Home
Class 11
CHEMISTRY
If N(2)+3H(2)overset(K)(hArr)2NH(3) the...

If `N_(2)+3H_(2)overset(K)(hArr)2NH_(3)` then `2N_(2)+6H_(2)overset(K)(hArr)4NH_(3)-K'` is equal to

A

`K^(2)`

B

`(K)^(1//3)`

C

`1//sqrt(K)`

D

`1//K^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
For reaction
`N_(2)+3H_(2)overst(K)(hArr)2NH_(3), `:. K=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
Multiplying this equation with `2`, we get
`2N_(2)+6H_(2)overset(K')(hArr)4NH_(3)`
and `K'=([NH_(3)]^(4))/([N_(2)]^(2)[H_(2)]^(6))`
or `K'=(K)^(2)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Concept Applicationexercise 7.1|53 Videos

  • CHEMICAL EQUILIBRIUM

    CENGAGE CHEMISTRY|Exercise Ex 7.2|40 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    CENGAGE CHEMISTRY|Exercise Archives Subjective|15 Videos

  • CLASSIFICATION AND NOMENCLATURE OF ORGANIC COMPOUNDS

    CENGAGE CHEMISTRY|Exercise Analytical and Descriptive Type|3 Videos

Similar Questions

Explore conceptually related problems

Consider the equilibrium reactions, H_(3)PO_(4)overset(K_(1))(hArr ) H^(+)+H_(2)PO_(4)^(-) H_(3)PO_(4)^(-)overset(K_(2))(hArr ) H^(+)+HPO_(4)^(-2) HPO_(4)^(-2)overset(K_(3))(hArr)H^(+)+PO_(4)^(-3) The equilibrium constant K for the following dissociation H_(3)PO_(4)hArr 3H^(+)+PO_(4)^(-) is

The equilibrium constant for the reversible reaction N_(2)+3H_(2) hArr 2NH_(3) is K and for the reaction (1)/(2) N_(2)+(3)/(2) H_(2) hArr NH_(3) , the equilibrium constant is K',.K and K' will be related as

The stepwise formation of [Cu(NH_3)_4]^(2+) is given below Cu^(2+)+NH_3overset(K_1)hArr[Cu(NH_3)]^(2+] [Cu(NH_3)]^(2+)+NH_3overset(K_2)hArr[Cu(NH_3)_2]^(2+) [Cu(NH_3)_2]^(2+)+NH_3overset(K_3)hArr[Cu(NH_3)_3]^(2+) [Cu(NH_3)_3]^(2+)+NH_3overset(K_4)hArr[Cu(NH_3)_4]^(2+) The value of stability constants K_1,K_2,K_3 and K_4 are 10^4,1.58xx10^3,5xx10^2 and 0^2 respectively. The overall equilibrium constant for dissociation of [Cu(NH_3)_4]^(2+)" is ""x"xx10^(-12) . The value of x is "______" . (Rounded off to the nearest integer)

If the value of equilibrium constant K_(c) for the reaction, N_(2)+3H_(2)hArr2NH_(3) is 7. The equilibrium constant for the reaction 2N_(2)+6H_(2)hArr4NH_(3) will be

One "mole" of N_(2) is mixed with three moles of H_(2) in a 4L vessel. If 0.25% N_(2) is coverted into NH_(3) by the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , calculate K_(c) . Also report K_(c) for 1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)

For the reaction N_(2)+3H_(2)hArr2NH_(3) and (1)/(2)N_(2)+(3)/(2)H_(2)hArrNH_(3) write down the expression for equilibrium constants K_(c ) and K_(c )^(') . How is K_(c ) related to K_(c )^(') ?