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Calculate the lower and higher calorific value of benzene, expressed in $\pu{kJ L^-1},$ if the standard enthalpy of formation is known:

$$ \begin{array}{lc} \hline \text{Compound} & \Delta_\mathrm{f}H^\circ/\pu{kJ mol^-1} \\ \hline \ce{CO2(g)} & -393.52 \\ \ce{H2O(l)} & -285.5 \\ \ce{H2O(g)} & -241.6 \\ \ce{C6H6(l)} & 48.8 \\ \hline \end{array} $$

The density of benzene is $\pu{0.879 g cm^-3}.$

I know the following formula:

$$q = \frac{1000}{\mu}|\Delta H^\circ|$$

Is this formula correct? I took it from the internet because my teacher did not teach it to us. We also know that for benzene $\mu(\ce{C6H6}) = 78.$

I do not know how to use this formula and how to find the lower and higher calorific value of benzene. What is the correct formula for lower and higher calorific values?

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    $\begingroup$ For better site experience, you can find useful How can I format math/chem expressions/equations/formulas on Chemistry SE. ( Not to be applied to titles ) See also upright vs italic $\endgroup$
    – Poutnik
    Oct 19, 2021 at 14:50
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    $\begingroup$ Ask yourself as the first step, how is calculated the reaction enthalpy from formation enthalpies of reactants and products. when you get it, you recalculate it per mass and then per volume. $\endgroup$
    – Poutnik
    Oct 19, 2021 at 14:57
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    $\begingroup$ What is your understanding of the terms "higher heating value" and "lower heating value?" $\endgroup$ Oct 19, 2021 at 17:19

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