For the reaction. `H_2+I_2hArr2HI,K=47.6`. If the initial number of moles of each reactant product is 1 mol then at equilibrium-

Options

A. [I2]=[H2],[I2]>[HI]
B. [I2]=[H2],[I2]<[HI]
C. [I2]<[H2],[I2]=[HI]
D. [I2]>[H2],[I2]=[HI]

Explanation

`A+2B initial : 2 2 0
At equilibrium `2-x 2-x 2x
`=1.5 =1 =1
As 1 mol of C is formed at equilibrium,
`2x=1to=0.5`
`:.` At equilibrium number of moles of A = 1.5 ,
B=1 and C=1
`:.` Equilibrium concentration of A = 1.5/ 10
`K_c=([C]^2)/([A][B]^2)=([1//10]^2)/([1.5//10][1//10]^2)=6.67`