Question

In the reversible reaction,A + B ⇌ C + D, concentration of C and D at equilibrium was 0.8 mol/litre, then the equilibrium constant will be (if initial concentration of A and B will be 1M each)-
(A) 6.4
(B) 0.64
(C) 1.6
(D) 16.0

Answer 1

Hint: Equilibrium constant is defined as the ratio of the concentration of products to the conversation of reactants both raised to the power of their stoichiometric coefficient (mole used to balance the reaction). The equilibrium constant is the number that gives the relationship between the number of reactants and products in a reversible reaction.

Formula Used: For general reaction,
\[aA\text{ }+\text{ }bB\rightleftharpoons cC\text{ }+\text{ }dD\], equilibrium constant is defined as
\[Kc=\text{ }{{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}/{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}},\text{ }Where\text{ }{{\left[ C \right]}^{c}}and\text{ }{{\left[ D \right]}^{d}}\]are the molar concentration of products, C and D both raised to the power of their stoichiometric coefficients, c and d (number of moles) and similarly for reactant.

Complete Step by Step Solution:
Given chemical equation is
\[A\text{ }+\text{ }B\rightleftharpoons C\text{ }+\text{ }D\]

Initially, the molar concentration of A and B (reactant) was 1M each, and the molar concentration of product (A and B) is zero such as
\[A\text{ }+\text{ }B\rightleftharpoons C\text{ }+\text{ }D\]

After some time, the molar concentration of products, C and D changed from 0 to 0.8 each. As the concentration of product increases thus to maintain equilibrium the molar concentration of reactant will get decrease by the same amount with which the molar concentration of product increases (0.8) such as
A + B ⇋ C

Now putting the molar concentration of product and reactant in the equilibrium constant formula, we can find its value. The equilibrium constant for the given equation is
\[A\text{ }+\text{ }B\rightleftharpoons C\text{ }+\text{ }D\]
$ K_c\text{ }=\text{ }0.8\text{ }\times \text{ }0.8\text{ }/\text{ }0.2\text{ }\times \text{ }0.2 \\$
$ K_c\text{ }=\text{ }16.0. \\$
$ K_c\text{ }=\text{ }0.8\text{ }\times \text{ }0.8\text{ }/\text{ }0.2\text{ }\times \text{ }0.2 \\$
$ K_c\text{ }=\text{ }16.0. \\$
Thus, the correct option is D.

Note: The equilibrium constant unit of this reaction is 0 or can say the equilibrium constant of this reaction is unit less. Unit is equilibrium constant can be found just by putting a unit of the molar concentration of products and reactants. Unit of molar concentration is always Mole/litre. So the unit of the molar concentration of products (numerator) in this reaction will be cancelled out by the unit of the molar concentration of reactant (denominator).