Question

At ${27}^o$C, the combustion of ethane takes place according to the reaction

$C_2{H}_6\left(g\right)+\frac{7}{2}{O}_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

$\Delta E-\Delta H$ for this reaction at ${27}^o$C will be:

• A. + 1247.1 J
• B. - 1247.1 J
• C. -6235.5 J
• D. +6235.5 J

Answer 1

The correct option is D +6235.5 J

$\Delta E-\Delta H$ $=$ $-\Delta nRT$

$\Delta n=$ number of moles of gaseous product - number of moles of gaseous reactants $=2-(1+7/2)=-2.5$

$\Delta E-\Delta H$ $=$ $2.5\times 8.314\times 300$

$\Delta E-\Delta H$ $=$ $+6235.5J$