Suppose a 250.0 mL flask is filled with 1.3 mol of I2 and 1.0 mol of HI. The following reaction becomes possible:
H2 (g) +I2 (g) ⇆ 2HI (g)
The equilibrium constant for this reaction is 0.983 at the temperature of the flask.
Calculate the equilibrium molarity of HI. Round your answer to two decimal places.

2.27 M

Explanation:

Given that :

volume = 250.0 mL = 0.250 L

Number of moles of = 1.3 mol

Number of moles of = 1.0 mole

Initial concentration of   =

=

= 5.20 M

Initial concentration of   =

=

= 4.0 M

Equation of the reaction is represented as:

The I.C.E Table is as follows:

                               +             ----->      

Initial(M)                  0.0              5.20                   4.0  

Change                    +x               +x                       - 2x

Equilibrium(M)           x             5.20+x                 4 - 2x

             where K = 0.983

multiplying through by (-) and rearranging in the order of quadratic equation; we have:

using the quadratic formula:

=

=

=     OR  

= 6.133      OR      0.864

since 6.133 is greater than K value then it is void, so we go by the lesser value which is 0.864

so x = 0.864 M

The equilibrium molarity of HI = (4.0- 2x)

= 4.0 - 2(0.864)

= 4.0 - 1.728

= 2.272 M

= 2.27 M to two decimal places

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8