Question

On complete combustion of $2g$ methane $26575Cal$ heat is generated. The heat of formation of methane will be (given heat of formation of $C{O}_2$ and $H_{2}O$ are $-97000$ and $-68000Cal$ respectively).

• A. $+20400$ cal
• B. $+20600$ cal
• C. $-20400$ cal
• D. $-2000$ cal

Answer 1

Answer: (C)

$-20400$ cal

Solution:- (C) $-20400cal$

Mol. wt. of methane $=16gm$

Heat generated during the combustion of $2gm$ methane $=26575cal$

Heat evolved during the combustion of $16gm$ methane $=\frac{26575}{2}\times 16=212600cal$

Hence the heat generated during the combustion of $1$ mole of methane is $212600cal$.

Reaction of combustion of methane-

${C{H}_4}_{\left(g\right)}+2{O_2}_{\left(g\right)}⟶{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}\Delta H_R=-212600cal$

${C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(g\right)}⟶{C{H}_4}_{\left(g\right)}+2{O_2}_{\left(g\right)}\Delta H=212600cal.....\left(1\right)$

Given that:-

$C_{\left(s\right)}+{O_2}_{\left(g\right)}⟶{C{O}_2}_{\left(g\right)}\Delta H=-97000cal.....\left(2\right)$

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H_{2}O}_{\left(g\right)}\Delta H=-68000cal$

$2\times [{H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H_{2}O}_{\left(g\right)}\Delta H=-68000cal]$

$2{H_2}_{\left(g\right)}+{O_2}_{\left(g\right)}⟶2{H_{2}O}_{\left(g\right)}\Delta H=-136000cal.....\left(3\right)$

Adding ${eq}^n\left(1\right),\left(2\right)\&\left(3\right)$, we have

$C_{\left(s\right)}+2{H_2}_{\left(g\right)}⟶{C{H}_4}_{\left(g\right)}\Delta H=-20400$

Hence the heat of formation of methane is $-20400cal$.