Question

Which is true for the combustion of sucrose ($C_{12}{H}_{22}{O}_{11}$) at ${25}^{o}C$?

• A. $\Delta H>\Delta E$
• B. $\Delta H<\Delta E$
• C. $\Delta H=\Delta E$
• D. $None$

Answer 1

Answer: (A)

$\Delta H>\Delta E$

(A) $\Delta H>\Delta E$

Combustion of sucrose-

${C_{12}{H}_{22}{O}_{11}}_{\left(s\right)}+12{O_2}_{\left(g\right)}⟶12{C{O}_2}_{\left(g\right)}+11{H_{2}O}_{\left(g\right)}$

Change in stoichiometry (in gaseous state); ${\Delta n}_g=n_P-n_R=(12+11)-12=11$

As we know that,

$\Delta H=\Delta E+{\Delta n}_{g}RT$

As ${\Delta n}_g=11$

$\therefore \Delta H=\Delta E+11RT⟶\left(1\right)$

As $11RT$ is a positive value, the ${eq}^n\left(1\right)$ clearly shows that $\Delta H>\Delta E$.