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The heat evolved in the combustion of 1...

The heat evolved in the combustion of 112 litre of water gas (mixture of equal volume of `H_(2)` and CO) is
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g), DeltaH=-241.8 kJ`
`CO(g) + 1/2O_(2)(g) to CO_(2)(g), DeltaH=-283 kJ`

A

`241.8` kJ

B

283 kJ

C

1312 kJ

D

1586 kJ

Text Solution

Verified by Experts

The correct Answer is:
C
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Find out the heat evolved in combustion if 112 litres ( at STP ) of water gas (mixture of eqal volume of H_(2)(g) and CO(g))) . {:(H_(2)(g)+1//2O_(2)(g)rarrH_(2)O(g),,,,DeltaH= -241.8 kJ),(CO(g)+1//2O_(2)(g)rarrCO_(2)(g),,,,DeltaH= -283 kJ):}

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

Equal volumes of C_(2)H_(2) and H_(2) are combusted under identical conditons. The ratio of evolved in i.e. for C_(2)H_(2) and H_(2) is: H_(2)(g) + 1/2O_(2) (g) to H_(2)O(g), DeltaH=-241.8 kJ C_(2)H_(2)(g) + 21/2O_(2)(g) to 2CO_(2)(g) + H_(2)O(g), DeltaH = -1300 kJ

H_(2)(g)+(1)/(2)O_(2)(g)rarr2H_(2)O(l), DeltaH =- 286 kJ 2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l)……………kJ(+-?)

Calculate the heat of combustion of benzene form the following data: a. 6C(s) +3H_(2)(g) rarr C_(6)H_(6)(l), DeltaH = 49.0 kJ b. H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l), DeltaH =- 285.8 kJ c. C(s) +O_(2) (g) rarr CO_(2)(g), DeltaH =- 389.3 kJ

The enthalpy of vaporisation of liquid water using the data H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l) , DeltaH=-285.77 kJ//mol H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g) , DeltaH=-241.84 kJ//mol

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