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Standard XII

Chemistry

Rate of Reaction

Kp for the re...

Question

Kp for the reaction NH4I(s)---->NH3(g)+HI(g) is 1/4 at a particular temperature. Above equilibrium is established by taking 4 moles of NH4I(s) in 100 litre container, then moles of NH4I(s) left in the container at equilibrium is

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Solution

Dear Student,

The relation of Kp and Kc is:
Kp = Kc(RT)Δn
Δn = 2
R = 0.082
T = 298 K
1/4 = Kc (0.082298)2
Kc = 0.00042
For reaction,

[NH4I] [NH3] [HI]

Initial 0.04 0 0

Change -x +x +x

Equilibirum 0.04-x x x

Kc = [NH3][HI]/[NH4I]
0.00042 = x2/(0.04-x)
0.0000168 - 0.00042x = x2
x2 + 0.00042x - 0.0000168 = 0
x = 0.004
At equilibrium, concentration of NH4I left = 0.04-0.004 = 0.036 M
Moles of NH4I left = concentrationvolume = 0.036*100 = 3.6 moles
3.6 moles of NH4I are left.

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