Question

Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -393.5kJ/mol, and the enthalpy of combustion for the reaction is -2220.1kJ/mol, what is the heat of formation of propane?

Answer 1

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$  $\Delta H=-2220.1kJ/mol$

The expression for enthalpy change is,

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

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]$

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

$-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]$

$\Delta H_{C_3H_8}=72.4kJ/mol$

Therefore, the heat of formation of propane is 72.4 kJ/mol