Question

For a reaction, \( A + B \rightleftharpoons 2C \), \( 1.0 \) mole of \( A \), \( 1.5 \) mole of \( B \) and \( 0.5 \) mole of \( C \) were taken in a \( 1\,L \) vessel. At equilibrium, the concentration of \( C \) was \( 1.0\,\text{mol L}^{-1} \). The equilibrium constant for the reaction is \( x/15 \). The value of \( x \) is:

• A. 32
• B. 22
• C. 18
• D. 16

Hint:

For equilibrium problems, make an ICE table and use stoichiometric coefficients carefully to calculate equilibrium concentrations.

Answer 1

The Correct Option is D

Step 1: Write the given reaction.
\[ A + B \rightleftharpoons 2C \]

Step 2: Write initial concentrations.
Since volume is \(1\,L\), moles are equal to molar concentrations.
\[ [A]_0 = 1.0,\quad [B]_0 = 1.5,\quad [C]_0 = 0.5 \]

Step 3: Let the change in concentration be \( y \).
For the reaction \( A+B \rightleftharpoons 2C \), if \( y \) mole/L of \( A \) and \( B \) react, then \( 2y \) mole/L of \( C \) is formed.
So at equilibrium:
\[ [A] = 1-y \]
\[ [B] = 1.5-y \]
\[ [C] = 0.5+2y \]

Step 4: Use given equilibrium concentration of \( C \).
\[ 0.5+2y=1.0 \]
\[ 2y=0.5 \]
\[ y=0.25 \]

Step 5: Find equilibrium concentrations of \( A \) and \( B \).
\[ [A] = 1-0.25 = 0.75 \]
\[ [B] = 1.5-0.25 = 1.25 \]
\[ [C] = 1.0 \]

Step 6: Write expression for equilibrium constant.
\[ K_c = \frac{[C]^2}{[A][B]} \]
\[ K_c = \frac{(1.0)^2}{(0.75)(1.25)} \]
\[ K_c = \frac{1}{0.9375} \]
\[ K_c = \frac{16}{15} \]

Step 7: Final conclusion.
Given:
\[ K_c = \frac{x}{15} \]
So:
\[ \frac{x}{15}=\frac{16}{15} \]
\[ x=16 \]
\[ \boxed{16} \]