Question

On complete of combustion of 2 gm methane 26575 cals 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 -68000 cals respectively):

• A. +20400 cals
• B. +20600 cals
• C. - 20400 cals
• D. -20000 cals

Answer 1

The correct option is C - 20400 cals

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O$

in $2g$ of $C{H}_4$, moles= $\frac{2}{16}=0.125$ moles

Heat generated for $0.125$ moles=$26575$ cal

For $1$ mole= $\frac{26575}{0.125}=212,600$ cal

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O⟶\left(1\right)$ $\Delta H_1=-212,600$ cal

$C+O_2⟶C{O}_2⟶\left(2\right)$ $\Delta H_2=-97000$ cal

$H_2+1/2O_2⟶H_{2}O⟶\left(3\right)$ $\Delta H_3=-68000$ cal

On multiplying equation $3$ by $2$;

$2H_2+O_2⟶2H_{2}O⟶\left(4\right)$, $\Delta H_4^{\prime }=2\times -68000$ cal

Now, add equation $2$ & $4$, subtract $1$ from it we get

$C+O_2+2H_2+O_2⟶C{O}_2+2H_{2}O$

$-C{H}_4+2O_2⟶C{O}_2+2H_{2}O$

$C+2H_2⟶C{H}_4$ $\Delta HF$

$\Delta HF=\Delta H_4+\Delta H_2-\Delta H_1$

$\Delta HF=2\times (-68000)+(-97000)-(-212,000)=-20,400$ cal