IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

sucess_seeker

At 550 degree celcius, 10 gm mole of HI are produced by the reaction of 15 gm mole of hydrogen and 5.2 gm mole of iodine vapour. Find the equilibrium constant

nick

i am not good in equi. but still is the ans. 3.87

shikharbaid

is it 1.28???

nunoxic

10 gm mole of HI are produced by the reaction of 15 gm mole of hydrogen and 5.2 gm mole of iodine vapour. Find the equilibrium constant

H2 + I2 >>>>>> 2 HI
K = [HI]^2 / [H2] [I2]
= 100 / 15 * 5.2
= 100 / 78
= 1.28

Please Rate If Correct

smriti.mathur

H₂ + I₂

2HI

a b

(a-x) (b-x) 2x

K = [HI]²

______

[H₂ ] [I₂ ]

2x² 4x²

________ = _________

(a-x) (b-x) (a-x) (b-x)

2x =10 (given)

x = 5

a = 15 (given)

b = 5.2 (given)

Therefore ,

K = 4 * 5²

___________

(15 - 5) (5.2 - 5)

K = 50

thevyzz

smriti's soln is correct

@nunoxic , u should see the final concentration of products/reactants while determining eq . constant

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