Question

calculating enthalpy of reaction
compound δh (kj/mol)
c6h6 (g) 82.90
co2 (g) -393.51
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)
-3109 kj/mol
-6339 kj/mol

Explanation:

Step1: Recall the formula for enthalpy of reaction

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

Step2: Identify the coefficients and enthalpies of formation

For the reaction $2C_6H_6(g)+15O_2(g)
ightarrow12CO_2(g) + 6H_2O(g)$, $\Delta H_f(O_2(g)) = 0$ (standard state). For $C_6H_6(g)$, $\Delta H_f = 82.90\ kJ/mol$, for $CO_2(g)$, $\Delta H_f=-393.51\ kJ/mol$ and for $H_2O(g)$, $\Delta H_f = - 241.82\ kJ/mol$.

Step3: Calculate enthalpy of reaction for 2 moles of $C_6H_6$

$\Delta H_{rxn}=[12\times(-393.51)+6\times(-241.82)]-[2\times82.90 + 15\times0]$
$=[-4722.12-1450.92]-[165.8]$
$=-6173.04 - 165.8=-6338.84\ kJ$ for 2 moles of $C_6H_6$.

Step4: Calculate enthalpy of reaction for 1 mole of $C_6H_6$

$\Delta H=\frac{-6338.84}{2}=-3169.42\approx - 3169\ kJ/mol$

Answer:

$-3169\ kJ/mol$