Home
Class 11
CHEMISTRY
13.8 g of N(2)O(4) was placed in 1 L rea...

13.8 g of `N_(2)O_(4)` was placed in 1 L reaction vessel at 400K and allowed to attain equilibrium :`N_(2)O_(4) (g) hArr 2NO_(2)(g).`
the total pressure at equilibrium was found to be 9.15 bar. Calculate `K_(c),K_(p)` and partial pressure at equilibrium .

Text Solution

Verified by Experts

`13*8 "g" N_(2)O_(4) = (13*8)/92" mol"=0*15" mol" "` `13*8 N_(2)O_(4) = (13*8)/92" mol"=0*15" mol " "(Molar mass of "N_(2)O_(4) = 92 g " mol"^(-1))`
PV= nRT
` :. P xx 1 L = 0* 15 mol xx 0* 083 " bar"L mol^(-1) K^(-1) xx 400 K or P = 4*98 " bar"`
` {:(,N_(2)O_(4)(g),hArr,2NO_(2)),("Initial pressures",4*98 "bar",,0),(" At equilibrium",(4*98-p),,2p):}`
` :. 4*98 - p+ 2 p = 9*15 " bar" or p= 4*17 " bar"`
`:. (p_(N_(2)O_(4)))eq = 4* 98 - 4*17 = 0*81 "bar", (p_(NO_(2)))eq = 2 xx 4*17 = 8*34 "bar"`
`K_(p) = p_(NO_(2))^(2)//_(P_(N_(2)O_(4)))=(8*34)^(2)//)*81=85*87,`
` K_(p) = K_(c) (RT)^(Delta n) :. 85*87 = K_(c) (0*083 xx400)^(1) or K_(c)= 2*586 = 2*6`
Promotional Banner

Topper's Solved these Questions

  • EQUILIBRIUM IN PHYSICAL AND CHEMICAL PROCESSES

    PRADEEP|Exercise PROBLEMS FOR PRACTICE|19 Videos

  • EQUILIBRIUM IN PHYSICAL AND CHEMICAL PROCESSES

    PRADEEP|Exercise PROBLEM FOR PRACTICE|1 Videos

  • EQUILIBRIUM

    PRADEEP|Exercise Competition Focus (VIII. Assertion-Reason Type Questions)|16 Videos

  • HYDROCARBONS

    PRADEEP|Exercise Competition Focus (JEE(main and advanced)/Medical Entrance) VIII. ASSERTION - REASON TYPE QUESTIONS|31 Videos

Similar Questions

Explore conceptually related problems

13.8 g of N_(2)O_(4) was placed in a 1L reaction vessel at 400 K and allowed to attain equilibrium N_(2)O_(4) (g) hArr 2NO_(2)(g) The total pressuers at equilibrium was found to be 9.15 bar. Calculate K_(c), K_(p) and partial pressure at equilibrium.

The degree of dissociation of N_(2)O_(4)(1) obeying the equilibrium, N_(2)O_(4) (g) hArr 2NO_(2)(g) , is approximately related to the pressure at equilibrium by :

18.4 g of N_(2)O_(4) is taken in a 1 L closed vessel and heated till the equilibrium is reached. N_(2)O_(4(g))rArr2NO_(2(g)) At equilibrium it is found that 50% of N_(2)O_(4) is dissociated . What will be the value of equilibrium constant?

For the following equilibrium N_(2)O_(4)(g)hArr 2NO_(2)(g) K_(p) is found to be equal to K_(c) . This is attained when :

The degree of dissociation 'alpha' of the reaction N_(2)O_(4)(g)hArr 2NO_(2)(g) Can be related of K_(p) as [Given : Total pressure at equilibrium = P]

For the reaction, N_(2)O_(4)(g)hArr 2NO_(2)(g) the degree of dissociation at equilibrium is 0.14 at a pressure of 1 atm. The value of K_(p) is

For the reaction N_(2)O_(4)(g) hArr 2NO_(2)(g) the degree of dissociation at equilibrium is 0.2 at 1 atmospheric pressure. The equilibrium constant K_(p) will be

For the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant K_(p) will be

The temperature at which K_(c) and K_(p) will have the same value for the equilibrium , N_(2) O_(4) (g) hArr 2 NO_(2) (g) is

For the following gases equilibrium, N_(2)O_(4) (g)hArr2N_(2) (g) , K_(p) is found to be equal to K_(c) . This is attained when: