Answer
$\Delta H^{\circ}rxn = - 84.6 kJ/mol$
Work Step by Step
Equation (1) $C_{(graphite)} + O_{2(g)} →CO_{2(g)}$
$\Delta H^{\circ}rxn(1) = - 393.5 kJ/mol$
Equation (2) $H_{2(g)} +(1/2) O_{2(g)} → H_{2}O_{(l)}$
$\Delta H^{\circ}rxn(2) = - 285.8 kJ/mol$
Equation (3) $2C_{2}H_{6(g)} + 7O_{2(g)} → 4CO2(g) + 6H2O(l)$
$\Delta H^{\circ}rxn(3) = - 3119.6 kJ/mol$
Multiply Equation (1) by 2 to get Equation (4)
Equation (4) $2C_{(graphite)} +2 O_{2(g)} →2CO_{2(g)}$
$\Delta H^{\circ}rxn(4) = - 787 kJ/mol$
Multiply Equation (2) by 3 to get Equation (5)
Equation (5) $3H_{2(g)} +(3/2) O_{2(g)} → 3H_{2}O_{(l)}$
$\Delta H^{\circ}rxn(5) = - 857.4 kJ/mol$
Divide Equation (3) by 2 and reverse to get Equation (6)
Equation (6) $ 2CO_{2(g)} + 3H_{2}O_{(l)} → C_{2}H_{6(g)} + (7/2)O_{2(g)} $$\Delta H^{\circ}rxn(6) = - (- 1559.8) = +1559.8 kJ/mol$
The required equation, $2C_{(graphite)} + 3H_{2(g)} → C_{2}H_{6(g)}$ is formed by adding Equation (4), Equation (5), and Equation (6).
$\Delta H^{\circ}rxn = \Delta H^{\circ}rxn(4) +\Delta H^{\circ}rxn(5) + \Delta H^{\circ}rxn(6) = (-787+ -857.4 +1559.8) kJ/mol = - 84.6 kJ/mol$
