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

Step 1: Set up the equilibrium picture.
The reaction is $A + B \rightleftharpoons C + D$. We start with $2$ mol of A and $1$ mol of B in a $1$ L flask. Since volume is $1$ L, the number of moles equals the concentration in mol/L.
Step 2: Use a change variable.
Let $x$ mol of A react. Each mole of A uses one mole of B and makes one mole of C and one mole of D. So the amounts that change are $-x$ for A and B, and $+x$ for C and D.
Step 3: Write equilibrium concentrations.
\[ [A] = 2-x, \quad [B] = 1-x, \quad [C] = x, \quad [D] = x. \]
Step 4: Apply the given condition.
We are told $[C] = 3[B]$ at equilibrium. So \[ x = 3(1-x) \ \Rightarrow \ x = 3 - 3x \ \Rightarrow \ 4x = 3 \ \Rightarrow \ x = \tfrac{3}{4}. \]
Step 5: Find each concentration.
\[ [A] = 2 - \tfrac34 = \tfrac54, \quad [B] = 1 - \tfrac34 = \tfrac14, \quad [C] = \tfrac34, \quad [D] = \tfrac34. \]
Step 6: Plug into the Kc expression.
\[ K_c = \frac{[C][D]}{[A][B]} = \frac{\tfrac34 \times \tfrac34}{\tfrac54 \times \tfrac14} = \frac{9/16}{5/16} = \frac{9}{5} = 1.8. \] So \[ \boxed{K_c = 1.8} \]