The heat evolved during the combustion of 112 litre of water gas (mixture of equal volume of H 2 and CO ) is H 2(g)+ O 2(g) arrow H 2 O (g) ; Δ H=-241.8 kJ CO (g)+ O 2(g) arrow CO 2(g) ; Δ H=-283 kJ

Question

The heat evolved during the combustion of 112 litre of water gas (mixture of equal volume of H 2 and CO ) is H 2(g)+ O 2(g) arrow H 2 O (g) ; Δ H=-241.8 kJ CO (g)+ O 2(g) arrow CO 2(g) ; Δ H=-283 kJ (A) 241.8 kJ (B) 283 kJ (C) -1312 kJ (D) 1586 kJ

Tardigrade Admin · Accepted Answer

112 litre of water gas contains 56 litre of H 2 and 56 litre of CO Heat evolved by combustion of 56 litre of H 2 =(-241.8 × 56/22.4)=-604.5 kJ Δ H for combustion of 56 litre of CO =(-283 × 56/22.4)=-707.5 kJ Thus Δ H for combustion of 112 litre of water gas =-604.5+(-707.5)=-1312 kJ

Q. The heat evolved during the combustion of $112 l i t re$ of water gas (mixture of equal volume of
$H_{2}$ and $CO$ ) is $H_{2 (g)} + O_{2 (g)} \to H_{2} O_{(g)}; Δ H = - 241.8 k J$
$C O_{(g)} + O_{2 (g)} \to C O_{2 (g)}; Δ H = - 283 k J$

$241.8 k J$

$283 k J$

$- 1312 k J$

$1586 k J$

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

241.8kJ

283kJ

−1312kJ

1586kJ

112 litre of water gas contains 56 litre of H2​ and 56 litre of CO Heat evolved by combustion of 56 litre of H2​ =22.4−241.8×56​=−604.5kJ ΔH for combustion of 56 litre of CO=22.4−283×56​=−707.5kJ Thus ΔH for combustion of 112 litre of water gas =−604.5+(−707.5)=−1312kJ

Q. The heat evolved during the combustion of $112 l i t re$ of water gas (mixture of equal volume of
$H_{2}$ and $CO$ ) is $H_{2 (g)} + O_{2 (g)} \to H_{2} O_{(g)}; Δ H = - 241.8 k J$
$C O_{(g)} + O_{2 (g)} \to C O_{2 (g)}; Δ H = - 283 k J$

$241.8 k J$

$283 k J$

$- 1312 k J$

$1586 k J$