Question

Calculate ΔH∘ in kilojoules for the reaction of acetylene (C2H2) (ΔH∘f=227.4kJ/mol) with O2 to yield carbon dioxide (CO2) (ΔH∘f=−393.5 kJ/mol) and H2O(g) (ΔH∘f=−241.8kJ/mol), a reaction which is supplied by the industrial gases industry for oxyacetylene gas welding and cutting due to the high temperature of the flame.

Answer 1

Answer: The value of $\Delta H^o$ for the reaction is, -2512.4 kJ

Explanation:

The chemical equation for the combustion of acetylene follows:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_2H_2(g))})+(5\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=227.4kJ$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(4\times (-393.5))+(2\times (-241.8))]-[(2\times (227.4)+(5\times (0))]\\\\\Delta H^o_{rxn}=-2512.4kJ$

Therefore, the value of $\Delta H^o$ for the reaction is, -2512.4 kJ