CO₂H (l) + H₂O (l)

\begin{array}{|c|c|c|c|c|}\hline & \begin{array}{c}\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \\(\mathrm{l}) \end{array} & \mathrm{O}_{2}(\mathrm{~g}) & \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{I}) & \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\\hline \Delta_{\mathrm{f}} H_{298}^{\circ} / \mathrm{kJ} \mathrm{mol}^{-1} & -277.6 & 0 & -484.5 & -285.8 \\\hline S_{298}^{\circ} / \mathrm{JK}^{-1} \mathrm{~mol}^{-1} & 159.9 & 205.1 & 158.0 & 69.9 \\\hline C_{p} / \mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1} & 111.5 & 29.4 & 123.1 & 75.3 \\\hline\end{array}

Quizplus · Accepted Answer

The correct answer is A because the Gibbs energy change of reaction can be calculated using the equation:$$\Delta G = \Delta H - T\Delta S$$where $\Delta H$ is the enthalpy change of reaction, $\Delta S$ is the entropy change of reaction, and $T$ is the temperature in Kelvin.The enthalpy change of reaction can be calculated using the equation:$$\Delta H = \sum \Delta_{\mathrm{f}} H_{298}^{\circ}(\text{products}) - \sum \Delta_{\mathrm{f}} H_{298}^{\circ}(\text{reactants})$$where $\Delta_{\mathrm{f}} H_{298}^{\circ}$ is the standard enthalpy of formation of the substance.The entropy change of reaction can be calculated using the equation:$$\Delta S = \sum S_{298}^{\circ}(\text{products}) - \sum S_{298}^{\circ}(\text{reactants})$$where $S_{298}^{\circ}$ is the standard entropy of the substance.Substituting the values given in the table into these equations, we get:$$\Delta H = (-484.5) + (-285.8) - (-277.6) - (0) = -451.7 \text{ kJ mol}^{-1}$$$$\Delta S = (158.0) + (69.9) - (159.9) - (205.1) = -147.1 \text{ J K}^{-1} \text{ mol}^{-1}$$$$\Delta G = -451.7 \text{ kJ mol}^{-1} - (310 \text{ K})(-147.1 \text{ J K}^{-1} \text{ mol}^{-1}) = -450.2 \text{ kJ mol}^{-1}$$

Using the Data Given Below Estimate the Value of the Gibbs