Question

At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation. If there has been no dissociation the pressure would have been 0.074 atm. Calculate ${K_P}$ for the reaction, ${I_2}(g) \Leftrightarrow 2I(g)$.
(A) 1.60 atm
(B) 16 atm
(C) 0.016 atm
(D) 0.160 atm

Answer 1

The ${K_P}$ is the equilibrium constant. It is equal to the partial pressure of products divided by the partial pressure of reactants. The partial pressures are raised with the power which is equal to the coefficient of the substance. Mathematically, it can be written as - ${K_P}$=$\dfrac{{P_I^2}}{{{P_{{I_2}}}}}$

Complete step by step solution:
For such types of questions, first, let us write what is given to us and what we need to find out.
Given :
The reaction is ${I_2}(g) \Leftrightarrow 2I(g)$.
Temperature = 1000 K
Initial pressure for ${I_2}$ = 0.074 atm
Initial pressure for I = 0
Total partial at equilibrium = 0.112 atm
To know : ${K_P}$ for the reaction
We have the values for initial pressure. We can say that -
Equilibrium pressure for ${I_2}$= 0.074 - P atm
Equilibrium pressure for I = 2P atm
Thus, the total pressure at equilibrium = $0.074 - P + 2P$ $= 0.112$
Thus, solving the equation for P, we get
$P =$ $0.038$
Now, we can find out the equilibrium constant by the formula -
${K_P}$=$\dfrac{{P_I^2}}{{{P_{{I_2}}}}}$
Where ${K_P}$= equilibrium constant
$P_I^2$ is the partial pressure of products
${P_{{I_2}}}$ is the partial pressure of the reactants
On putting the values, we get
${K_P}$=$\dfrac{{(2 \times 0.038)}}{{(0.074 - 0.038)}}$
${K_P}$ $= 0.160 atm$

So, the option (D) is the correct answer.

Note: It must be noted that the partial dissociation means the substrate is not completely used up to make products. The reaction is in between. Some of the reactants have been dissociated to products. Because iodine is a gas. So, the partial pressure is being taken. If we had any liquid, then the terms concentration would have been used.