
In the reaction, $A + B \rightleftharpoons 2C$, at equilibrium, the concentration of A and B is $0.20mol/{{L}^{-1}}$ each, and that of C was found to be $0.60mol/{{L}^{-1}}$. Calculate the equilibrium constant of the reaction .
(A) 2.4
(B) 18
(C) 4.8
(D) 9
Answer
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Hint: An equilibrium constant (K) is the relationship between the concentration of reactants and products present at equilibrium in a reversible chemical process at a given temperature. It is the ratio of the concentration of products and reactants.
Formula Used: For the reaction: $A + B \rightleftharpoons 2C$
The equilibrium constant is $K=\frac{{{[C]}^{2}}}{[A][B]}$
Complete Step by Step Answer:
The given reaction is-
The concentration of the reactants and products are $\left[ A \right]=0.20$ ,$\left[ B \right]=0.20$,
$\left[ C \right]=0.60$
The equilibrium constant of this reaction is given by $K=\frac{{{[C]}^{2}}}{[A][B]}$
$K=\frac{{{(0.60)}^{2}}}{(0.20)(0.20)}$
$K=9$
Thus, the equilibrium constant of this reaction is 9.
Correct Option: (D) 9.
Additional Information: Chemical equilibrium is a state in which the rates of both the forward and backward reactions are equal and the concentration of both the reactants and products is constant. At equilibrium, there is no net change in the number of moles, although conversion from reactants to products or products to reactants is still occurring. According to the law of mass action, the value of the equilibrium constant, Kc, is constant at a constant temperature. The concentrations of reactants and products may vary, but the value for Kc remains the same.
Note: The units of equilibrium constant depend upon the number of moles of reactant and product. Its units can be given by ${{\left( mol/l \right)}^{\Delta n}}$ . Here, $\Delta n={{n}_{P}}-{{n}_{R}}=2-2=0$ .So, K has no units. To calculate the equilibrium constant, you must first understand the entire reaction and its stoichiometric coefficients. If K>1, the equilibrium favours products, and if K<1, then the equilibrium favours reactants. But, if K = 1, then both reactants and products are present in the mixture in significant amounts.
Formula Used: For the reaction: $A + B \rightleftharpoons 2C$
The equilibrium constant is $K=\frac{{{[C]}^{2}}}{[A][B]}$
Complete Step by Step Answer:
The given reaction is-
The concentration of the reactants and products are $\left[ A \right]=0.20$ ,$\left[ B \right]=0.20$,
$\left[ C \right]=0.60$
The equilibrium constant of this reaction is given by $K=\frac{{{[C]}^{2}}}{[A][B]}$
$K=\frac{{{(0.60)}^{2}}}{(0.20)(0.20)}$
$K=9$
Thus, the equilibrium constant of this reaction is 9.
Correct Option: (D) 9.
Additional Information: Chemical equilibrium is a state in which the rates of both the forward and backward reactions are equal and the concentration of both the reactants and products is constant. At equilibrium, there is no net change in the number of moles, although conversion from reactants to products or products to reactants is still occurring. According to the law of mass action, the value of the equilibrium constant, Kc, is constant at a constant temperature. The concentrations of reactants and products may vary, but the value for Kc remains the same.
Note: The units of equilibrium constant depend upon the number of moles of reactant and product. Its units can be given by ${{\left( mol/l \right)}^{\Delta n}}$ . Here, $\Delta n={{n}_{P}}-{{n}_{R}}=2-2=0$ .So, K has no units. To calculate the equilibrium constant, you must first understand the entire reaction and its stoichiometric coefficients. If K>1, the equilibrium favours products, and if K<1, then the equilibrium favours reactants. But, if K = 1, then both reactants and products are present in the mixture in significant amounts.
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