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Question

If A+BC+D; KC=K1 and E=aV; 2A+2B2C+2D; KC=K2 and E=bV then:


A
a=b
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B
K2=K21
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C
a=2b
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D
b=a2
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Solution

The correct options are
A a=b
B K2=K21
K1=[C][D][A][B] and K2=[C]2[D]2[A]2[B]2

Also, E is independent of stoichiometry.

Hence, a=b

Also, K2=K21 (from the expression given above)

Hence, options A and B are correct.

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