Questions: Calculate ΔHmin° at 25°C Express your answer in kilojoules to three significant figures. Liquid ethanol burns in oxygen to form gaseous carbon dioxide and gaseous water Substance H°(kJ / mol) S°(J /(mol · K)) C2H5OH(l) -277.6 160.7 O2(g) 0 205.2 CO2(g) -393.5 213.8 CO2(aq) -413.8 117.6 H2O(g) -241.8 188.8 H2O(l) -187.8 70.0 ΔHmin°= □ kJ Be sure to consider the states of the participants of the reaction and use ΔHf° values from the table in your calculations.

Calculate ΔHmin° at 25°C

Express your answer in kilojoules to three significant figures. Liquid ethanol burns in oxygen to form gaseous carbon dioxide and gaseous water Substance H°(kJ / mol) S°(J /(mol · K)) C2H5OH(l) -277.6 160.7 O2(g) 0 205.2 CO2(g) -393.5 213.8 CO2(aq) -413.8 117.6 H2O(g) -241.8 188.8 H2O(l) -187.8 70.0 ΔHmin°= □ kJ

Be sure to consider the states of the participants of the reaction and use ΔHf° values from the table in your calculations.

Transcript text: Calculate $\Delta H_{\text {min }}^{\circ}$ at $25^{\circ} \mathrm{C}$ Express your answer in kilojoules to three significant figures. Liquid ethanol burns in oxygen to form gaseous carbon dioxide and gaseous water \begin{tabular}{|c|c|c|} \hline Substance & $H^{\circ}(\mathrm{kJ} / \mathrm{mol})$ & $S^{\circ}(\mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}))$ \\ \hline $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$ & -277.6 & 160.7 \\ \hline $\mathrm{O}_{2}(\mathrm{~g})$ & 0 & 205.2 \\ \hline $\mathrm{CO}_{2}(\mathrm{~g})$ & -393.5 & 213.8 \\ \hline $\mathrm{CO}_{2}(\mathrm{aq})$ & -413.8 & 117.6 \\ \hline $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ & -241.8 & 188.8 \\ \hline $\mathrm{H}_{2} \mathrm{O}(\mathrm{I})$ & -187.8 & 70.0 \\ \hline \end{tabular} $\Delta H_{\mathrm{min}}^{\mathrm{o}}=$ $\square$ kJ Be sure to consider the states of the participants of the reaction and use $\Delta H_{\mathrm{f}}^{\circ}$ values from the table in your calculations.

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the combustion of liquid ethanol ($\mathrm{C_2H_5OH(l)}$) in oxygen ($\mathrm{O_2(g)}$) to form carbon dioxide ($\mathrm{CO_2(g)}$) and water ($\mathrm{H_2O(g)}$) is: \[ \mathrm{C_2H_5OH(l)} + 3\mathrm{O_2(g)} \rightarrow 2\mathrm{CO_2(g)} + 3\mathrm{H_2O(g)} \]

Step 2: Identify the Enthalpy of Formation Values

From the given table, we have the standard enthalpy of formation ($H^{\circ}$) values:

$\mathrm{C_2H_5OH(l)}$: $-277.6 \, \mathrm{kJ/mol}$
$\mathrm{O_2(g)}$: $0 \, \mathrm{kJ/mol}$
$\mathrm{CO_2(g)}$: $-393.5 \, \mathrm{kJ/mol}$
$\mathrm{H_2O(g)}$: $-241.8 \, \mathrm{kJ/mol}$

Step 3: Calculate the Enthalpy Change of the Reaction

The enthalpy change of the reaction ($\Delta H_{\text{rxn}}^{\circ}$) can be calculated using the formula: \[ \Delta H_{\text{rxn}}^{\circ} = \sum \Delta H_{\text{f, products}}^{\circ} - \sum \Delta H_{\text{f, reactants}}^{\circ} \]

Step 4: Substitute the Values into the Formula

Substitute the enthalpy of formation values into the formula: \[ \Delta H_{\text{rxn}}^{\circ} = \left[2(-393.5) + 3(-241.8)\right] - \left[1(-277.6) + 3(0)\right] \]

Step 5: Perform the Calculations

Calculate the enthalpy change for the products: \[ 2(-393.5) + 3(-241.8) = -787.0 - 725.4 = -1512.4 \, \mathrm{kJ} \]

Calculate the enthalpy change for the reactants: \[ -277.6 + 3(0) = -277.6 \, \mathrm{kJ} \]

Calculate the overall enthalpy change: \[ \Delta H_{\text{rxn}}^{\circ} = -1512.4 - (-277.6) = -1512.4 + 277.6 = -1234.8 \, \mathrm{kJ} \]