Question

Standard molar heat of formation of ethane $C{O}_2$ and liquid water are $-21.1$, $-94.1$ and $-68.3$ kilo calorie. The value of standard molar heat of combustion of ethane will be:

• A. $-372$ kcal
• B. $-272$ kcal
• C. $-172$ kcal
• D. $-472$ kcal

Answer 1

The correct option is A $-372$ kcal

Solution:- (A) $-372kcal$

Combustion of ethane-

${C_2{H}_6}_{\left(s\right)}+\frac{7}{2}{O_2}_{\left(g\right)}⟶2{C{O}_2}_{\left(g\right)}+3{H_{2}O}_{\left(l\right)}$

$\Delta H^0={\sum \Delta {H^0}_f}_{\left(product\right)}-{\sum \Delta {H^0}_f}_{\left(reactant\right)}$

$\therefore \Delta H^0=(2\times (-94.1)+3\times (-68.3))-(-21.1+0)$

$\Rightarrow \Delta H^0=-393.1+21.1=-372kcal$

Hence the standard molar heat of combustion will be $-372kcal$.