Question

1.00 sample of a mixture of $C{H}_4\left(g\right)$ & $O_2\left(g\right)$ measured at ${25}^{\circ }C$ & 740 torr was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of $1260cal/K$. The complete combustion of methane to $C{O}_2$ & $H_{2}O$ caused a temperature rise, in the calorimeter, of 0.667 K. Then the mole precent of $C{H}_4$ in the original mixture was:
Given that : $\Delta H_{comb}^{\circ }\left(C{H}_4\right)=-215$ $kcal.mo{l}^{-1};volume=1lit.$

• A. 9.82 mol % $C{H}_4$
• B. 6.36 mol % $C{H}_4$
• C. 4.52 mol % $C{H}_4$
• D. 8.54 mol % $C{H}_4$

Answer 1

Answer: (A)

9.82 mol % $C{H}_4$

Given that :

$\Delta T=0.667K$; $C_V=1260cal/K$; $P=740torr$

$760torr=1atm\Rightarrow 740torr=0.973atm$

$T={25}^{o}C=298K$

${CH}_4\left(g\right)+2O_2\to {CO}_2\left(g\right)+2H_{2}O\left(g\right)$

Let $x=$ quantity of heat from reaction mixture, $x=C_V\Delta T$

Heat lost by system (reaction mixture) is equal to heat gained by calorimeter.

So $x=-C_V\times \Delta T=-1260cal/K\times 0.667=-840.42cal$

Negative sign$\Rightarrow$ Exothermic nature

Molar mass of ${CH}_4=12+4=16g/mol$

Empathy of combustion of $16g{CH}_4=-215kcal/mol=-215000cal/mol$

Thus enthalpy of $-840.42cal$ would corrospond to $\frac{16}{21500}\times 840.42=0.0625gm$

So,$0.0625gm{CH}_4$ is present in mixture.

Moles of ${CH}_4$ in mixture $=\frac{0.0625}{16}=3.91\times {10}^{-3}$

Now, $PV=nRT$

$\Rightarrow 0.973\times 1=n\times 0.082\times 298\Rightarrow n=3.98times{10}^{-2}moles$

Mole % of ${CH}_4=\frac{3.91\times {10}^{-3}}{3.92\times {10}^{-2}}\times 100=9.97$%