If C ( s ) + 2 1 O 2 ( g ) → CO ( g ) ; Δ H = − 110 kJ C ( s ) + H 2 O ( g ) → CO ( g ) H 2 ; Δ H = 132 kJ The mole composition of mixture of O & steam on being passed over coke at 1

Question

If C ( s ) ​ + 2 1 ​ O 2 ( g ) ​ → CO ( g ) ​ ; Δ H = − 110 kJ C ( s ) ​ + H 2 ​ O ( g ) ​ → CO ( g ) ​ H 2 ​ ; Δ H = 132 kJ The mole composition of mixture of O & steam on being passed over coke at 1273 K, so that temperature remains constant is :

TestHub · Accepted Answer

Question Explanation: We have to find out mole composition of mixture.

Concept: This question is based on enthalpy of reaction.

Solution: energy released w. r. t. $\frac{1}{2}$ mole $O_{2} = 110 KJ$

energy released w. r. t. $n_{1}$ mole $O_{2} = (220 \times n_{1})$

energy consumed w.r.t. 1 mole steam $= 132 KJ$

energy consumed w.r.t. $n_{2}$ mole steam $= 132 X n_{2}$

$220 X n_{1} = 132 X n_{2}$

$\frac{n _{1}}{n _{2}} = \frac{132}{220} = \frac{0.6}{1}$

Final Answer: option (A)

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