Home
Class 12
CHEMISTRY
In the reaction, H(2)(g)+I(2)(g)hArr2H...

In the reaction,
`H_(2)(g)+I_(2)(g)hArr2HI(g)`
The concentration of `H_(2), I_(2)`, and `HI` at equilibrium are `8.0, 3.0` and `28.0` mol per `L` respectively. Determine the equilibrium constant.

Text Solution

Verified by Experts

Applying law of mass action, `K_(c)=([HI]^(2))/([H_(2)][I_(2)])=((28)^(2))/(8xx3)=32.6`
Promotional Banner

Similar Questions

Explore conceptually related problems

At 783 K in the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) , the molar concentrations ("mol"^(-1))"of "H_(2),I_(2)and HI at some instant of time are 0.1, 0.2 and 0.4 , respectively. If the equilibrium constant is 46 at the same temperature, then as the reaction proceeds

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

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

The equilibrium constant for the reaction H_(2)(g)+I_(2)(g)hArr 2HI(g) is 32 at a given temperature. The equilibrium concentration of I_(2) and HI are 0.5xx10^(-3) and 8xx10^(-3)M respectively. The equilibrium concentration of H_(2) is

For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g) , the equilibrium constant will change with

In a reaction between H_(2) and I_(2) at a certain temperature, the amounts of H_(2), I_(2) and HI at equilibrium were found to be 0.45 mol, 0.39 mol, and 3.0 mol respectively. Calculate the equilibrium constant for the reaction at the given temperature.

The equilibrium constant K_c for the reaction H_2(g)+I_2(g)hArr 2HI(g) is 54.3 at 400^@C . If the initial concentrations of H_2 , I_2 and HI are 0.00623M , 0.00414M , and 0.0224M , respectively, calculate the concentrations of these species at equilibrium.