Question

One mole of $A(g)$ was taken in a 1L closed flask and heated to T(K). At equilibrium, the concentration of $A(g)$ is equal to four times the equilibrium concentration of $B(g)$. What is $K_c$ for $A(g) \rightleftharpoons B(g) + C(g)$?

• A. 0.02
• B. 0.04
• C. 0.05
• D. 0.06

Hint:

Always define the equilibrium expression clearly from the balanced equation.

Answer 1

The Correct Option is C

Step 1: Set up an ICE table.
For $A(g) \rightleftharpoons B(g) + C(g)$ in a $1$ L flask, start with $1$ mol of A. Let $x$ mol of A react. At equilibrium $[A] = 1-x$, $[B] = x$, $[C] = x$.

Step 2: Use the given condition.
We are told $[A]$ is four times $[B]$, so \[ 1 - x = 4x \]
Step 3: Solve for x.
\[ 5x = 1 \Rightarrow x = 0.2 \] So $[A] = 0.8$, $[B] = 0.2$, $[C] = 0.2$ (all in mol/L since volume is $1$ L).

Step 4: Write the equilibrium expression.
\[ K_c = \frac{[B][C]}{[A]} \]
Step 5: Plug in the values.
\[ K_c = \frac{0.2 \times 0.2}{0.8} = \frac{0.04}{0.8} = 0.05 \]
Step 6: Conclusion.
So the equilibrium constant is $0.05$. \[ \boxed{K_c = 0.05} \]