Home
Class 12
CHEMISTRY
Dissociation equilibrium constant of HI ...

Dissociation equilibrium constant of HI is `2.06xx10^(-2)` at `458^(@)C`. At equilibrium, concentrations of HI and `I_(2)` are `0.36M` and `0.15M` respectively. What is the equilibrium concentration of `H_(2)` at `458^(@)C`.

Text Solution

Verified by Experts

The correct Answer is:
`[H_(2)]=1.78xx10^(-2)M`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM - II

    NCERT TAMIL|Exercise SELF EVALUATION (C. Answer not exceeding 60 words)|3 Videos

  • CHEMICAL KINETICS

    NCERT TAMIL|Exercise EXAMPLE|10 Videos

  • CHEMICAL KINETICS-II

    NCERT TAMIL|Exercise SELF EVALUATION ((D) Solve the problems :)|7 Videos

Similar Questions

Explore conceptually related problems

For the equilibrium 2SO_(3(g))hArrSO_(2(g))+O_(2(g)) , the value of equilibrium constant is 4.8xx10^(-3) at 700^(@)C . At equilibrium, if the concentration of SO_(3) and SO_(2) are 0.60M and 0.15M respectively. Calculate the concentration of O_(2) in the equillibrium mixture.

An equilibrium constant of 3.2xx10^(-6) for a reaction means, the equilibrium is

In the equilibrium, 2A(g)hArr2B(g)+C_(2)(g) the equilibrium concentrations of A, B and C_(2) at 400 K are 1xx10^(-4)M,2.0xx10^(-3)M, 1.5xx10^(-4)M respectively. The value of K_(C) for the equilibrium at 400 K is