Question

The standard enthalpies of formation of $C{O}_2\left(g\right)$, $H_{2}O\left(l\right)$ and glucose(s) at ${25}^o$C are $-400$ kJ/mol, $-300$ kJ/mol and $-1300$ kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at ${25}^o$C is:

• A. $+2900$ kJ
• B. $-2900$ kJ
• C. $-16.11$ kJ
• D. $+16.11$ kJ

Answer 1

The correct option is C $-16.11$ kJ

Solution:- (C) $-16.11kJ$

Combustion of glucose-

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

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

$\therefore \Delta H^0=(6\times (-400)+6\times (-300))-(-1300+0)$

$\Rightarrow \Delta H^0=-4200+1300=-2900kJ/mol$

Now,

$\Delta H^0\left(kJ/gm\right)=\frac{\Delta H^0\left(kJ/mol\right)}{\text{Mol. wt.}\left(\text{in gm}\right)}$

Molecular weight of glucose $=180gm$

$\therefore \Delta H^0\left(kJ/gm\right)=\frac{-2900}{180}=-16.11kJ/gm$

Hence the standard enthalpy of combustion per gram of glucose at $25℃$ is $-16.11kJ$.