2.00 mol of H2(g) and 1.00 mol of I2(g) are placed in a 1.00 L container, and they react to form HI(g). At equilibrium, it is found that 1.80 moles of HI(g) are present in the container. Calculate K for the reaction: H2(g) + I2(g) ⇄ 2 HI(g)

The value of equilibrium constant is 29.45.

Explanation:

Moles of hydrogen gas = 2.00 mol

Concentration of  hydrogen gas =

Moles of iodine gas = 1.00 mol

Concentration of  iodine gas =

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 =

The expression of an equilibrium constant is given by ;

Putting x equal to 1.80 M.

The value of equilibrium constant is 29.45.

the correct answer to this question is 2