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:

• A. 0.342
• B. 2.92
• C. 0.292
• D. 3.42

Answer 1

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
$aA+bB\rightleftharpoons cC+dD{K}_{eq}=\frac{K_f}{K_b}=\frac{[C{]}^c\times [D{]}^d}{[A{]}^a\times [B{]}^b}$
We have, $K_{eq}=\frac{K_f}{K_b}$

Now, $K_f=2.38\times {10}^{-4};K_b=8.15\times {10}^{-5}$
$K_{eq}=\frac{2.38\times {10}^{-4}}{8.15\times {10}^{-5}}$
$K_{eq}=2.92$



Theory:

Equilibrium constant :
Equilibrium constant is the ratio of rate constants of the forward and
the backward reaction.
$K_{eq}=\frac{K_f}{K_b}$
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 $mol{L}^{-1})$
For ex :

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

$K_c=\frac{\left[PC{l}_3{]}_{eq}\right[C{l}_2{]}_{eq}}{\left[PC{l}_5{]}_{eq}\right]}$


The equilibrium constant for a general reaction:

$aA+bB\rightleftharpoons cC+dD$
$K_c=\frac{\left[C{]}_{eq}^c\right[D{]}_{eq}^d}{\left[A{]}_{eq}^a\right[B{]}_{eq}^b}$

where $\left[C{]}_{eq}^c\right[D{]}_{eq}^d\left[A{]}_{eq}^{a}and\right[B{]}_{eq}^b$ are the equilibrium concentrations of the reactants and products.

Units of Equilibrium constant $K_c$:

$K_c=\frac{\left(mol/L{)}^c\right(mol/L{)}^d}{\left(mol/L{)}^a\right(mol/L{)}^b}$

Unit of $K_c(mol/L{)}^{(c+d)-(a+b)}$