Question

ITEM 18 The reaction X2(g)?2X(g) occurs in a closed reaction vessel at constant volume and temperature. Initially, the vessel contains only X2at a pressure of 1.90atm . After the reaction reaches equilibrium, the total pressure is 2.65atm .What is the value of the equilibrium constant, Kp, for the reaction?

Answer 1

Answer 2

	$X_2$	X
Initial pressure (atm)	1.90	0
Equilibrium pressure (atm)	1.90-x	2x

The equilibrium total pressure is 2.65 atm.

Hence, $1.90-x+2x=2.65$

$x=0.75$

Hence, the equilibrium pressures of X₂ and X are $1.90-x=1.90-0.75=1.15 \quad atm$ and $2x=2(0.75)=1.50 \quad atm$ respectively.

The expression for the equilibrium constant is $K=\frac {P_X^2}{P_{X2}}$

Substitute values in the above expression.

$K=\frac {(1.50)^2}{1.15}=1.956$