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Question: What is the amount of work done when \[0.5\] mole of methane \[C{H_4}\] is subjected to combustion a...

What is the amount of work done when 0.50.5 mole of methane CH4C{H_4} is subjected to combustion at 300 K300{\text{ K}} , given that R = 8.314 JK−1mol−1R{\text{ = 8}}{\text{.314 J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}
A. −2494 J - 2494{\text{ J}}
B. −4988 J - 4988{\text{ J}}
C. +4988 J + 4988{\text{ J}}
D. +2494 J + 2494{\text{ J}}

Explanation

Solution

For better understanding of combustion of methane we will first write a balanced combustion reaction of methane. We will find the change in the number of gaseous moles when 0.50.5 mole of methane undergoes combustion. With the help of a change in the number of moles we will find the change in volume at S.T.P. Thus we can find the work with the help of S.T.P pressure and this change in volume.
Formula Used:
Work done  = −P ΔV{\text{ = }} - P{\text{ }}\Delta V

Complete Answer:
The combustion of hydrocarbons usually produces carbon dioxide gas and some amount of water. Therefore the combustion reaction of methane can be represented as,
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)C{H_4}(g){\text{ + 2}}{{\text{O}}_2}(g){\text{ }} \to {\text{ C}}{{\text{O}}_2}(g){\text{ + 2}}{{\text{H}}_2}{\text{O(l)}}
Let us assume that one mole of methane undergoes combustion, then the change in number of gaseous moles will be depicted as,
Δn = 1−(1+2)\Delta n{\text{ = 1}} - \left( {1 + 2} \right)
Δn = −2\Delta n{\text{ = }} - {\text{2}}
Therefore when 0.50.5 mole of methane undergoes combustion then the change in number of gaseous moles will be:
Δn = −2 × 0.5\Delta n{\text{ = }} - {\text{2 }} \times {\text{ 0}}{\text{.5}}
Δn = −1\Delta n{\text{ = }} - 1
At S.T.P the value of temperature in kelvin is 273 K{\text{273 K}}, and the given temperature is 300 K300{\text{ K}} and the volume is 22.4 L{\text{22}}{\text{.4 L}}, thus we can find the change in volume at S.T.P as:
ΔV = 22.4 L × Δn × 300 K273 K\Delta V{\text{ = 22}}{\text{.4 L }} \times {\text{ }}\Delta n{\text{ }} \times {\text{ }}\dfrac{{300{\text{ K}}}}{{273{\text{ K}}}}
ΔV = 22.4 L × ( - 1) × 300 K273 K\Delta V{\text{ = 22}}{\text{.4 L }} \times {\text{ ( - 1) }} \times {\text{ }}\dfrac{{300{\text{ K}}}}{{273{\text{ K}}}}
ΔV = 24.6 L\Delta V{\text{ = 24}}{\text{.6 L}}
Therefore the amount of work done on combustion of methane can be calculated by using formulae,
Work done  = −P ΔV{\text{ = }} - P{\text{ }}\Delta V
We know that at S.T.P the amount of pressure is taken to be 1 atm1{\text{ atm}}, therefore on substituting the values we get the amount of work done as,
Work done  = −P ΔV{\text{ = }} - P{\text{ }}\Delta V
Work done  = −1 atm × (−24.6 L){\text{ = }} - 1{\text{ atm }} \times {\text{ }}\left( { - 24.6{\text{ L}}} \right)
Work done  = + 24.6 atm L{\text{ = + 24}}{\text{.6 atm L}}
For finding work done in joules we multiply it by 101.33 J atm−1L−1101.33{\text{ J at}}{{\text{m}}^{ - 1}}{L^{ - 1}} , thus we get the work done in joules as,
Work done  = + 2494 J{\text{ = + 2494 J}}

Hence the correct option is D.

Note:
On combustion of methane we obtained a water molecule in a liquid state. Hence for finding change in gaseous moles we do not count mole of water molecules. Moles of only gaseous state are counted while finding the change in gaseous moles. The combustion reaction must be balanced for finding the accurate value of work done.