Predicting the Direction and Extent of Reaction

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The magnitude of the equilibrium constant, K of a reaction indicates how far a reaction can go in the direction as written. The larger the value of K, the greater will be the equilibrium concentration of the components on the right hand side of the reaction (products) relative to those on the left hand side (reactants).

Thus, the value of K provides information about the extent of reaction in a particular direction.

When K » 1

When the value of K is very high such as, 10⁷-10¹⁵ or more, the reaction proceeds to almost completion. In such reactions almost the whole of the reacting substance gets converted into products.

When K > 1

When the value of K is greater than one (but not too large), the reaction in the forward direction is favoured more than the reaction in the backward direction. In such cases, the equilibrium concentration of products is higher than that of the reactants.

When K = 1

When the value of K is equal to one, both the directions of the reaction are almost equally favoured. In such cases, the equilibrium concentration of the reactants and products are comparable.

When K < 1

When the value of K is smaller than one, the reaction in the backward direction is favoured. In such cases, the equilibrium concentration of reactants will be much higher than the concentration of the products.

For example, the equilibrium constant for the reaction,

at 298 K is,

The value of K = 2.0 x 10¹⁵ is very high. So, the reaction in the forward direction is highly favoured, and may proceed to almost completion. Thus, from very high value of the equilibrium constant for this reaction we can conclude that when a copper rod is dipped in a solution containing Ag⁺, silver metal (Ag) gets precipitated almost completely. That is, copper displaces the whole of silver from a solution containing Ag⁺ ions.

The equilibrium constant for the reaction

at 298 K is,

The value of K = 2.0 x 10^-19 is very small. So, the reaction in the backward direction is highly favoured. The forward reaction will take place to a very small extent. Thus, we can conclude that a copper rod dipped into a solution of Zn²⁺_(aq) is not able to precipitate zinc metal. As the reaction in the backward direction will proceed to a large extent Cu will be precipitated. Zinc displaces copper from any solution containing Cu²⁺ions almost completely.

The equilibrium constant for the reaction,

at 720 K is 48, i.e.,

The value of K = 48 for this reaction is not very high. So, at equilibrium both the reactants and the products should be present in almost comparable amounts.

Problems

12. The equilibrium concentrations of NO₂ and N₂O₄ at 142°C for the reaction:

are 0.710 moles litre^-1 and 0.145 moles litre^-1 respectively. What is the equilibrium constant of the reaction?

Solution

The equilibrium constant for the reaction

Substituting the given values, we can write

= 0.204 mol^-1 litre

13. At a certain temperature 1 mole of PCl_3(g) and 2 moles of Cl_2(g) were placed in a 3.0 litre container. At equilibrium, only 0.7 mole of PCl_3(g) was left. Calculate the value of the equilibrium constant for the reaction:

Solution

Let the initial amounts of PCl₃ and Cl₂ placed in a container of volume V, be a and b moles respectively. According to the balanced chemical equation, x moles of PCl₃ combine with x moles of Cl₂ to form x moles of PCl₅

PCl_3(g) + Cl_2(g) g PCl_5(g)

1 mol 1 mol 1 mol

x mol x mol x mol

At equilibrium, 0.7 mole of PCl₃ remained. Therefore,

(a-x)=(Number of moles of PCl₃ initially present)-(Number of moles that have reacted)

0.7 = 1 - x

or x = 0.3 mol

At equilibrium, the equilibrium concentration in mol litre^-1 are,

= 0.77 mol^-1 L or simply as 0.77.

14. For the reaction N₂(g) + O₂(g)  « 2NO(g), the value of K_c at 2000°C is 0.1. At equilibrium [N₂] = [O₂] = 0.1 mol litre^-1. Calculate the concentration of NO.

Solution

Let the concentration of NO be x mol litre^-1

Therefore [NO] = 0.032 mole litre^-1.

15. constants for the forward and backward reactions are found to be 4.2 x 10^-2mol min^-1 and 3.6 x 10³ mol min^-1 respectively. Calculate the equilibrium constant for the reaction.

Solution