The equilibrium constant for the following reaction will be:
$3 A + 2 B ⇋ C$

\frac{[3 A] [2 B]}{[C]}

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\frac{[C]}{[3 A] [2 B]}

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\frac{[C]}{{[A]}^{2} {[B]}^{2}}

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\frac{[C]}{{[A]}^{3} {[B]}^{2}}

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Solution

The correct option is D $\frac{[C]}{{[A]}^{3} {[B]}^{2}}$
$3 A + 2 B ⇋ C$

Equilibrium constant $K = \frac{[C]}{{[A]}^{3} {[B]}^{2}}$

Hence, option D is correct

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Similar questions

For the system 3A + 2B $⇌$ C, the expression for equilibrium constant is

3 A + 2 B

⇌

C

[\to a \to b \to c] = - 2

[\to a + 2 \to b, \to b - 2 \to c, \to c + 3 \to a] =

3 a (a - b + c), 2 b (a - b + c)

3 a - 2 b + 4 c

- 2 a + b - 5 c

3 a - 2 b + c

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