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Question

# The equilibrium constant for the reaction CO(g)+H2O(g)⇌CO2(g)+H2(g) is 3 at 500K. In a 2 litre vessel 60gm of water gas [equimolar mixture of CO(g) and H2(g)] and 90 gm of steam is initially taken. What is the equilibrium concentration of H2(g) at equilibrium (mole/L)?

A
1.75
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B
3.5
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C
1.5
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D
0.75
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Solution

## The correct option is B 3.560 g of water gas correspond to 6030=2 moles of CO and 2 moles of H290 g of steam corresponds to 9018=5 moles of H2The initial moles of CO,H2O,CO2 and H2 are 2,5,0 and 2 moles respectively.The moles of CO,H2O,CO2 and H2 present at equilibrium are 2−x,5−x,x and 2+x moles respectively.The expression for the equilibrium constant is Kc=[CO2][H2][CO][H2O]Substitute values in the above expression.3=(x2)(2+x2)(2−x2)(5−x2)Thus x=1.5 and [H2]=2+1.5=3.5M

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