Question

The difference in $\Delta H$ and $\Delta E$ for the combustion of methane at ${27}^o$C in calories would be:

• A. $-1800$
• B. $-162$
• C. $-1200$
• D. $-100$

Answer 1

The correct option is C $-1200$

Combustion of methane means the burning of methane in the presence of oxygen to give carbon dioxide and water.

The balanced reaction for combustion of methane is as follows:

$CH{}_4\left(g\right)+2O{}_2\left(g\right)\to CO{}_2\left(g\right)+2H_{2}O$

In such reactions, the change in the number of moles is calculated for gaseous products only.

The number of moles of $C{H}_4,O_2$ ans $CO$ are $1,1$ and $2$ respectively as seen from the balanced reaction given above.

Therefore, the change in the number of moles is given by $=\Delta n_g=n_{\text{products}}-n_{\text{reactants}}=1-(2+1)$ $=-2$

We know the relation: $\Delta H=\Delta E+\Delta n_{g}RT$, where $\Delta H$ is the enthalpy change, $\Delta E$ is the internal energy change and $\Delta n_g$ is the change in the number of moles (as per the law of thermodynamics).

The value of gas constant $\left(R\right)$ is $2$ cal and temperature is ${27}^{o}C$ or $300K$ (given).

$\therefore \Delta H-\Delta E=\Delta n_{g}RT$$=-2\times 2\times 300$ $=-1200$ calories

Hence, the difference in $\Delta H$ and $\Delta U$ for the combustion of methane at ${27}^{o}C$ is $-1200$ cal.