Home
Class 11
CHEMISTRY
Starting from ‘a’ moles of H2 and ...

Starting from ‘a’ moles of `H_2` and ‘b’ moles of `I_2` an equilibrium `H_2+I_2

A

`(4x^2)/(ab)`

B

`(4x^2)/((a-x)(b-x))`

C

`(2x)/((a-x)(b-x))`

D

`(4x^2)/((a-2x)(b-2x))`

Text Solution

Verified by Experts

The correct Answer is:
B

`K_(C)=([HI]^(2))/([H_(2)][I_(2)])=((2x)^(2))/((a-x)(b-x))`
Promotional Banner

Similar Questions

Explore conceptually related problems

A mixture of 2 moles of N_(2) and 8 moles of H_(2) are heated in a 2 lit vessel, till equilibrium is established. At equilibrium, 04 moles of N_(2) was present. The equilibrium concentration of H_(2) will be

I_(2(aq)) + I_((aq))^(-) hArr I_(3(aq))^(-). We started with I mole of I_(2) and 0.5 mole of l^(-) in one litre flask. After equilibrium is reached, excess of AgNO_(2) gave 0.25 mole of yellow precipitate. Equilibrium constant is

4.5 mol each of H_(2) and I_(2) are present in a 10 lit vessel . At equilibrium the mixture contains 3 mole . HI , Kp for H_(2) + I_(2) hArr 2 HI and mass of I_(2) at equilibrium (in g) respectively are

Consider the following reaction equilibrium N_(2)(g)+ 3H_(2)(g) hArr 2NH_(3) Initially, 1 mole of N_(2) and 3 moles of H_(2) are taken in a 2 2L flask. At equilibrium state, if the number of moles of N_(2) is 0.6, what is the total number of moles of all gases present in the flask?

When H_2 and I_2 are mixed and equilibrium is attained, then

Starting from one mole each of anol and acetic acid, at equilibrium 2//3 mole of ester is formed . Calculate the equilibrium constant.

x mole HI is taken in 10 lit vessel and heated . If the equilibrium concentration of H_(2 (g)) is 0.1 mol - lit^(-1) , value of x and mass of I_(2) at equilibrium respectively are

A gaseous phase reaction taking place in 1 lit at 400K is given N_(2) (g) + 3H_(2) (g) hArr 2NH_(3) (g) starting with 1 mole N_(2) and 3 moles H_(2) equilibrium mixture required 250 ml of 1 M H_(2) SO_(4) . Thus , Ke is

For the reaction A+ 2B to C: 5 moles of A and 8 moles of B will produce