How much energy is released when 6 mole of octane is burnt in air ? Given `DeltaH_(f)^(@)` for `CO_(2)(g),H_(2)O(g)` and `C_(8)H_(18)(l)` respectively are `-490,-240` and `+160KJ//mol`

Options

A. `- 6.2 MJ`
B. `- 37.4 MJ`
C. `-35.5 MJ`
D. `- 20 MJ`

Answer

B

Explanation

Thermochemical equation for the combustion of one mole octane is
`C_(8) H_(18) (1) + (15)/(2) O_(2) (g) rarr 8 CO_(2) (g) + 9 H_(2) O (1)`
Accroding to Eq., we have
`Delta_(C ) H^(@) = [(8 mol) Delta_(f) H^(@) (C_(8) H_(18), l) + ((15)/(9) mol) Delta_(f) H^(@) (O_(2), g)]`
`[8(-490) + 9 (-240)] - (160)`
`= - 3920 - 2160 - 160`
`= - 6240 kJ`
This is the heat energy released when 1 mol of octane burns. Thus, the amount of heat energy released on burning 6 mol of octane will be
`- (6) (6240) kJ = - 37400 kJ`
`= - 37.440 xx 10^(3) kJ`
`= - 37.440 MJ (1 MJ = 10^(6) J = 10^(3) kJ)`