Question

3. Given the following thermochemical equations,
C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) ∆H = -1299.5 KJ
C(s) + O2(g) → CO2(g) ∆H = -393.5 KJ
H2(g) + ½ O2(g) → H2O(l) ∆H = -285.8 KJ
Calculate ∆H for the decomposition of one mole of acetylene, C2H2(g), to its elements in their stable state at 25oC and 1 atm.

Answer 1

From the equations given;

$\Delta_{H_a}=\Delta_{H_b}+\Delta_{H_c}$

The equation for the end reaction is;

$C_2H_2(-1299.5kJ)\to 2CO_2(-393.5kJ)+H_2O(-285.8kJ)$

$\Delta H°_D=\Sigma \Delta H°_{products}-\Sigma \Delta H°_{reactants}$

$=(-1299.5kJ)-[(2\times -393.5kJ)+(-285.8kJ)]$

$=(-1299.5kJ)-(-1072.8kJ)$

$=-226.7kJ$

$\Delta H°_D$ For the reaction is therefore $-226.7kJ$