Question

At a certain temperature, a $10L$ vessel contains $0.4moles$ of ${H_2}$ . $0.4moles$ of ${I_2}$ and $0.1moles$ of $HI$At equilibrium.
Then ${K_p}$ for
${H_2} + {I_2} \rightleftharpoons 2HI$ is:
A. $16$
B. $0.0625$
C. $4$
D. Data insufficient

Answer 1

Hint:
The equilibrium constant is equal to the ratio of partial pressure of the product to the product of partial pressure of reactant. Partial pressure of any component is also directly proportional to the number of moles.Relationship given by the equilibrium constant and number of moles can be used.
Formula used:
${K_p} = \dfrac{{{{({n_{HI}})}^2}}}{{{n_{{H_2}}} \times {n_{{I_2}}}}}$
${n_{HI}}$ is the number of moles for hydrogen iodide
 ${n_{{H_2}}}$ is the number of moles for dihydrogen
${n_{{I_2}}}$ is the number of moles for diiodine
${K_p}$ is the equilibrium constant

Complete step by step answer:
Let us understand what partial pressure would mean for our equation:
${H_2} + {I_2} \rightleftharpoons 2HI$
We know that there is an equilibrium here,so
${K_p} = \dfrac{{{{({p_{HI}})}^2}}}{{{p_{{H_2}}} \times {p_{{I_2}}}}}$
Where,
${K_p}$ is the equilibrium constant
${p_{HI}}$ is the partial pressure for hydrogen iodide
 ${p_{{H_2}}}$is the partial pressure for dihydrogen
${p_{{I_2}}}$ is the partial pressure for diiodine
Now, we also know than
Partial pressure is directly proportional to the number of moles.
Hence, if I substitute this value in the above equation, I get:
${K_p} = \dfrac{{{{({n_{HI}})}^2}}}{{{n_{{H_2}}} \times {n_{{I_2}}}}}$
${n_{HI}}$ is the number of moles for hydrogen iodide
 ${n_{{H_2}}}$is the number of moles for dihydrogen
${n_{{I_2}}}$ is the number of moles for diiodine

As it has been given in the question:
${n_{HI}} = 0.1$ ,${n_{{H_2}}} = 0.4$, ${n_{{I_2}}} = 0.4$
Substituting these values in the above equation, we get
${K_p} = \dfrac{{{{(0.1)}^2}}}{{0.4 \times 0.4}}$
Solving this, we get
${K_p} = 0.0625$
Hence option B is correct.

Note:The equilibrium constant tells us about the relationship between the reactant and the products when they attain equilibrium. For an ideal solution, we can also use concentration instead of moles, but when we consider the solution to be real we have to take into consideration the activities of the components and not just the concentration.