For an equilibrium reaction- the rate constants for the forward and backward reactions are 2-38 - 10-4 s -1 and 8-15 - 10-5 s -1 respectively- The equilibrium constant for the reaction is-A- 0-342B- 2-92C- 0-292D- 3-42

The correct option is B 2.92The nbsp;equilibrium constant nbsp;of a chemical reaction is the value of its reaction quotient at chemical nbsp;equilibrium. ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

For an equili...

Question

For an equilibrium reaction, the rate constants for the forward and backward reactions are $2.38 \times 10^{- 4} s^{- 1}$ and $8.15 \times 10^{- 5} s^{- 1}$ respectively. The equilibrium constant for the reaction is:

0.342

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2.92

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0.292

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3.42

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Solution

The correct option is B 2.92
The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium.
For a given reaction
$a A + b B ⇌ c C + d D K_{e q} = \frac{K_{f}}{K_{b}} = \frac{[C]^{c} \times [D]^{d}}{[A]^{a} \times [B]^{b}}$
We have, $K_{e q} = \frac{K_{f}}{K_{b}}$

Now, $K_{f} = 2.38 \times 10^{- 4}; K_{b} = 8.15 \times 10^{- 5}$
$K_{e q} = \frac{2.38 \times 10^{- 4}}{8.15 \times 10^{- 5}}$
$K_{e q} = 2.92$

Theory:

Equilibrium constant :
Equilibrium constant is the ratio of rate constants of the forward and
the backward reaction.
$K_{e q} = \frac{K_{f}}{K_{b}}$
Types of equilibrium constant :
$K_{p}$ ( It is the equilibrium constant calculated from the partial pressures of a reaction equation.)
$K_{c}$ (A constant in terms of concentration, where all the concentrations are at equilibrium and are expressed in $m o l L^{- 1})$
For ex :

$P C l_{5} ⇌ P C l_{3} + C l_{2}$

$K_{c} = \frac{[P C l_{3}]_{e q} [C l_{2}]_{e q}}{[P C l_{5}]_{e q}]}$

The equilibrium constant for a general reaction:

$a A + b B ⇌ c C + d D$
$K_{c} = \frac{[C]_{e q}^{c} [D]_{e q}^{d}}{[A]_{e q}^{a} [B]_{e q}^{b}}$

where $[C]_{e q}^{c} [D]_{e q}^{d} [A]_{e q}^{a} a n d [B]_{e q}^{b}$ are the equilibrium concentrations of the reactants and products.

Units of Equilibrium constant $K_{c}$ :

$K_{c} = \frac{(m o l / L)^{c} (m o l / L)^{d}}{(m o l / L)^{a} (m o l / L)^{b}}$

Unit of $K_{c} (m o l / L)^{(c + d) - (a + b)}$

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Similar questions

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For an equilibrium reaction, the rate constants for the forward and backward reactions are 2.38×10−4s−1 and 8.15×10−5s−1 respectively. The equilibrium constant for the reaction is:

For an equilibrium reaction, the rate constants for the forward and backward reactions are $2.38 \times 10^{- 4} s^{- 1}$ and $8.15 \times 10^{- 5} s^{- 1}$ respectively. The equilibrium constant for the reaction is: