What is the amount of work done when 0.5 mole of methane, `CH_(4)`(g) is subjected to combustion at 300 K? (given, `R=8.314 Jk^(-1) mol^(-1))`

Options

A. `-2494 J`
B. `-4988 J`
C. `+4988 J`
D. `+2494 J`

Answer

D

Explanation

`CH_(4)(g) + 2CO_(2)(g) to CO_(2)(g) + 2H_(2)O(l)`
`W =Deltan RT`
1 mole of reacts with 2 moles of oxygen to produce 2 moles of `CO_(2)`
Hence,
`0.5` mole of `CH_(4)` reacts with 1 mole of `O_(2)` to produce 1 mole of `CO_(2)`
`therefore n_(1)`= moles of reactants =1.5
`n_(2)` =moles of product =1
`Deltan=n_(2) -n_(1) = 1 - 1.5 =-0.5`
`W= -(-0.5) xx 300 xx 8.314 J K^(-1) mol^(-1)`
`=+2494` J