Question

The standard enthalpies of formation, at 25.00degrees celsius, of methanol (CH4O (l)), water (H2O (l)), and carbon dioxide (CO2 (g)) are respectively -238.7 kJ / mol, - 285.8 kJ / mol, and -393.5 kJ / mol. Calculate the change in the surrounding entropy (in J / K) when burning 15.4 g of methanol under a constant pressure of 1,000 atm at 25.00 degrees celsius (combustion is the reaction of a substance with oxygen molecular to produce water and carbon dioxide).

Answer 1

Answer on Question #81926, Chemistry/ General Chemistry

The standard enthalpies of formation, at 25.00 degrees celsius, of methanol (CH4O (l)), water (H2O (l)), and carbon dioxide (CO2 (g)) are respectively -238.7 kJ/mol, -285.8 kJ/mol, and -393.5 kJ/mol. Calculate the change in the surrounding entropy (in J/K) when burning 15.4 g of methanol under a constant pressure of 1,000 atm at 25.00 degrees celsius (combustion is the reaction of a substance with oxygen molecular to produce water and carbon dioxide).

Solution

$$2 \mathrm{CH_3OH} (l) + 3 \mathrm{O_2(g)} \rightarrow 2 \mathrm{CO_2(g)} + 4 \mathrm{H_2O(l)}$$

Calculate the standard enthalpy of this reaction

$$\Delta H_f^0(\mathrm{CH_4O(l)}) = -238.7\ \mathrm{kJ/mol}$$ $$\Delta H_f^0(\mathrm{CO_2(g)}) = -393.5\ \mathrm{kJ/mole}.$$ $$\Delta H_f^0(\mathrm{H_2O(l)}) = -285.8\ \mathrm{kJ/mol}$$ $$\Delta H_f^0(\mathrm{O_2(g)}) = 0\ \mathrm{kJ/mol}$$ $$\Delta H_{rxn}^0 = \sum H_{f(\text{products})}^0 - \sum H_{f(\text{reactants})}^0$$ $$\begin{aligned} \Delta H_{rxn}^0 &= \left[2\ \mathrm{mol} \times H_f^0(\mathrm{CO_2(g)}) + 4\ \mathrm{mol} \times H_f^0(\mathrm{H_2O(l)})\right] \\ &\quad - \left[2\ \mathrm{mol} \times H_f^0(\mathrm{CH_4O(l)}) + 3\ \mathrm{mol} \times H_f^0(\mathrm{O_2(g)})\right] \\ &= \left[2\ \mathrm{mol} \times \left(-393.5\ \frac{\mathrm{kJ}}{\mathrm{mol}}\right) + 4\ \mathrm{mol} \times \left(-285.8\ \frac{\mathrm{kJ}}{\mathrm{mol}}\right)\right] \\ &\quad - \left[2\ \mathrm{mol} \times \left(-238.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\right) + 1\ \mathrm{mol} \times \left(0\ \frac{\mathrm{kJ}}{\mathrm{mol}}\right)\right] = -1452.8\ \mathrm{kJ} \end{aligned}$$

Calculate the standard enthalpy of this reaction when 15.4 g of methanol is burned:

$$n = m/M$$ $$M(\mathrm{CH_4O}) = 32\ \mathrm{g/mol}$$ $$n(\mathrm{CH_4O}) = \frac{15.4\ \mathrm{g}}{32\ \frac{\mathrm{g}}{\mathrm{mol}}} = 0.48\ \mathrm{mol}$$

Solve the proportion:

When 2 mol of methanol is burned $\Delta H_{rxn}^0 = -1452.8\ \mathrm{kJ}$

When 0.48 mol of methanol is burned $\Delta H_{rxn}^0 = x\ \mathrm{kJ}$

$$\frac{2}{0.48} = \frac{-1452.8}{x}$$ $$x = -348.7 \, \text{kJ}$$ $$\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}$$ $$\Delta S_{\text{surroundigs}} = -\frac{-348700 \, \text{J}}{(273 + 25) \, \text{K}} = 1170 \, \frac{\text{J}}{\text{K}}$$

Answer: 1170 J/K

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