I) Using standard enthalpy of formation values from the appendix in your textbook, calculate the enthalpy of combustion, Delta H^circ comb, of one mole of ethane at 25^circ C C_(2)H_(6)(g)+3.5O_(2)(g)arrow 2CO_(2)(g)+3H_(2)O(l)

To calculate the enthalpy of combustion, $\Delta H^{\circ }$ comb, of one mole of ethane, we need to use the standard enthalpy of formation values for the reactants and products in the given chemical equation.<br /><br />The balanced chemical equation for the combustion of ethane is:<br /><br />$C_{2}H_{6}(g) + 3.5O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(l)$<br /><br />The standard enthalpy of formation values for the reactants and products are:<br /><br />$C_{2}H_{6}(g)$: $-84.7 \, kJ/mol$<br />$O_{2}(g)$: $0 \, kJ/mol$ (since it is the standard state)<br />$CO_{2}(g)$: $-393.5 \, kJ/mol$<br />$H_{2}O(l)$: $-285.8 \, kJ/mol$<br /><br />Now, we can calculate the enthalpy of combustion using the formula:<br /><br />$\Delta H^{\circ }_{\text{comb}} = \sum \Delta H^{\circ }_{f,\text{products}} - \sum \Delta H^{\circ }_{f,\text{reactants}}$<br /><br />Substituting the values, we get:<br /><br />$\Delta H^{\circ }_{\text{comb}} = [2(-393.5) + 3(-285.8)] - [(-84.7) + 3(0)]$<br /><br />$\Delta H^{\circ }_{\text{comb}} = [-787 + (-857.4)] - [-84.7]$<br /><br />$\Delta H^{\circ }_{\text{comb}} = -1644.4 + 84.7$<br /><br />$\Delta H^{\circ }_{\text{comb}} = -1559.7 \, kJ/mol$<br /><br />Therefore, the enthalpy of combustion, $\Delta H^{\circ }$ comb, of one mole of ethane at $25^{\circ }C$ is $-1559.7 \, kJ/mol$.