Question

The standard heats of formation of ${CH}_4\left(g\right),{CO}_2\left(g\right)$ and $H_{2}O\left(g\right)$ are $-76.2,-398.8$ and $-241.6$ $kJ{mol}^{-1}$ respectively. Calculate the amount of heat evolved by burning $1m^3$ of methane measured under normal conditions.

Answer 1

The required equation for the combustion of methane is:

${CH}_4+2O_2⟶{CO}_2+2H_{2}O;\Delta H=?$

$\Delta H^{}=\Delta H_{f\left(products\right)}-\Delta H_{f\left(reactants\right)}$

$=[\Delta H_{\left({CO}_2\right)}+2\times \Delta H_{\left(H_{2}O\right)}]-[\Delta H_{\left({CH}_4\right)}+2\times \Delta H_{\left(O_2\right)}]$

$=\left[\right(-398.8)-2\times 241.6]-\left[\right(-76.2)-2\times 0]=-805.8kJ{mol}^{-1}$

Heat evolved by burning $22.4$ litre (1 mole) methane$=-805.8kJ$.

So, heat evolved by burning 1000 litre ($1m^3$) methane$=\frac{805.8}{22.4}\times 1000=-35973.2kJ$