The equilibrium constant for, H2(g)+CO2(g)⇌H2O(g)+CO(g) is 1'80 at 1000°c. If 1.0 mole of H2 and 1.0 mole of CO2 are placed in one litre flask, the final equilibrium concentration of CO at 1000°c will be:

CO21(1-x)+H211-x⇌H2O0x+CO0x∴Kc=x2(1−x)2or x1−x=Kc=1.8∴x1−x=1.34∴x=0.573M

The equilibrium constant for, H2(g)+CO2(g)⇌H2O(g)+CO(g) is 1'80 at 1000°c. If 1.0 mole of H2 and 1.0 mole of CO2 are placed in one litre flask, the final equilibrium concentration of CO at 1000°c will be:

CO21(1-x)+H211-x⇌H2O0x+CO0x∴Kc=x2(1−x)2or x1−x=Kc=1.8∴x1−x=1.34∴x=0.573M

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The equilibrium constant for, $H_{2} (g) + {CO}_{2} (g) ⇌ H_{2} O (g) + CO (g)$ is 1'80 at 1000°c. If 1.0 mole of H₂ and 1.0 mole of CO₂ are placed in one litre flask, the final equilibrium concentration of CO at 1000°c will be:

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0 .385 M

0.295 M

5 .73 M

0 .573 M

answer is A.

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Detailed Solution

$\underset{(1 - x)}{\underset{1}{{CO}_{2}}} + \underset{(1 - x)}{\underset{1}{H_{2}}} ⇌ \underset{x}{\underset{0}{H_{2} O}} + \underset{x}{\underset{0}{CO}}$
$\begin{array}{l} ∴ K_{c} = \frac{x^{2}}{(1 - x)^{2}} \\ or \frac{x}{1 - x} = \sqrt{K_{c}} = \sqrt{1 .8} \\ ∴ \frac{x}{1 - x} = 1 .34 \\ ∴ x = 0 .573 M \end{array}$