Question

At certain temperature and total pressure of 1 atm, equilibrium mixture contains 20% by volume of a atoms a2-----> 2a the kp is

Answer 1

Answer:

Total Pressure, P = 1 atm  

It is given that there is 20% by volume of “a” atoms, therefore “a2” atoms will 80% by volume.

Now,  

The partial pressure of “a”,

Pa = 20/100 * 1 = 0.2 atm = 0.2/101325 Pa = 20265 Pa [∵ 1atm = 101325 Pa]

And,

The partial pressure of “a2”,

Pa2 = 80/100 * 1 = 0.8 atm =  0.8/101325 Pa = 81060 Pa

Thus, for the given reaction in the question, the value for kp is given by

kp

= [a]² / [a2]

= [20265]² / [81060]

= [5066.25/10³] * (10³)  ......[dividing and multiplying by 10³]

= 5.066 * 10³

≈ 5 * 10³