In a 1 lit. Container following equilibrium is estabilished with equal moles of `NO_(2)(g)` & `N_(2)O_(4)(g)`. `N_(2)O_(4)(g) hArr 2NO_(2)(g) hArr 2NO_(2)(g)` at equilibrium `M_(avg.)=(184)/(3)`, then ratio of `K_(c)` & total initial mole is .
A
3
B
`3//2`
C
`27//4`
D
6
Text Solution
Verified by Experts
The correct Answer is:
2
`{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g),,,),(t=0,b,,b,,,"Conservation of mass law"),(t=t_(1),b-x,,b+2x,,,92(b-x)+(b+2x)46=(184)/(3)(2b+x)),(,=3b//4,,3b//2,,,x=b//4):}` So, `K_(c)=(9b^(2)//4)/(3b//4)=3b` `(K_(C))/(2b)=(3)/(2)`
In the given reaction N_(2)(g)+O_(2)(g) hArr 2NO(g) , equilibrium means that
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,
Write the relationship between equilibrium constants for the following reactions: 2NO(g)+O_2(g) hArr 2NO_2(g) 4NO_2(g) hArr 4NO(g) + 2O_2(g)
One mole of SO_(3) was placed in a two litre vessel at a certain temperature. The following equilibrium was established in the vessel 2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g) The equilibrium mixture reacts with 0.2 mole KMnO_(4) in acidic medium. Hence, K_(c) is :
At a given temperature, the equilibrium constant for the reactions NO(g)+1//2O_(2)(g)hArr NO_(2)(g) and 2NO_(2)(g)hArr 2NO(g)+O_(2)(g) are K_(1) and K_(2) respectivel. If K_(1) is 4xx10^(-3) then K_(2) will be
The equilibrium constant for the reaction NH_(4)NO_(2)(s)hArr N_(2)(g)+2H_(2)O(g) , is given by
Write the relation between K_(p)and K_(c) for the following reactions (i) N_(2)(g)+3H_(2)(g)hArr2H_(2)(g)+O_(2)(g) (ii) 2H_(2)O(g)hArr2H_(2)(g)+O_(2)(g)
9.2gm of N_(2)O_(4) (g) is taken is a closed one litre vessel and heated till the following equilibrium is reached N_(2)O_(4) (g) harr 2 NO_(2) (g) . At equilibrium 50% of N_(2)O_(4)(g) is dissociated. What is the equilibrium constant (in mole lit^(-1) ) ?.(M.wt.of N_(2)O_(4) is 92)
