Question

Find the molar enthalpy of formation for paraffin wax (C2H526)) given the following reaction

C2H526) + 38 026) ► → 25 CO26) + 26 H,O)
A H = -14 800 kJ

Answer 1

Answer: The molar enthalpy of formation for paraffin wax is -2460.5 kJ

Step-by-step explanation:

The balanced chemical reaction is,

`C_(25)H_(52)(g)+38O_2(g)\rightarrow 25CO_2(g)+26H_2O(g)`

The expression for enthalpy change is,

`\Delta H=\sum [n* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

`\Delta H=[(n_(CO_2)* \Delta H_(CO_2))+(n_(H_2O)* \Delta H_(H_2O))]-[(n_(O_2)* \Delta H_(O_2))+(n_{C_(25)H_(52)}* \Delta H_{C_(25)H_(52)})]`

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

`-14800=[(25* -393.5)+(26* -285.5)]-[(38* 0)+(1* \Delta H_{C_(25)H_(52)})]`

`\Delta H_{C_(25)H_(52)}=-2460.5kJ/mol`

Therefore, the molar enthalpy of formation for paraffin wax is -2460.5 kJ