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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.

Respuesta :

Answer: 48.6 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)[/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_6H_6}\times \Delta H_{C_6H_6})][/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]-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})][/tex]

[tex]\Delta H_{C_6H_6}=48.6kJ/mol[/tex]

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

The standard enthalpy of formation, [tex]\Delta H^\circ_{f}[/tex], for benzene is 48.6 KJ/mol

Standard enthalpy of a reaction

From the question, we are to calculate the standard enthalpy of formation, [tex]\Delta H^\circ_{f}[/tex], for benzene

The standard enthalpy of a reaction is given by

[tex]\Delta H^\circ_{rxn} = \sum \Delta H^\circ _{f}(products) - \sum \Delta H^\circ _{f}(reactants)[/tex]

Therefore,

[tex]\Delta H^\circ_{rxn} = 12 ( \Delta H^\circ _{f} \ of\ CO_{2}) + 6(\Delta H^\circ _{f} \ of\ H_{2}O) - [2(\Delta H^\circ _{f} \ of\ C_{6}H_{6})+ 15 (\Delta H^\circ _{f} \ of\ O_{2})][/tex]

From the given information,

[tex]\Delta H^\circ_{rxn} = -6534.0\ KJ[/tex]

[tex]\Delta H^\circ _{f} \ of\ CO_{2} = -393.5 \ KJ/mol[/tex]

[tex]\Delta H^\circ _{f} \ of\ H_{2}O = -285.8 \ KJ/mol[/tex]

Putting the parameters into the equation, we get

[tex]-6534.0= 12 ( -393.5) + 6(-285.8) - [2(\Delta H^\circ _{f} \ of\ C_{6}H_{6})+ 15 (0)][/tex]

[tex]-6534.0= -4722 -1714.8 - 2(\Delta H^\circ _{f} \ of\ C_{6}H_{6})[/tex]

[tex]2(\Delta H^\circ _{f} \ of\ C_{6}H_{6})= -6436.8+6534.0[/tex]

[tex]2(\Delta H^\circ _{f} \ of\ C_{6}H_{6}) = 97.2[/tex]

[tex]\Delta H^\circ _{f} \ of\ C_{6}H_{6} = \frac{97.2}{2}[/tex]

[tex]\Delta H^\circ _{f} \ of\ C_{6}H_{6} = 48.6\ KJ/mol[/tex]

Hence, the standard enthalpy of formation, [tex]\Delta H^\circ_{f}[/tex], for benzene is 48.6 KJ/mol

Learn more on Standard enthalpy of reaction here: https://brainly.com/question/13866554

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