Consider the gas reaction: h2 (g)2(g) 2hi (g) the equilibrium constant at 731 k is 50.3. equal amounts of all three gases (0.100 m) are introduced in a container, calculate the concentration of each gas after the system reaches equilibrium. express your results with the right number of significant figures.

Answer : The concentration of and at equilibrium 0.033 M, 0.033 M and 0.234 M respectively.

Solution :  Given,

Concentration = 0.100 M

The given equilibrium reaction is,

Initially conc.       0.100     0.100       0.100

At equilibrium  (0.100-x)   (0.100-x)  (0.100+2x)

The expression of will be,

Now put all the given values in this expression, we get:

By solving the term x, we get

From the values of 'x' we conclude that, x = 0.158 can not more than initial concentration. So, the value of 'x' which is equal to 0.158 is not consider.

Thus, the concentration of and at equilibrium = (0.100-x) = 0.100 - 0.067 = 0.033 M

The concentration of at equilibrium = (0.100+2x) = 0.100 + 2(0.067) = 0.234 M

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).