Question

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

• A. 0.3
• B. 0.6
• C. 1.2
• D. 1.5

Hint:

Always use equilibrium concentrations in the \(K_c\) expression, not initial concentrations.

Answer 1

The Correct Option is A

Step 1: Set up an ICE table.
For $A + B \rightleftharpoons C$ in a $1$ L flask, let $x$ moles of $C$ form. Start: $A = 2$, $B = 1$, $C = 0$. At equilibrium: $A = 2-x$, $B = 1-x$, $C = x$. Because volume is $1$ L, moles equal concentrations.
Step 2: Use the given condition.
We are told $[B] = 2[C]$ at equilibrium, so $1 - x = 2x$.
Step 3: Solve for x.
$1 = 3x$ gives $x = \tfrac{1}{3}$.
Step 4: Find all equilibrium concentrations.
$[A] = 2 - \tfrac{1}{3} = \tfrac{5}{3}$, $[B] = 1 - \tfrac{1}{3} = \tfrac{2}{3}$, $[C] = \tfrac{1}{3}$.
Step 5: Write the equilibrium expression.
\[ K_c = \frac{[C]}{[A][B]} = \frac{1/3}{(5/3)(2/3)} = \frac{1/3}{10/9} \]
Step 6: Simplify.
$K_c = \tfrac{1}{3} \times \tfrac{9}{10} = \tfrac{3}{10} = 0.3$, which is option 1.
\[ \boxed{0.3} \]