Question

A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is ${{K}_{p}}$ for the given equilibrium?
$2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)$
A. 6
B. 16
C. 4
D. 2

Answer 1

Hint: The formula for finding out the ${{K}_{p}}$ is to be needed in this question. Form the equations and find out the pressure before and after equilibrium is attained; then use that data to find out the value of ${{K}_{p}}$

Complete step by step answer:
In order to answer the question, we need to learn about chemical equilibrium. Let us get to know about the characteristics of chemical equilibrium:
(i) At equilibrium both the forward and the backward reactions proceed at the same rate and hence, the equilibrium is dynamic in nature.
(ii) At equilibrium all properties that are generally macroscopic such as concentration, pressure, density and colour of the system remain unchanged thereafter and stay constant.
(iii) A chemical equilibrium can be established only when none of the products is allowed to escape out
(iv) Chemical equilibrium can be approached from either direction.
(v) A catalyst affects the forward and the backward reactions equally. So, a catalyst only helps in attaining the equilibrium earlier.
For a reaction $A+B\rightleftharpoons C+D$ the equilibrium constant is given by ${{K}_{c}}=\dfrac{[C][D]}{[A][B]}$ and ${{K}_{p}}=\dfrac{(pC)(pD)}{(pA)(pB)}$, where p is partial pressure. - Now, let us come to our question. It is given that the initial pressure is given to be 0.2 atm. Now, we have: \[2HI(g)\rightleftharpoons {{H}_{2}}(g)+{{I}_{2}}(g)\]
 - At t = 0, we have the concentration of HI as 0.2 and ${{H}_{2}}$ and ${{I}_{2}}$ have concentration 0. However, when when equilibrium is reached, then at that time, the concentration of HI is $0.2 - 2x$ and ${{H}_{2}}$ and ${{I}_{2}}$ both have concentrations $x$.
- Now, it is given in the question that the value of equilibrium pressure of HI is 0.04 atm, so we can use the equation: $0.2 - 2x = 0.04\,\,\Rightarrow x = 0.08$. Now, as discussed above, we can use the formula for writing the value of ${{K}_{p}}$ for the reaction that is:
\[{{K}_{p}}=\dfrac{(p{{H}_{2}})(p{{I}_{2}})}{{{(pHI)}^{2}}}=\dfrac{0.08\times 0.08}{{{(0.04)}^{2}}}=4\]
So we get the value of ${{K}_{p}}$ as 4, So the correct answer is “C”:

Note: It is to be noted that the equilibrium constant does not depend upon initial concentration and has a definite value for every chemical reaction at a given temperature. Also, pure solids and liquids are not taken into consideration while calculating the equilibrium constant.