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In the reaction between the hydrogen and iodine, 6.34 moles of hydrogen and 4.02 moles of iodine are found to be present equilibrium with 42.85 moles of hydrogen iodine at 35c. Calculate the equilibrium constant?
(A) 72.04
(B) 65.38
(C) 52.08
(D) 45.20


Answer
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Hint: We should know the chemical equation of the reaction between hydrogen and iodine. The relationship between the amount of reactant and product when it is at equilibrium is equal to the equilibrium constant.

Complete step by step answer:
The reaction between the hydrogen and iodine is given by:
H2+I22HI
Number of moles in equilibrium: 6.34 4.02 42.85

We need to calculate molar concentration,
Molar concentration=No.of.molesvolume
Molar concentration is given by the ratio of moles of a solute and liters in volume
Let the volume of H2, I2 and HI be V,
Molar concentration ofHI=42.85V
Molar concentration ofI2=4.02V
Molar concentration ofH2=6.34V
The formula to calculate equilibrium constant:

kc=[product][reactant]
Equilibrium constant is the ratio between molar concentration of product and molar concentration of reactant.
kc=[HI]2[H2][I2]
kc=[42.85V]2[6.34V][4.02V]
kc=[42.85]225.4868
kc=72.04

Thus, the correct option is A

Note: Firstly we should be thorough with formula, kc=[product][reactant] Here, kc=[HI]2[H2][I2]
We should make sure that we don't forget the power of the product in this case while calculating. If we forget squares, that will result in wrong answers.