The value of k p for the reaction at 27ºC
Br 2 ( λ ) + Cl 2 (g) 
2BrCl(g)
is ‘1 atm’. At equilibrium in a closed container partial pressure of BrCl gas is 0.1 atm and at this temperature the vapour pressure of Br 2 ( λ ) is also 0.1 atm. Then what will be minimum moles of Br 2 ( λ ) to be added to 1 mole of Cl 2 , initially, to get above equilibrium situation :
moles
moles
moles
Text Solution
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Br 2 ( λ ) + Cl 2 (g) 
2BrCl(g)
t = 0 1 0
(1 – x) 2x
k p = 
= 1 so, 
= 
= 0.01 atm
then at equilibrium, 
= 
= 10 = 
So, 10 – 10x = 2x or x = 
= 
moles
Moles of Br 2 ( λ ) required for maintaining vapour pressure of 0.1 atm
= 2 × 
moles = 
moles = moles of BrCl(g).
Moles required for taking part in reaction = moles of Cl 2 used up = 
moles.
Hence total moles required = 
= 
moles.
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