Calculate the standard enthalpy of combustion for benzene, C6H6, in kJ/mol: C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l) CO2(g) ΔHf° = -393.509 kJ/mol H2O(l) ΔHf° = -285.83 kJ/mol C6H6(l) ΔHf° = 48.95 kJ/mol
Added by Phillip C.
Step 1
We have 6 moles of CO2 and 3 moles of H2O, so we multiply the enthalpy of formation for each by the number of moles: 6 * -393.509 kJ/mol (for CO2) = -2361.054 kJ 3 * -285.83 kJ/mol (for H2O) = -857.49 kJ Then we add these together to get the total enthalpy of Show more…
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The thermochemical equation for the combustion of benzene is shown below: 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3909.9 kJ/mol-rxn What is the enthalpy change for the combustion of 156.2 g C6H6? a) -3909.9 kJ b) -7819.8 kJ c) +7819.8 kJ d) +1955.0 kJ e) +3909.9 kJ
Adi S.
Calculate $\Delta H_{f}^{\circ}$ in $\mathrm{kJ} / \mathrm{mol}$ for benzene, $\mathrm{C}_{6} \mathrm{H}_{6},$ from the following data: $$ \begin{aligned} 2 C_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-6534 \mathrm{kJ} \end{aligned} $$ $$ \begin{array}{l}{\Delta H^{\circ}_{\mathrm{f}}\left(\mathrm{CO}_{2}\right)=-393.5 \mathrm{kJ} / \mathrm{mol}} \\ {\Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)=-285.8 \mathrm{kJ} / \mathrm{mol}}\end{array} $$
Calculate the enthalpy change when gaseous benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)$ dissociates into gaseous atoms at $298 \mathrm{K}$. Carry out the calculation by two different methods using the data in (a) and (b) below. Comment on the difference in the values you obtain by the two methods. (Section 13.3 ) (a) Assume benzene molecules contain three single and three double carbon-carbon bonds, and use mean bond enthalpy data. (Mean bond enthalpies/kJmol- $^{-1}:$ C-C $, 347$; $\mathrm{C}=\mathrm{C}, 612 ; \mathrm{C}-\mathrm{H}, 412 .$ (b) The enthalpy change of combustion of liquid benzene at $298 \mathrm{K}$ is $-3267.4 \mathrm{kJmol}^{-1}$. The enthalpy change of vaporization of benzene at $298 \mathrm{K}$ is $+33.9 \mathrm{kJ} \mathrm{mol}^{-1}$ \[ \begin{array}{l} \left(\mathrm{A}_{1} \mathrm{H}_{298}^{\circ} / \mathrm{kJmol}^{-1}: \mathrm{CO}_{2}(\mathrm{g}),-393.5 ; \mathrm{H}_{2} \mathrm{O}(0,-285.8 ; \mathrm{C}(\mathrm{g})\right. \\ 716.7 ; \mathrm{H}(\mathrm{g}), 218 .) \end{array} \]
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