Question

The equilibrium constant for the reaction, $H_2+I_2\rightharpoonup 2HI$ at 700 K is 56. If 0.5 mole of hydrogen and 1 mole of iodine are added to the system at equilibrium, the value of the equilibrium constant will be ________.

• A. 25
• B. 100
• C. 56
• D. 0.25

Answer 1

Answer: (A)

25

$H_2+I_2⟶2HI$

At t=0 :- $a$ $a$ $0$

At equilibrium:- (a-x) (a-x) 2x

Equilibrium constant(k)=$\frac{{2x}^2}{(a-x{)}^2}$

$\therefore 56=\frac{{4x}^2}{a^2-2x+x^2}$

$\therefore 56a^2-112x+56x^2-4x^2=0$

$\therefore 52x^2-112x+56a^2=0$

$\therefore 13x^2-28x+14a^2=0$

Assuming, a=1

$\therefore 13x^2-28x+14=0$

$\therefore x=0.789$

$\therefore$ If 0.5 mole of Hydrogen of 1 mole of Iodine is added

at equilibrium then,

$k^{{}^{\prime }}=\frac{2x}{(a-x+0.5)(a-x+1)}$

$k^{{}^{\prime }}=\frac{2\times 0.789}{(1-x+0.5)(-x)}$

$k^{{}^{\prime }}=\frac{2\times 0.789}{(1-0.789+0.5)(2-0.789)}$

$k^{{}^{\prime }}$=25

The correct answer is $k^{{}^{\prime }}$=25