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For the reversible reaction, N(2)(g) +...

For the reversible reaction,
`N_(2)(g) + 3H_(2)(g)hArr2NH_(3)(g)` + heat
The equilibrium shifts in forward direction

A

by increasing the concentration of `NH_3(g)`

B

by decreasing the pressure

C

by decreasing the concentrations of `N_2(g)` and `H_2(g)`

D

by increasing pressure and decreasing temperature

Text Solution

Verified by Experts

The correct Answer is:
D

Any change in the concentration, pressure and temperature of the reaction results in change in the direction of equilibrium This change in the direction of equilibrium is governed by le-chateller's principle. According to this equilibrium shifts in the opposite direction to undo the change.
`N_2(g)+3H_2(g)hArr2NH_3(g)+Heat`
(a) Increasing the concentration of `NH_3(g)`: On increasing the concentration of `NH_3(g)` , the equilibrium shifts in the backward direction where concentration of `NH_3(g)` decreases.
(b) Decreasing the pressure: Since `p prop n` (numgber of moles) , therefore equilibrium shifts in the backward direction where number of moles are increasing.
(c) Decreasing the conentraion of `N_2(g)` and `H_2(g)` : Equilibrium shifts in the backward direction when concentration of `N_2(g) and H_(2) (g)` decreases.
(d) Increasing pressure and decreasing temperature : On increasing pressure, equilibrium shits int he forward direction where number of moles decreases. it is an eaxample of exothermic reaction therefore decreasing temperature favours the forward directions.
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