If the equilibrium constant for a reverse reaction is 9-0- 10 -4 is the equilibrium constant for the forward reaction-

The correct option is D 1/9.0× 10 ^ 4 #160; K_eq(forward)×K_eq(reverse) =1 K_eq(forward)× #160; 9.0 × 10^ 4= 1 K_eq(forward) = 1 9.0 × ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

If the equili...

Question

If the equilibrium constant for a reverse reaction is $9.0 \times 10^{- 4}$ is the equilibrium constant for the forward reaction?

3.0 \times 10^{- 2}

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- 3.0 \times 10^{- 2}

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- 9.0 \times 10^{- 2}

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1 / 9.0 \times 10^{- 4}

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1 / - 9.0 \times 10^{- 4}

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Solution

The correct option is D $1 / 9.0 \times 10^{- 4}$
$K_{e q} (f o r w a r d) \times K_{e q} (r e v e r s e) = 1$
$K_{e q} (f o r w a r d) \times 9.0 \times 10^{- 4} = 1$
$K_{e q} (f o r w a r d) = \frac{1}{9.0 \times 10^{- 4}}$

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If the equilibrium constant for a reverse reaction is 9.0×10−4 is the equilibrium constant for the forward reaction?

The correct option is D 1/9.0×10−4Keq(forward)×Keq(reverse)=1Keq(forward)×9.0×10−4=1Keq(forward)=19.0×10−4

If the equilibrium constant for a reverse reaction is $9.0 \times 10^{- 4}$ is the equilibrium constant for the forward reaction?

The correct option is D $1 / 9.0 \times 10^{- 4}$
$K_{e q} (f o r w a r d) \times K_{e q} (r e v e r s e) = 1$
$K_{e q} (f o r w a r d) \times 9.0 \times 10^{- 4} = 1$
$K_{e q} (f o r w a r d) = \frac{1}{9.0 \times 10^{- 4}}$