1 mol `N_2` and 3 mol `H_2` are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the

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Q 1118112909.     1 mol `N_2` and 3 mol `H_2` are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature

when the following equilibrium is attained.

`N_(2(g)) + 3 H_(2(g)) ⇌ 2NH_(3(g))`. The equilibrium constant `K_P` for dissociation of `NH_3` is:
A

`1/0.5 × (1.5)^3 atm^-2`

B

`(0.5×(1.5)^3)/(3×3) atm^2`

C

`(3×3)/(0.5×(1.5)^3) atm^-2`

D

`0.5 × (1.5)^3 atm^2`

HINT

Use formula: `n_2/n_1=P_2/P_1`

Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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