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?

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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?

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Answer 1 · 2015-12-09T13:55:28

The equilibrium will shift to the right.

Explanation:

The equilibrium in question is the following:

$I_{2} + C l_{2} ⇌ 2 I C l K_{P} = 81.9$ $I_2+Cl_2rightleftharpoons2ICl" " " "K_P=81.9$

The expression of $K_{P}$ $K_P$ is: $K_{P} = \frac{P_{I C l}^{2}}{P_{I_{2}} \cdot P_{C l_{2}}}$ $K_P=(P_(ICl)^2)/(P_(I_2)*P_(Cl_2)$

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

The expression of $Q_{P}$ $Q_P$ is: $Q_{P} = \frac{{(P_{0})}_{0}^{2}}{{(P_{0})}_{I_{2}} \cdot {(P_{0})}_{C l_{2}}}$ $Q_P=((P_0)_(ICl)^2)/((P_0)_(I_2)*(P_0)_(Cl_2))$

$\Rightarrow Q_{P} = \frac{{(0.1)}^{2}}{(0.1) \times (0.1)} = 1 < K_{P}$ $=>Q_P=(0.1)^2/((0.1)xx(0.1))=1 < K_P$

Since $Q_{P} < K_{P}$ $Q_P < K_P$ then the equilibrium will shift to the right to increase the pressure of the product (numerator) and decrease the pressure of reactants (denominator).

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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?

1 Answer

Explanation:

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