Question

At constant pressure and at $290K$ , the heatnof combustion of glucose (s) was found to be $-2723.78kJ$ . Calculate the heat of combustion of glucose at constant volume. (Assume water to be in gaseous state)

Answer 1

Combustion of glucose-

${C_6{H}_{12}{O}_6}_{\left(s\right)}+6{O_2}_{\left(g\right)}⟶6{C{O}_2}_{\left(g\right)}+6{H_{2}O}_{\left(g\right)}$

$\Delta n_g=n_P-n_R=(6+6)-6=6$

Given:- $\Delta H=-2723.78kJ$

$T=290K$

As we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$

$\Rightarrow \Delta U=-2723.78-(6\times 8.314\times {10}^{-3}\times 290)$

$\Rightarrow \Delta U=-2723.78-14.46=-2738.24kJ$

Hence the heat of combustion of glucose at constant volume is $-2738.24kJ$.