Question

Consider the following equilibrium reaction in gaseous state at T(K): \[ A + 2B \rightleftharpoons 2C + D \] The initial concentration of B is 1.5 times that of A. At equilibrium, the concentrations of A and B are equal. The equilibrium constant for the reaction is:

• A. 6
• B. 16
• C. 12
• D. 4

Hint:

Always express all concentrations in terms of a single variable when solving equilibrium problems, and simplify the ratios before canceling variables.

Answer 1

The Correct Option is D

Step 1: Let initial concentration of A be \(x\)
Then, initial concentration of B is \(1.5x\) Step 2: Let change in concentration of A be \(-a\)
Then, \[ \begin{aligned} \text{Change in B} &= -2a
\text{Change in C} &= +2a
\text{Change in D} &= +a \end{aligned} \] Step 3: Use equilibrium condition
Given: \text{[A] = [B]} at equilibrium So, \[ x - a = 1.5x - 2a \Rightarrow x - a = 1.5x - 2a
\Rightarrow a = 0.5x \] Step 4: Calculate equilibrium concentrations
\[ \begin{aligned} [A] &= x - a = 0.5x
[B] &= 1.5x - 2a = 0.5x
[C] &= 2a = x
[D] &= a = 0.5x \end{aligned} \] Step 5: Apply equilibrium expression
\[ K = \frac{[C]^2[D]}{[A][B]^2} = \frac{(x)^2(0.5x)}{(0.5x)(0.5x)^2} = \frac{x^2 . 0.5x}{0.5x . 0.25x^2} = \frac{0.5x^3}{0.125x^3} = 4 \] Step 6: Conclusion
The equilibrium constant \(K = 4\)