The heat evolved in combustion of 112 L of water gas (mixture of equal volume of H 2( g ) and .CO ( g )) is H 2( g )+(1/2) O 2( g ) leftharpoons H 2 O ( g ) ; Δ H =-241.8 kJ CO ( g )+(1/2) O 2( g ) leftharpoons CO 2( g ) ; Δ H =-283 kJ

Question

The heat evolved in combustion of 112 L of water gas (mixture of equal volume of H 2( g ) and .CO ( g )) is H 2( g )+(1/2) O 2( g ) leftharpoons H 2 O ( g ) ; Δ H =-241.8 kJ CO ( g )+(1/2) O 2( g ) leftharpoons CO 2( g ) ; Δ H =-283 kJ (A) 241.8 kJ (B) 296.5 kJ (C) 1312 kJ (D) 1586 kJ

Tardigrade Admin · Accepted Answer

On adding both equations we get H 2+ CO + O 2 arrow H 2 O + CO 2 ; Δ H =-524.8 kJ 2 × 22.4 litre ( H 2+ CO ) evolve 524.8 kJ heat ∴ 112 L water gas evolved heat =(524.8/44.8) × 112 =1312 kJ

Q. The heat evolved in combustion of $112 L$ of water gas (mixture of equal volume of $H_{2} (g)$ and $CO (g))$ is
$H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g); Δ H = - 241.8 k J$
$CO (g) + \frac{1}{2} O_{2} (g) ⇌ C O_{2} (g); Δ H = - 283 k J$

241.8 kJ

296.5 kJ

1312 kJ

1586 kJ

On adding both equations we get
$H_{2} + CO + O_{2} \to H_{2} O + C O_{2}; Δ H = - 524.8 k J$
$2 \times 22.4$ litre $(H_{2} + CO)$ evolve $524.8 k J$ heat
$∴ 112 L$ water gas evolved heat
$= \frac{524.8}{44.8} \times 112$
$= 1312 k J$

Q. The heat evolved in combustion of 112L of water gas (mixture of equal volume of H2​(g) and CO(g)) is H2​(g)+21​O2​(g)⇌H2​O(g);ΔH=−241.8kJ CO(g)+21​O2​(g)⇌CO2​(g);ΔH=−283kJ