Question

Given the reaction between 2 gases represented by $A_2$ and $B_2$ to give the compound ${AB_{(g)}}$. ${ A_{2(g)} + B_{2(g)} <=> 2 \, AB_{(g)}}$ At equilibrium, the concentration of ${A_2 = 3.0 \times 10^{-3} \, M}$ of ${B_2 = 4.2 \times 10^{-3}\, M }$ of ${AB = 2.8 \times 10^{-3} \, M}$ If the reaction takes place in a sealed vessel at $527^{\circ} C$, then the value of $K_C$ will be :

• A. 1.9
• B. 0.62
• C. 4.5
• D. 2

Answer 1

The Correct Option is B

To find the equilibrium constant $K_C$ for the given reaction: \[ A_{2(g)} + B_{2(g)} \leftrightarrow 2 \, AB_{(g)} \] we use the expression for the equilibrium constant for a reaction involving gases: \[ K_C = \frac{{[AB]^2}}{{[A_2][B_2]}} \] Given equilibrium concentrations are:

• $[A_2] = 3.0 \times 10^{-3} \, M$
• $[B_2] = 4.2 \times 10^{-3} \, M$
• $[AB] = 2.8 \times 10^{-3} \, M$

Substitute these values into the $K_C$ expression:

\[ K_C = \frac{{(2.8 \times 10^{-3})^2}}{{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}} \]

First, calculate the numerator:

\[ (2.8 \times 10^{-3})^2 = 7.84 \times 10^{-6} \]

Then calculate the denominator:

\[ (3.0 \times 10^{-3})(4.2 \times 10^{-3}) = 12.6 \times 10^{-6} \]

Now, divide the numerator by the denominator:

\[ K_C = \frac{{7.84 \times 10^{-6}}}{{12.6 \times 10^{-6}}} = 0.622 \]

Therefore, the equilibrium constant $K_C$ is approximately 0.62.

Thus, the correct answer is: 0.62.