Home
Class 12
CHEMISTRY
N(2)(g) +3H(2)(g) hArr 2NH(3)(g), In ...

`N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g),`
In the reaction given above, the addition of small amount of an inert gas at constant pressure will shift the equilibrium towardss which side of

A

LHS(Left hand side)

B

RHS(Right hand side)

C

Neither side

D

Either side

Text Solution

Verified by Experts

The correct Answer is:
A
Addition of inert gas at constant pressure drives the reaction to the side with higher number of gaseous moles
Promotional Banner

Similar Questions

Explore conceptually related problems

Addition of inert gas at constant pressure in equlibrium 2SO_2(g)+O_2(g) hArr 2SO_3(g) , will

Consider the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) The equilibrium constant of the above reaction is K_(p) If pure ammonia is left to dissociated, the partial pressure of ammonia at equilibrium is given by : (Assume that P_(NH_(3)) lt lt P_(total) at equilibrium)

In the reaction N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g) , the value of the equlibrium constant depends on

On addition of an inert gas at constant volume to the reaction N_(2) + 3H_(2)hArr2NH_(3) at 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 :

For the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the units of K_(p) are …………