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Chemistry Question on Thermodynamics

How much energy is released when 6 moles of octane is burnt in air ? Given ΔHf∘\Delta H_{f}{ }^{\circ} for CO2(g),H2O(g)CO _{2}(g), H _{2} O (g) and C8H18(l)C _{8} H _{18}(l) respectively are −490,−240-490,-240 and +160 J/mol:+160\, J / mol :

A

- 6.2 kJ

B

- 37.4 kJ

C

- 35.5 kJ

D

- 20.0 kJ

Answer

- 37.4 kJ

Explanation

Solution

C+O2⟶CO2;ΔHf∘=−490 kJ/mol×8C + O_2 \longrightarrow CO_2; \Delta H_f^{\circ} = -490 \, kJ /mol \times 8 H2+12O2⟶H2O;ΔHf∘=−240 kJ/mol×9H_2 + \frac{1}{2} O_2\longrightarrow H_2 O ; \Delta H_f^{\circ} = -240\, kJ /mol \times 9 8C+18H⟶C8H18;ΔHf∘=+160 kJ/mol8C + 18H \longrightarrow C_8 H_{18} ; \Delta H_f^{\circ} = +160 \, kJ /mol 8C+8O2+9H2+92O2−8C−18H⟶8CO2+9H2O−C8H18;8C + 8O_2 + 9H_2 + \frac{9}{2} O_2 - 8C - 18 H \longrightarrow 8CO_2 + 9 H_2 O - C_8 H_{18} ; ΔHf∘=−3920−2160−160\Delta H_f^{\circ} = -3920 - 2160 - 160 C8H18+252O2⟶8CO2+9H2OC_8 H_{18} + \frac{25}{2} O_2 \longrightarrow 8CO_2 + 9H_2O; ΔHf∘=6240 kJ/mol\Delta H_f^{\circ} = 6240 \, kJ /mol ΔH∘\Delta H^{\circ} for 66 moles of octane = 6240×6=37440 kJ/mol 6240 \times 6 = 37440 \, kJ / mol.