Question

What is the amount of work done when 0.5 mole of methane, $C{H}_{4\left(g\right)}$, is subjected to combustion at 300 K? (given, R = 8.314 J $K^{-1}$ mol${}^{-1}$)

• A. $-2494J$
• B. $-4988J$
• C. $+4988J$
• D. $+2494J$

Answer 1

Answer: (D)

$+2494J$

The balanced chemical equation for combustion reaction is
$C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
When 1 mole of methane undergoes combustion
$\Delta n=1-(1+2)=-2$

When 0.5 mole of methane undergoes combustion
$\Delta n=-2\times 0.5=-1$
At STP (1 atm and 273 K) , 1 mole of a gas occupies a volume of $22.4L$.

The volume change $\Delta V=22.4L\times \Delta n\times \frac{300K}{273K}$

$\Delta V=-24.6L$

The work done $W=-P\Delta V$

$W=-1atm\times (-24.6L)$

$W=+24.6Latm$

$1L$ atm $=-101.33$ J

$W=-24.6Latm\times -101.33J/Latm=+2494J$

Hence, the amount of work done is $+2494J$.