The equilibrium constant K for a reaction is related to the equilibrium constant of the corresponding reaction, whose equation is obtained by multiplying or dividing the equation for the original reaction by a small integer.

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Solution

$A + 2 B \leftrightarrow 3 C$

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

$n A + 2 n B \leftrightarrow 3 n C$

$K^{'} = \frac{[C]^{3 n}}{[A]^{n} [B]^{2 n}}$

$= {(\frac{[C]^{3}}{[A] [B]^{2}})}^{n}$

$= K^{n}$

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Similar questions

Q. I.

A + B ⇌ 2 C + D

II.

2 C + D ⇌ A + B

If the equilibrium constant for the reaction represented by equation I above is

4.0 \times 10^{- 2}

. What is the value of the equilibrium constant for the reaction represented by equation II?

For the reaction $N_{2}$ + $O_{2}$ $⇋$ 2 NO equilibrium constant is $K_{c}$

Now if we multiply across the given equation by a constant say 3, the new equilibrium constant is ${K_{c}}^{3}$ (previous equilibrium raised to the power of the constant 3)

Q. The value of equilibrium constant for the reaction represented by the equation :

A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)

is ____.

The equilibrium constant expression for a gas reaction is,
$K_{c} = \frac{[N H_{3}]^{4} [O_{2}]^{5}}{[N O]^{4} [H_{2} O]^{6}}$
Write the balanced chemical equation corresponding to this expression.

Q. Assertion :The value of equilibrium constant depends on the stoichiometry of the equation. BECAUSE Reason: The value of equilibrium constant does not change when the equation is multiplied or divided by a number.

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The equilibrium constant K for a reaction is related to the equilibrium constant of the corresponding reaction, whose equation is obtained by multiplying or dividing the equation for the original reaction by a small integer.

A+2B↔3CK=[C]3[A][B]2nA+2nB↔3nCK′=[C]3n[A]n[B]2n=([C]3[A][B]2)n=Kn

$A + 2 B \leftrightarrow 3 C$

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

$n A + 2 n B \leftrightarrow 3 n C$

$K^{'} = \frac{[C]^{3 n}}{[A]^{n} [B]^{2 n}}$

$= {(\frac{[C]^{3}}{[A] [B]^{2}})}^{n}$

$= K^{n}$