Home
Class 11
CHEMISTRY
A mixture of N(2)and H(2) in the molare ...

A mixture of `N_(2)and H_(2)` in the molare ratio `1:3` attains equlibrium when `50%` of mixture has reacted. If P is the total pressue of the mixture, the partial pressure of `NH_(3)` formed is `P//y.` The value of y is

Text Solution

Verified by Experts

The correct Answer is:
3
`{:(,N_(2) + 3H_(2) hArr 2NH_(3)),("Initial mole", " 1 3 0"),("At eq.","1 - x 3 - 3x 2 x "):}`
`:.` Out of 4 mole of mixture, 2 mole have reacted and hence 2 mole are left
`:.` 1 - x + 3 - 3x = 2 or 4x = 2 or x = 0.5
`:.` Total mole at equilibrium = 1 - x + 3 - 3 x + 2 x = 4 - 2 x = 4 - 2 0.5 = 3
No. of mole of ` NH_(3)` at equilibrium
`= 2 x = 2 0.5 = 1`
`:. P_(NH_(3)) = (1)/(3) xx P = (P)/(3) :. y = 3 `
Promotional Banner

Similar Questions

Explore conceptually related problems

A mixture of nitrogen and hydrogen in the ratio of 1: 3 reach equilibrium with ammonia, when 50 % of the mixture has reacted. If the total pressure is P , the partial pressure of ammonia in the equilibrium mixture was :

In a mixture of N_(2) and O_(2) gases, the mole fraction of N_(2) is found to be 0.700. the total pressure of the mixture is 1.42atm. What is the partial pressure of O_(2) in the mixture?

A mixture of nitrogen and hydrogen are initially in the molar ratio of 1:3 related equilibrium to form ammonia when 25% of the material had reacted. If the total pressure of the system is 28 atm , calculate the partial pressure of ammonia at the equilibrium.

N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) For the reaction intially the mole ratio was 1:3 of N_(2):H_(2) .At equilibrium 50% of each has reacted .If the equilibrium pressure is P, the parial pressure of NH_(3) at equilibrium is :

If a mixture of CO and N_(2) in equal amount have total 1 atm pressure, then partial pressure of N_(2) in the mixture is

A mixture of N_2 and Ar gases in a cylinder contains 7g of N_2 & 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27bar, the partial pressure of N_2 is: