# I2 + Cl2 →2ICl Kp=81.9. A reaction mixture initially P(I2)=P(Cl2)=P(ICl)= 0.1 atm. How to determine the direction of the reaction will proceed?

Dec 9, 2015

The equilibrium will shift to the right.

#### Explanation:

The equilibrium in question is the following:

${I}_{2} + C {l}_{2} r i g h t \le f t h a r p \infty n s 2 I C l \text{ " " } {K}_{P} = 81.9$

The expression of ${K}_{P}$ is: K_P=(P_(ICl)^2)/(P_(I_2)*P_(Cl_2)

Since we have the initial pressures, we will determine the direction of the equilibrium from the reaction quotient ${Q}_{P}$.

The expression of ${Q}_{P}$ is: ${Q}_{P} = \frac{{\left({P}_{0}\right)}_{I C l}^{2}}{{\left({P}_{0}\right)}_{{I}_{2}} \cdot {\left({P}_{0}\right)}_{C {l}_{2}}}$

$\implies {Q}_{P} = {\left(0.1\right)}^{2} / \left(\left(0.1\right) \times \left(0.1\right)\right) = 1 < {K}_{P}$

Since ${Q}_{P} < {K}_{P}$ then the equilibrium will shift to the right to increase the pressure of the product (numerator) and decrease the pressure of reactants (denominator).