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Sagot :
Answer: The molar enthalpy of formation for paraffin wax is -2460.5 kJ
Explanation:
The balanced chemical reaction is,
[tex]C_{25}H_{52}(g)+38O_2(g)\rightarrow 25CO_2(g)+26H_2O(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
[tex]\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}})][/tex]
where,
n = number of moles
[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
[tex]-14800=[(25\times -393.5)+(26\times -285.5)]-[(38\times 0)+(1\times \Delta H_{C_{25}H_{52}})][/tex]
[tex]\Delta H_{C_{25}H_{52}}=-2460.5kJ/mol[/tex]
Therefore, the molar enthalpy of formation for paraffin wax is -2460.5 kJ
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