Answer:
The value of equilibrium constant is 29.45.
Explanation:
Moles of hydrogen gas = 2.00 mol
Concentration of hydrogen gas =[tex][H_2]= \frac{2.00 mol}{1.00 L}=2.00 M[/tex]
Moles of iodine gas = 1.00 mol
Concentration of iodine gas =[tex][I_2]= \frac{I.00 mol}{1.00 L}=1.00 M[/tex]
[tex]H2(g) + I2(g)\rightleftharpoons 2 HI(g)[/tex]
initially
2.00 M 1.00 M 1.00 M
At equilibrium:
(2.00-x/2) (1.00-x/2) x
Moles of HI at equilibrium = 1.80 M
Concentration of HI at equilibrium =[tex][HI]=\frac{1.80 mol}{1.00L} = 1.80M= x[/tex]
The expression of an equilibrium constant is given by ;
[tex]K_c=\frac{[HI]^2}{[H_2][I_2]}[/tex]
[tex]K_c=\frac{x^2}{(2.00-\frac{x}{2})(1.00-\frac{x}{2})}[/tex]
Putting x equal to 1.80 M.
[tex]K_c=\frac{(1.80)^2}{(2.00-\frac{1.80}{2})(1.00-\frac{1.80}{2})}=29.45[/tex]
The value of equilibrium constant is 29.45.
