Question

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

• A. $2.0$
• B. $1.9$
• C. $0.62$
• D. $4.5$

Answer 1

The correct option is C $0.62$

The equilibrium reaction is as given below.
$A_2+B_2\rightleftharpoons 2AB$
The expression for the equilibrium constant is as given below:
$K_c=\frac{[AB{]}^2}{\left[A_2\right]\left[B_2\right]}$
Substitute the values in the above expression
$K_c=\frac{2.8\times 2.8\times {10}^{-6}}{3\times {10}^{-3}\times 4.2\times {10}^{-3}}=0.62$
Hence, the value of the equilibrium constant is $0.62$.