Chemistry - Chemical Equilibrium Question with Solution | TestHub
ChemistryChemical EquilibriumProblems involving KcMedium2 minQB
ChemistryMediumnumerical
When 2.0 moles of H2(g) and 6.0 moles of I2 vapours are allowed to react, 'x' moles of HI(g) is formed at equilibrium. Addition of a further 4.0 moles of H2(g) gave an additional 'x' moles of HI(g) at the new equilibrium. The value of 'x' is
Answer:
3.00
Solution:
First equilibrium: HI = x, H2 = 2 - x/2, I2 = 6 - x/2 K = x2/[(2-x/2)(6-x/2)] = 4x2/[(4-x)(12-x)]
Second equilibrium: HI = 2x, H2 = 6 - x, I2 = 6 - x
K = (2x)2/(6-x)2 = 4x2/(6-x)2
Equating: (4-x)(12-x) = (6-x)2
48 - 16x + x2 = 36 - 12x + x2, so 4x = 12
x = 3
Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kc
⏱ 2mℹ️ Source: QB
Doubts & Discussion
Loading discussions...
