Question

$4.5\,moles$ each of hydrogen and iodine are heated in a sealed $10\,litre$ vessel. At equilibrium 3 moles of hydrogen iodide was found. The equilibrium constant for
${H_{2(g)}} + {I_{2(g)}} \rightleftarrows 2H{I_{(g)}}$
A. 1
B. 10
C. 5
D. 0.33

Answer 1

Hint : According to the question, if we have the quantity or the given moles of both the reactants and when equilibrium acquires, again we have also the measure of the products. Then according to the ICE table, we can conclude the Equilibrium Constant.

Complete Step By Step Answer:
Initially, we have, $4.5\,moles$ each of hydrogen and iodine which is heated in a sealed $10\,litre$ vessel.
So, the formation of hydrogen iodide are as below:
${H_{2(g)}} + {I_{2(g)}} \rightleftarrows 2H{I_{(g)}}$
Since, $1mole$ of ${H_2}$ and $1mole$ of ${I_2}$ combine to give $2moles$ of $HI$ then $3moles$ of $HI$ would be formed by ICE table,
${K_c}$ for the reaction at equilibrium or the Equilibrium Constant is:
${K_c} = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}}$
${K_c} = \dfrac{{{{[3]}^2}}}{{[3][3]}}$
${K_c} = 1$
Hence, the correct option is (A).

Additional Information:
In other way, we can also solve this by supposing the value;
Formation of Hydrogen Iodide:
${H_2} + {I_2} \to 2HI$

Initially	4.5	4.5	0
At Equilibrium	4.5-x	4.5-x	2x

Given that: $[HI] = 2x = 3mol,\,\therefore x = 1.5mol$
$[{H_2}] = 4.5 - x = 4.5 - 1.5 = 3mol$
$[{I_2}] = 4.5 - x = 4.5 - 1.5 = 3mol$
Equilibrium constant for the reaction:-
${K_c} = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}}$
${K_c} = \dfrac{{{{[3]}^2}}}{{[3][3]}}$
${K_c} = 1$ .

Note :
If the value for the equilibrium constant is small, then the equilibrium favours the reaction to the left, and there are more reactants than products. If the value of ${K_c}$ approaches zero, the reaction may be considered not to occur.