Given R = `8.314 J K^(-1) "mol"^(-1)`, the work done during combustion of 0.090 kg of ethane(molar mass = 30) at 300 K is

Options

A. `-18.7kJ`
B. 18.7kJ
C. 6.234kJ
D. `-6.234kJ`

Answer

B

Explanation

`2C_(2)H_(6(g)) + 7O_(2(l)) rarr 4CO_(2(g)) + 6H_(2)O_((l))`
`C_(2)H_(6) + (7)/(2) O_(2) rarr 2CO_(2) + 3H_(2)O`
`Delta n_(g)= n_(p)- n_(r)= (2- 1- (7)/(2))= -2.5`
`-P Delta v= Delta n_(g)RT`
`W= -Delta n_(g)RT = -2.5 xx 8.314 xx 300`
`W= -(-2.5 xx 8.314 xx 300) = 6235.5J`
For one mole combustion we get work done =6235.5J
`therefore` for 0.090 kg = 0.090 `xx =(1000)/(30)` moles = 3 moles
For 3 moles the work done = 3 `xx` 6235.5 J = 18706.5 = 18.7 kJ