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Question:

For the reaction, H2 + I2 ⇌ 2HI, K = 47.6. If the initial number of moles of each reactant and product is 1 mole, then at equilibrium:

[I2]=[H2],[I2]>[HI]

[I2]>[H2],[I2]=[HI]

[I2]=[H2],[I2]<[HI]

[I2]<[H2],[I2]=[HI]

Solution:

For the given reaction ,K=[HI]2/[H2][I2]
As one mole of H2 reacts with 1 mole of I2, even at equilibrium ,[H2]=[I2]
Hence,K=[HI]2/[I2]2
or √K=[HI]/[I2]=√47.6
i.e ,[HI]>[I2]
Hence, option B is correct.