Question

4 moles of A are mixed with 4 moles of B when 2 moles of C are formed at equilibrium. According to the equation $A+B\rightleftharpoons C+D$. Find the equilibrium constant.
(a) 1
(b) 25
(c) 10
(d) 12

Answer 1

Hint: The equilibrium constant is always equal to the ratio of the concentration of the product to the concentration of the reactant. And each concentration is raised to some power that is equal to the number of moles in the chemical reaction.

Complete step by step answer:
-The equilibrium constant is always equal to the ratio of the concentration of the product to the concentration of the reactant. And each concentration is raised to some power that is equal to the number of moles in the chemical reaction.
-So the given reaction is:
$A+B\rightleftharpoons C+D$
-Now the question says 4 moles of A and B are taken and they form 2 moles of C. So in the reaction, both the reactants have the same stoichiometric coefficient i.e., 1, therefore, there will be the formation of 2 moles of D.
-So, before equilibrium, the number of moles of A and B is 4 and the number of moles of C and D is 0.
-After equilibrium the number of moles of C and D is 2, so the number of moles of A and B will be:
$4-2=2$
-Let the volume of the solution V.
So the concentration of A will be: $\dfrac{2}{V}$
Concentration of B is: $\dfrac{2}{V}$
Concentration of C is: $\dfrac{2}{V}$
Concentration of D is: $\dfrac{2}{V}$
-The equilibrium constant:
${{K}_{c}}=\dfrac{[D][C]}{[A][B]}$
$\Rightarrow {{K}_{c}}=\dfrac{\dfrac{2}{V}\times \dfrac{2}{V}}{\dfrac{2}{V} \times \dfrac{2}{V}}$
$\Rightarrow {{K}_{c}}=\dfrac{2 \times 2}{2 \times 2}$
$\Rightarrow {{K}_{c}}=1$
So the value of the equilibrium constant is 1.

Therefore, the correct answer is option (a) 1.

Note: For calculating the equilibrium constant all the components in the equilibrium must be at the same temperature. Don't forget to raise the concentration of the molecule to the number of moles in the chemical reaction.