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![Equilibriurn Constant H2(g) + 12(g)-→ 2HI(g) 0%][12] Equilibrium achieved [HI] In the equilibrium region [H2] [I21 2 [HI]= 55.3 = K [H2 10I2] K equilibrium constan , [HI] Reactants proceeding toward equilibrium](https://d2vlcm61l7u1fs.cloudfront.net/media%2F17e%2F17e97c7f-0505-46f3-99c6-81d4139b0dd6%2Fimage)
Please help…I need step by step demonstration of how K=55.3 came about. I understand some stoichiometry was involved in achieving that constant. However, I am totally lost.
Equilibriurn Constant H2(g) + 12(g)-→ 2HI(g) 0%][12] Equilibrium achieved [HI] In the equilibrium region [H2] [I21 2 [HI]= 55.3 = K [H2 10I2] K equilibrium constan , [HI] Reactants proceeding toward equilibrium
Expert Answer
Ans. H2(g) + I2(g) <——-> 2HI(g)
Equilibrium constant in terms of concentration, Kc = [HI]2 / ([H2] [I2]) –
all concentrations are at equilibrium.
Also,
Equilibrium constant in terms of partial pressure, Kp = (pHI)2 / [(pH2) (pI2) – all partial pressures are at equilibrium.
In the picture, the region of parallel running green and red lines indicate ‘equilibrium’.
Note that expression of equilibrium constant as written above- we must have the values of concentration or partial pressure of the respective chemical species.
The picture does not show any concentration values. So, the value of K shown in picture is NOT based on the picture. It has been calculated elsewhere.
# To brief the calculation, let the equilibrium concentration of the chemical species are as follow-
[H2] = 6.9 x 10-5 M ; [I2] = 1.00 x 10-3 M ; 1.91 x 10-3 M
Putting the values in above equation we get,
K = (1.91 x 10-3)2 / [(6.9 x 10-5) (1.00 x 10-3)] = 52.87 = 53
Again, note that K is not calculated on basis of the picture shown. It’s a pre-calculated mentioned to demonstrate the look of reactant and product curves at equilibrium.
