Home
Class 12
CHEMISTRY
If K(1) represents the equilibrium const...

If `K_(1)` represents the equilibrium constant for reaction: `H_(2)+I_(2)hArr2HI & K_(2)" for "(1)/(2)H_(2)+(1)/(2)I_(2)hArr HI` the relation between `K_(1)& K_(2)` would be

A

`K_(1)xxK_(2)=1`

B

`sqrt(K_(1))xxK_(2)=1`

C

`sqrt(K_(1))=K_(2)`

D

`sqrt(K_(1))=sqrt(K_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise ENABLE|49 Videos

  • CHEMICAL EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise EFFICIENT|50 Videos

  • CHEMICAL EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|28 Videos

  • CHEMICAL BONDING-I & II

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|98 Videos

  • CHEMICAL KINETICS

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|52 Videos

Similar Questions

Explore conceptually related problems

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

The equilibrium constant for N_(2)(g)+O_(2)(g)hArr2NO "is" K_(1) and that for NO(g)hArr(1)/(2)N_(2)(g)+(1)/(2)O_(2)(g) is K_(2). K_(1) "and" K_(2) will be related as

At 25^(@) C the equilibrium constant K,and K_(2) of two reaction are 2NH_(3)hArrN_(2)+3H_(2) ,K_(1) (1)/(2)N_(2)+(3)/(2)H_(2)hArrNH_(3),K_(2) The relation b//w two equilibrium constant is :-

The value of equilibrium constant of the reaction. HI(g)hArr(1)/(2)H_2(g)+(1)/(2)I_2(g) is 8.0 The equilibrium constant of the reaction. H_2(g)+I_2(g)hArr2HI(g) will be

Equilibrium constant for the reaction: H_(2)(g) +I_(2) (g) hArr 2HI(g) is K_(c) = 50 at 25^(@)C The standard Gibbs free enegry change for the reaction will be:

If the equilibrium constant of the reaction 2HIhArr H_(2)+I_(2) is 0.25, then the equilibrium constant for the reaction, H_(2)(g)+I_(2)(g)hArr 2HI(g) would be

K_(1) and K_(2) are equilibrium constants for reaction (i) and (ii) N_(2)(g)+O_(2)(g) hArr 2NO(g) …(i) NO(g) hArr 1//2 N_(2)(g)+1//2O_(2)(g) …(ii) then,

K_(1) and K_(2) are equilibrium constants for reaction (i) and (ii) N_(2)(g)+O_(2)(g) hArr 2NO(g) …(i) NO(g) hArr 1//2 N_(2)(g)+1//2O_(2)(g) …(ii) then,

If K_(1) and K_(2) are respective equilibrium constants for two reactions : XeF_(6)(g) +H_(2)O hArr XeOF_(4)(g) +2HF(g) XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g) Then equilibrium constant for the reaction XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)+H_(2)O(g) will be

If K_(1) and K_(2) are respective equilibrium constants for two reactions : XeF_(6)(g) +H_(2)O hArr XeOF_(4)(g) +2HF_(g) XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g) Then equilibrium constant for the reaction XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)+H_(2)O(g) will be

Home
Profile