Chemistry: Calculate δh∘f (in kilojoules per mole) for benzene, c6h6, from the following data: 2c6h6(l) + 15o2(g)→12co2(g)+6h2o(l) δh∘=−6534.0 kjδh∘f co2=−393.5 kj/molδh∘f h2o=−285.8 kj/mol express the enthalpy change in kilojoules per mole to three significant digits.

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Calculate δh∘f (in kilojoules per mole) for benzene,

Beres7

10.01.2022

48.6 kJ/mol

Explanation:

The balanced chemical reaction is,

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)$

The expression for enthalpy change is,

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

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_6H_6}\times \Delta H_{C_6H_6})]$

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

$-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})]$

$\Delta H_{C_6H_6}=48.6kJ/mol$

Therefore, the enthalpy change for benzene is 48.6 kJ/mol

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