Home
Class 11
CHEMISTRY
The rate of the elementary reaction H(...

The rate of the elementary reaction
`H_(2)(g) +I_(2)(g) hArr 2HI(g) `at `25^(@)C` is given by `:`
Rate `=1.7xx10^(-18) [H_(2)][I_(2)]`
The rate of decomposition of gaseous HI to `H_(2)(g)` and `I_(2)(g)` at `25^(@)C` is Rate`=2.4xx10^(-21)[HI]^(2)`
Equilibrium constant for the formation of one mole of gaseous HI from the `H_(2)(g)` and `I_(2)(g)` is `:`

Text Solution

Verified by Experts

The correct Answer is:
`26.6`

` (1)/(2) H_(2) (g) + (1)/(2) I_(2) (g) hArr HI(g)`
`K_(c) = ([HI])/([H_(2)]^((1)/(2)) [I_(2)]^((1)/(2)))`
For the reaction
`H_(2) (g) + I_(2) (g) hArr 2HI(g)`
`R_(f) = 1.7 xx 10 ^(-18) [H_(2) ] [I_(2)]`
`R_(b) = 2.4 xx 10 ^(-21) [HI]^(2)`
At eqm. `R_(f) = R_(b)`
`1.7 xx 10 ^(-18) [H_(2) ] [I_(2)] = 2. 4 xx 10 ^(-2) [HI]^(2)`
`(1. 7 xx 10 ^(-18))/( 2 . 4 xx 10 ^(-21)) = ([HI]^(2))/([H_(2)][I_(2)])`
`(170)/(24) xx 10 ^(2) = K_(c)^(2)`
`K_(c) = sqrt((170)/(24)) xx 10 = sqrt(7.08) xx 10`
`= 2.66 xx 10 = 26.6`
Promotional Banner

Similar Questions

Explore conceptually related problems

For reaction H_(2)(g) +I_(2)(g) hArr 2HI (g) The value of K_(p) changes with

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

For the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) , the rate of reaction is expressed as

For the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) The equilibrium constant K_(p) changes with

If DeltaG^(@)[HI(g)=-1.7kJ] , the equilibrium constant for the reaction 2HI(g)hArr H_(2)(g)+I_(2)(g) at 25^(@)C is

For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g) the equilibrium constant K_(p) changes with

In the reversible reaction, 2HI(g) hArr H_(2)(g)+I_(2)(g), K_(p) is