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Reaction Quotient

K1 and K2 a...

Question

$K_{1}$ and $K_{2}$ are the velocity constants of the forward and backward reactions. The equilibrium constant K of the reaction is:

k_{1} \times k_{2}

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k_{1} - k_{2}

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\frac{k_{1}}{k_{2}}

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\frac{k_{2}}{k_{1}}

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Solution

The correct option is D $\frac{k_{1}}{k_{2}}$
The equilibrium constant of a reaction is given by the ratio of its forward rate to its backward rate, i.e,
$K = \frac{k_{f}}{k_{b}}$
Therefore, for the given reaction, the rate constant will be given as:
$K = \frac{k_{1}}{k_{2}}$

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Similar questions

Equilibrium constants $K_{1}$ and $K_{2}$ for the following equilibria are

$N O (g) + 0.5 O_{2} (g) \to K_{1} N O_{2} (g)$ and
$2 N O_{2} (g) \to K_{2} 2 N O (g) + O_{2} (g)$

Two gaseous equilibria $S O_{2} (g) + 0.5 O_{2} (g) ⇌ S O_{3} (g)$ and

$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ have equilibrium constants $K_{1}$ and $K_{2}$ respectively

at 298K. Which of the following relationships between $K_{1}$ and $K_{2}$ is correct

S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

K_{1}

K_{2}

298 K

K_{1}

K_{2}

S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

K_{1}

K_{2}

K_{1}

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K_{2}

N_{2} + 2 O_{2} ⇋ 2 N O_{2}, K

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2 N O + O_{2} ⇋ 2 N O_{2}, K_{2}

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