1.

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-

Answer»

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

Solution :`A+2B initial : 220
At equilibrium`2-x2-x2x
`=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`


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