Question

$A{B}_2$ dissociates as $A{B}_2\left(g\right)\iff AB\left(g\right)+B\left(g\right)$. When the initial pressure of $A{B}_2$ is 600 mm of Hg, the total equilibrium pressure is 800 mm of Hg.Calculate $K_p$ for the reaction, assuming that the volume of the system remains unchanged.

• A. 50
• B. 100
• C. 166.8
• D. 400

Answer 1

The correct option is B 100

The equilibrium reaction is $A{B}_2\left(g\right)\iff AB\left(g\right)+B\left(g\right)$.

	$A{B}_2$	$AB$	$B$
Initial	600	0	0
Change	-x	x	x
Equilibrium	600-x	x	x

But total pressure $=600-x+x+x=800$ or $x=200$.

The equilibrium concentrations of $A{B}_2,AB$ and $B$ are 400, 200 and 200 respectively.

The expression for the equilibrium constant becomes $K_p=\frac{p_{AB}\times p_B}{p_{A{B}_2}}=\frac{200\times 200}{400}=100mmofHg$.