EQUILIBRIUM CONSTANT AND ITS UNITS - PUNJAB-CURRICULUM-AND-TEXTBOOK-BOARD-PCTB / 10TH-CLASS / CHEMISTRY / CHEMISTRY-10TH-CLASS-NBF-2024 / CHEMICAL-EQUILIBRIUM / EQUILIBRIUM-CONSTANT-AND-ITS-UNITS

9.3 Equilibrium Constant and its Units

Equilibrium constant is a ratio of the product of concentration of products raised to the power of coefficient to the product of concentration of reactants raised to the power of coefficient as expressed in the balanced chemical equation.

K_{c} =

It is conventional to write the products as numerator and reactants as denominator. By knowing, the balanced chemical equation for a reversible reaction we can write the equilibrium expression. Thus, we can calculate the numerical value of by putting actual equilibrium concentrations of the reactants and products into equilibrium expression. The value of $K_{c}$ depends only on temperature, it does not depend on the initial concentrations of the reactants and the products. A few problems have been solved to make the concept clear. $K_{c}$ has no units in reactions with equal number of moles on both sides of the equation. This is because concentration units cancel out in the expression for $K_{c}$ e.g., for the reaction:

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]} U n i t s = \frac{(m o l \cdot {d m}^{- 3})^{2}}{(m o l \cdot {d m}^{- 3}) (m o l \cdot {d m}^{- 3})} = n o u n i t s

For reactions in which the number of moles of reactants and product are not equal in the balanced chemical equation, K of course, have units, e.g., for the reaction

K_{c} = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}} = \frac{(m o l \cdot {d m}^{- 3})^{2}}{(m o l \cdot {d m}^{- 3}) (m o l \cdot {d m}^{- 3})^{3}} = \frac{1}{(m o l \cdot {d m}^{- 3})^{2}} = {m o l}^{- 2} {d m}^{6}

Problem 9.1

When hydrogen reacts with iodine at $25^{\circ}C$ to form hydrogen iodide by a reversible reaction as follows:

H_{2 (g)} + I_{2 (g)} 2 H_{I (g)}

The equilibrium concentrations are:

$[H_{2}] = 0.05 m o l {d m}^{- 3};$ $[I_{2}] = 0.06 m o l {d m}^{- 3};$ and $[H I] = 0.49 m o l {d m}^{- 3} .$

Calculate the equilibrium constant for this reaction.

Solution

Given equilibrium concentrations are;

$[H_{2}] = 0.05 m o l {d m}^{- 3}; [I_{2}] = 0.06 m o l {d m}^{- 3};$ $[H I] = 0.49 m o l {d m}^{- 3} .$

Write the equilibrium constant expression as

K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]}