Question

The heat evolved in the combustion of $112 \,L$ of water gas $\left( CO + H _{2}\right.$ in the ratio of 1 : 1 volume $)$ is (given that heats of combustion of $H _{2} \,\&\, CO$ are $-241.8 \,\&\,-283 \,kJ$ $mol ^{-1}$ respectively). Assume STP condition throughout:

• A. -1414 kJ
• B. -1209 kJ
• C. -1312 kJ
• D. -524.8 kJ

Answer 1

Answer: (C)

$CO +\frac{1}{2} O _{2} \rightarrow CO _{2} ;$

$ \Delta H =-283\, kJ / mol$

$H _{2}+\frac{1}{2} O _{2} \rightarrow H _{2} O ;$

$ \Delta H =-241.8\, kJ / mol$

In the above rxns, $56 L$ of $CO \& 56 L$ of $H _{2}$ reacts (since water gas has $CO \, \&\, H _{2}$ in 1: 1 volume ratio)

At STP, $56 L$ of $CO =\frac{56}{22.4}$ mole of $CO =2.5$ moles

At STP, $56 L$ of $H _{2}=\frac{56}{22.4}$ mole of $H _{2}=2.5$ moles

Energies released by burning 2.5 moles of $C O=2.5 \times 283$

$=707.5 \, kJ$

Energies released by burning 2.5 mole of $H_{2}=2.5 \times 241.8$

$=604.5\, kJ$

Total energy released $=707.5+604.5=-1312 kJ$