From the following data of ΔH, of the following reactions,C(s)+12O2(g)⟶CO(g) ΔH=−110 kJC(s)+H2O(g)⟶CO(g)+H2(g) ΔH=132 kJWhat is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, to maintain constant temperature :

The reactions we have,C(s)+12O2a mole⟶COC(s)+H2O(g)b mole⟶CO(g)+H2(g)220×a=132×b ab=132220 ab=0.61

From the following data of ΔH, of the following reactions,C(s)+12O2(g)⟶CO(g) ΔH=−110 kJC(s)+H2O(g)⟶CO(g)+H2(g) ΔH=132 kJWhat is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, to maintain constant temperature :

The reactions we have,C(s)+12O2a mole⟶COC(s)+H2O(g)b mole⟶CO(g)+H2(g)220×a=132×b ab=132220 ab=0.61

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From the following data of $ΔH$ , of the following reactions,
$\begin{array}{lr} C (s) + \frac{1}{2} O_{2} (g) ⟶ CO (g) ΔH = - 110 kJ \\ C (s) + H_{2} O (g) ⟶ CO (g) + H_{2} (g) ΔH = 132 kJ \end{array}$
What is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, to maintain constant temperature :

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0.5 : 1

0.6 : 1

0.8 : 1

1 : 1

answer is B.

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Detailed Solution

The reactions we have,

$C (s) + \underset{a mole}{\frac{1}{2} O_{2}} ⟶ CO$

$C (s) + \underset{b mole}{H_{2} O (g)} ⟶ CO (g) + H_{2} (g)$

$220 \times a = 132 \times b \frac{a}{b} = \frac{132}{220} \frac{a}{b} = \frac{0.6}{1}$

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From the following data of ΔH, of the following reactions,C(s)+12O2(g)⟶CO(g) ΔH=−110 kJC(s)+H2O(g)⟶CO(g)+H2(g) ΔH=132 kJWhat is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, to maintain constant temperature :

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