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Derivation of Kp and Kc

KI + I2⇌ KI3....

Question

$K I + I_{2} ⇌ K I_{3}$ . Given that the initial mass of $K I$ is 1.326. Mass of $K I_{3}$ at equilibrium is $0.105 g$ and the number of mole of free $I_{2}$ at equilibrium is 0.0025. The volume of solution is one litre. The value of $K_{c}$ of the reaction is _______.

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Solution

The equilibrium reaction, the initial number of moles and the equilibrium number of moles are as given below.

$K I (a q) + I_{2} (s) ⇌ K I_{3} (a q)$
Initial mole $\frac{1.326}{166}$

Mole at eq. $[\frac{1.326}{166} - \frac{0.105}{420}] 0.0025 \frac{0.105}{420}$

The volume of solution = 1 litre.

Thus, one mole represents concentrations.
The equilibrium constant expression is as given below:

$K_{c} = \frac{[K I_{3}]}{[K I]} = \frac{2.5 \times 10^{- 4}}{7.738 \times 10^{- 3}} = 0.032$

Hence, the equilibrium constant of the reaction has value of 0.032.

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K_{C}

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