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2 moles of N2 is mixed with 6 moles of H2 ina closed vessel of 1 litre capacity .If 50%N2 is converted into NH3 at equilibrium,the value of kc for the reaction:-N2(g)+3H2(g)------>2NH3(g)

2 moles of N2 is mixed with 6 moles of H2 ina closed vessel of 1 litre capacity .If 50%N2 is converted into NH3 at equilibrium,the value of kc for the reaction:-N2(g)+3H2(g)------>2NH3(g)

Grade:12th pass

5 Answers

Arun
25750 Points
6 years ago
Dear student
 

N2 (g) + 3H2 (g) → 2 NH3(g)

Initial 1mole (given) 3 mole (given) 0

At equilibrium 1-(0.25 * 1)/100 3--(0.25 * 1)/100 2(0.25 * 1)/100

0.9975 moles 2.9925 moles 0.005 moles

 

Given that 0.25% of nitrogen is converted to ammonia (given) therefore 1-0.25% is done

Kc = [NH3]2/[N2][H2]3 = (0.005/4)2/(2.9925/4)(0.9975/4)3 = 1.8 * 10-8

 

Regards

Arun (askIITians forum expert)

Berjees shah
19 Points
6 years ago
N2 + H2 2NH3. Initial moles = 2. 6. 0At equilibrium 2(1-x). 6(1-3x). 2xKc =[1]÷[1]×[3]=1÷27 ans....
nishant kumar
36 Points
6 years ago
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nishant kumar
36 Points
6 years ago
 

N2 (g) + 3H2 (g) → 2 NH3(g)

Initial 1mole (given) 3 mole (given) 0

At equilibrium 1-(0.25 * 1)/100 3--(0.25 * 1)/100 2(0.25 * 1)/100

0.9975 moles 2.9925 moles 0.005 moles

 

Given that 0.25% of nitrogen is converted to ammonia (given) therefore 1-0.25% is done

Kc = [NH3]2/[N2][H2]3 = (0.005/4)2/(2.9925/4)(0.9975/4)3 = 1.8 * 10-8

ankit singh
askIITians Faculty 614 Points
3 years ago
1 mole of n2 reacts with 3 moles of h2
2 moles of N2 will react with 6 moles of H2
Now, 1 mole of N2 give = 2 moles of NH3
Then, 2moles of N2 give = 4 moles of NH3
But 50% of N2 (1 mole) reacts to give = 2 moles of NH3
N2 + 3H2 ---- 2NH3
Initial 2 6 0
At equilibrium 1 3 2
Kc = [ NH3]2 / [N2][H]3 = 1 /[ (1) (3)3] = 1/ 27 = 3.7 * 10-2

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