Question

Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given ΔH o f of CO2(g) = −393.5 kJ/mol and ΔH o f of H2O(g) = −241.8 kJ/mol, find ΔH o f of C2H2(g).

Answer 1

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

$C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]$

$-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]$

$\Delta H_{C_2H_2}=-227kJ$

Therefore, the enthalpy change for $C_2H_2$ is -227 kJ.