Question

The difference in $\Delta H$ and $\Delta E$ for the combustion of methane at ${25}^{o}C$ would be:

• A. $zero$
• B. $-2\times 298\times 2cal.$
• C. $-2\times 298\times 3cal.$
• D. $-2\times 25\times 3cal.$

Answer 1

The correct option is B $-2\times 298\times 2cal.$

Combustion of methane-

${C{H}_4}_{\left(g\right)}+2{O_2}_{\left(g\right)}⟶{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}$

As we know that,

$\Delta H-\Delta E=\Delta n_{g}RT$

whereas,

$\Delta n_g=\text{diff. between the stoichimetric coeff. of product and reactant}=n_P-n_R=(1+2)=1-3=-2$

$R=\text{molar gas constant}=2cal/(mol.K)$

$T=\text{temperature}=25℃=(273+25)=298K\left[\text{given}\right]$

$\therefore \Delta H-\Delta E=(-2\times 298\times 2)\text{calories}$