Question

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6. The ammonia in equilibrium with a 1 : 3 N2-H2 mixture at 20 atm and 427°C
amounts to 16%. Calculate Kp and Kc for
1/2N2+3/2H2= NH3
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Answer 1

Answer:

hey mate

Explanation:

The ammonia in equilibrium with 1:3 N2-H2 mixture at 20atm and 427 degree celsius amounts to 16%. Calculate Kp and Kc for 1/2 N2 + 1/2 H2 gives NH3.

3 years ago

Answers : (1)

For,

1/2 N2 + 1/2 H2 --------> NH3

at eqm.=> x moles 3x moles 2x moles

Now given, (2x) = 0.16

solving for x we get x = 0.08 .

Kc = (2x)/root3x => 2/root3

Kp = Kc*(RT)^n

n = 1 – (0.5 + 0.5) = 0

Hence Kp = Kc =2/root3...hope uhh like it dear

Answer 2

Answer:

I will give u hint...

Explanation:

For,

N2 + 3H2 --------> 2NH3

Initial moles. 1 moles. 3 moles. o moles

at eqm.=> 1-x moles 3-3x moles 2x moles

Now given, (2x) = 0.16 therefore x = 0.08 .

at equilibrium [N2] = 0.92

[H2] = 2.76

[NH3] = 0.16

Kc = [NH3]²/[N2][H2]³

Kp = Kc*(RT)^∆n

∆n = 2 – 4 = -2

put the value n find the answer......