Question

The compound carbon suboxide, C3O2, is a gas at room temperature. Use the data supplied to calculate the heat of formation of carbon suboxide. (Data: 2CO(g) + C(s) → C3O2(g) ΔH° = 127.3 kJ/mol and: ΔHf° of CO(g) = –110.5 kJ/mol)

Answer 1

Answer: The standard enthalpy of formation of $C_3O_2(g)$ is -92.7 kJ/mol

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+C(s)\rightarrow C_3O_2(g)$

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_3O_2(g))})]-[(2\times \Delta H^o_f_{(CO(g))})+(1\times \Delta H^o_f_{(C(s))})]$

We are given:

$\Delta H^o_f_{(CO(g))}=-110kJ/mol\\\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_{rxn}=127.3kJ$

Putting values in above equation, we get:

$127.3=[(1\times \Delta H^o_f_{(C_3O_2(g))})]-[(2\times (-110))+(1\times (0))]\\\\\Delta H^o_f_{(C_3O_2(g))}=-92.7kJ/mol$

Hence, the standard enthalpy of formation of $C_3O_2(g)$ is -92.7 kJ/mol