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Question

Consider the general hypothetical reaction:
A(s)2B(g)+3C(g)
If the concentration of C at equilibrium is doubled, then after the equilibrium is re-established, the concentration of B will be :

A
Twice of its original value
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B
Half of its original value
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C
22 times of original value
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D
122 times the original value
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Solution

The correct option is D 122 times the original value
The given reaction is,
A(s)2B(g)+3C(g)
So the equilibrium constant will be,

Kc=[B]2[C]3....(1)

Now, after doubling the concentration of C, let new concentrations are [C]/ and [B]/.
Given, [C]/=2[C]
So,
Kc=([B]/)2(2[C])3
Kc=([B]/)28[C]3 ...(2)

From (1) and (2),
[B]2[C]3=([B]/)28[C]3
[B]/=122[B]

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