Question

Observe the following reaction \[ A(g) + B(g) \rightleftharpoons C(g) + D(g) \] In a \(1\,L\) closed flask, \(2\) moles of \(A(g)\) and \(1\) mole of \(B(g)\) were taken and heated to temperature \(T(K)\). At equilibrium, the concentration of \(C(g)\) is thrice the concentration of \(B(g)\). What is the value of \(K_c\)?

• A. \(7.2\)
• B. \(3.6\)
• C. \(0.9\)
• D. \(1.8\)

Hint:

In equilibrium problems, always use an ICE table: \[ \text{Initial} \rightarrow \text{Change} \rightarrow \text{Equilibrium} \] This avoids mistakes while calculating equilibrium concentrations.

Answer 1

The Correct Option is D

Concept: For a reversible reaction: \[ aA+bB \rightleftharpoons cC+dD \] the equilibrium constant is: \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] To solve equilibrium problems:

• Write initial concentrations

• Assume change using variable \(x\)

• Write equilibrium concentrations

• Use given condition to find \(x\)

• Substitute into equilibrium expression

Step 1: Write initial concentrations. Since volume of flask is: \[ 1\,L \] Therefore concentration equals number of moles. Initially: \[ [A]_0 = 2 \] \[ [B]_0 = 1 \] \[ [C]_0 = 0 \] \[ [D]_0 = 0 \]

Step 2: Assume \(x\) moles react. Reaction: \[ A + B \rightleftharpoons C + D \] Thus changes are: \[ A \rightarrow -x \] \[ B \rightarrow -x \] \[ C \rightarrow +x \] \[ D \rightarrow +x \] Therefore equilibrium concentrations become: \[ [A] = 2-x \] \[ [B] = 1-x \] \[ [C] = x \] \[ [D] = x \]

Step 3: Use the given condition. Given: \[ [C] = 3[B] \] Substituting equilibrium concentrations: \[ x = 3(1-x) \] \[ x = 3 - 3x \] \[ 4x = 3 \] \[ x = \frac{3}{4} \]

Step 4: Calculate equilibrium concentrations. \[ [A] = 2-\frac34 = \frac54 \] \[ [B] = 1-\frac34 = \frac14 \] \[ [C] = \frac34 \] \[ [D] = \frac34 \]

Step 5: Substitute into equilibrium constant expression. \[ K_c = \frac{[C][D]}{[A][B]} \] Substituting: \[ K_c = \frac{\left(\frac34\right)\left(\frac34\right)} {\left(\frac54\right)\left(\frac14\right)} \] \[ = \frac{\frac{9}{16}}{\frac{5}{16}} \] \[ = \frac95 \] \[ K_c = 1.8 \]

Step 6: Write the final answer. Hence: \[ \boxed{K_c = 1.8} \] Therefore, the correct option is: \[ \boxed{(D)} \]