Find out the heat evolved in combustion if 112 litre (at 1 atm, 273 K) of water gas (mixture of equal volume of `H_(2)(g)` and CO(g)) is combusted with excess oxygen.
`H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(g), Delta=-241.8 kJ` `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g), Delta=-283 kJ`

Options

A. `241.8` KJ
B. 283 KJ
C. 1321 KJ
D. 1586 KJ

Answer

C

Explanation

112 litre water gas contains 56 litre CO and 56 litre `H_(2)`.
`H_(2(g))+(1)/(2)O_(2(g))rarr H_(2)O_((g)) , Delta H =-241.8 kJ`
In the combustion of mole. `H_(2)=22.4` litre `H_(2)` the evolved heat = 241.8 kJ
`therefore` In the combustion of 56 litre `H_(2)` the
`Delta H = (241.8)/(22.4)xx56`
`therefore Delta H = 604.5 kJ`
`CO_((g))+(1)/(2)O_(2(g))rarr CO_(2(g)) , Delta H=-283 kJ`
By the combustion of 1 mole. CO = 22.4 litre CO teh evolved heat = 283 kJ
`therefore` By the combustion of 56 litre CO the
`Delta H=(283)/(22.4)xx56 =707.5 kJ`
So, the total heat evolved = 604.5+707.5=1312 kJ .