what is the enthalpy of combustion when 1 mol c₆h₆(g) completely reacts with oxygen? 2c₆h₆(g) + 15o₂(g) →…

what is the enthalpy of combustion when 1 mol c₆h₆(g) completely reacts with oxygen? 2c₆h₆(g) + 15o₂(g) → 12co₂(g) + 6h₂o(g)\n-6339 kj/mol\n-3169 kj/mol\n1268 kj/mol\n6339 kj/mol\ndone\n| compound | δhᵢ (kj/mol) |\n| ---- | ---- |\n| c₆h₆(g) | 82.90 |\n| co₂(g) | -393.50 |\n| h₂o(g) | -241.82 |

what is the enthalpy of combustion when 1 mol c₆h₆(g) completely reacts with oxygen? 2c₆h₆(g) + 15o₂(g) → 12co₂(g) + 6h₂o(g)\n-6339 kj/mol\n-3169 kj/mol\n1268 kj/mol\n6339 kj/mol\ndone\n| compound | δhᵢ (kj/mol) |\n| ---- | ---- |\n| c₆h₆(g) | 82.90 |\n| co₂(g) | -393.50 |\n| h₂o(g) | -241.82 |

Answer

Explanation:

Step1: Recall the formula for $\Delta H_{rxn}$

$\Delta H_{rxn}=\sum n\Delta H_{f}(products)-\sum m\Delta H_{f}(reactants)$

Step2: Calculate $\sum n\Delta H_{f}(products)$

For products: $n_{CO_2} = 12$, $\Delta H_{f}(CO_2)=- 393.50$ kJ/mol; $n_{H_2O}=6$, $\Delta H_{f}(H_2O)=-241.82$ kJ/mol. $\sum n\Delta H_{f}(products)=12\times(-393.50)+6\times(-241.82)$ $=-4722-1450.92=-6172.92$ kJ/mol

Step3: Calculate $\sum m\Delta H_{f}(reactants)$

For reactants: $m_{C_6H_6}=2$, $\Delta H_{f}(C_6H_6)=82.90$ kJ/mol; $\Delta H_{f}(O_2) = 0$ (element in standard - state). $\sum m\Delta H_{f}(reactants)=2\times82.90+15\times0 = 165.8$ kJ/mol

Step4: Calculate $\Delta H_{rxn}$

$\Delta H_{rxn}=-6172.92 - 165.8=-6338.72\approx - 6339$ kJ/mol This is for 2 moles of $C_6H_6$. For 1 mole of $C_6H_6$, $\Delta H=\frac{-6339}{2}=-3169.5\approx - 3169$ kJ/mol

Answer:

-3169 kJ/mol