
A reaction is, $A + B \to C + D$ initially we start with equal concentrations of A and B. At equilibrium, we find the moles of C is two times of A. What is the equilibrium constant of the reaction?
(A) $2$
(B) $4$
(C) $\dfrac{1}{2}$
(D) $\dfrac{1}{4}$
Answer
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Hint: In a reversible reaction, at equilibrium the product of the molar concentration of products, each raised to the power equal to its coefficient, divided by the product of the molar concentration of the reactant each raised to the power equal to its coefficient, is constant at a constant temperature and is called equilibrium constant.
Complete answer:
Given equation, $A + B \rightleftharpoons C + D$
According to question, At equilibrium, $[C] = 2A$
Let consider the no. of moles of [C] At equilibrium be $x$
$x = 2(1 - x) = 2 - 2x$
$3x = 2 \Rightarrow x = \dfrac{2}{3}$
Now, the equilibrium constant is given as:
${K_c} = \dfrac{{[C][D]}}{{[A][B]}}$
${K_c} = $Putting the values of final no. Of moles in this
${K_c} = \dfrac{{[0.67][0.67]}}{{[0.33][0.33]}}$
${K_c} = 4$
Now we will match these calculations with the given options:-
Out of all the given options, only option (B) matches our calculated value of reaction constant, hence, it is the right answer.
Therefore option B is the correct answer .
ADDITIONAL INFORMATION:
The law of mass action states that the rate at which a substance reacts is proportional to active mass and the rate of a chemical reaction is proportional to the product of the active masses of reacting substances. It is used in the above formula to find equilibrium constant. In a chemical reaction, the equilibrium constant is defined as the ratio of the amount of reactant and the amount of product formed which determines the behaviour of the reaction. Change in concentration, pressure, temperature does not alter the reaction constant. It remains constant.
Note:
Based upon the experimental results of many workers, the Norwegian scientists Goldberg and Wager in 1864 gave a general statement about the influence of masses of reactants on the reaction rates, known as the law of mass action.
Complete answer:
Given equation, $A + B \rightleftharpoons C + D$
According to question, At equilibrium, $[C] = 2A$
Let consider the no. of moles of [C] At equilibrium be $x$
$x = 2(1 - x) = 2 - 2x$
$3x = 2 \Rightarrow x = \dfrac{2}{3}$
| Initial Concentration | $1$ | $1$ | $0$ | $0$ | |
| At Equilibrium | $1 - x$ | $1 - x$ | $x$ | $x$ | |
| Final Concentration | $1 - \dfrac{2}{3}$ | $1 - \dfrac{2}{3}$ | $\dfrac{2}{3}$ | $\dfrac{2}{3}$ |
Now, the equilibrium constant is given as:
${K_c} = \dfrac{{[C][D]}}{{[A][B]}}$
${K_c} = $Putting the values of final no. Of moles in this
${K_c} = \dfrac{{[0.67][0.67]}}{{[0.33][0.33]}}$
${K_c} = 4$
Now we will match these calculations with the given options:-
Out of all the given options, only option (B) matches our calculated value of reaction constant, hence, it is the right answer.
Therefore option B is the correct answer .
ADDITIONAL INFORMATION:
The law of mass action states that the rate at which a substance reacts is proportional to active mass and the rate of a chemical reaction is proportional to the product of the active masses of reacting substances. It is used in the above formula to find equilibrium constant. In a chemical reaction, the equilibrium constant is defined as the ratio of the amount of reactant and the amount of product formed which determines the behaviour of the reaction. Change in concentration, pressure, temperature does not alter the reaction constant. It remains constant.
Note:
Based upon the experimental results of many workers, the Norwegian scientists Goldberg and Wager in 1864 gave a general statement about the influence of masses of reactants on the reaction rates, known as the law of mass action.
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