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timurjin [86]
3 years ago
9

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

Chemistry
1 answer:
pogonyaev3 years ago
3 0

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

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