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

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,

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

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

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

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

$\Delta H_{C_{25}H_{52}}=-2460.5kJ/mol$

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