Guest

If the pressure of N2/H2 mixture in a closed apparatus is 100atm and 20% of the mixture reacts then the pressure at the same temperature would be 1)100 2)90 3)85 4)80

If the pressure of N2/H2 mixture in a closed apparatus is 100atm and 20% of the mixture reacts then the pressure at the same temperature would be
1)100 
2)90
3)85
4)80

Grade:12th pass

6 Answers

Harshvardhan
13 Points
7 years ago
N2+3H2=2NH3; As we know N2+H2=100atm then; Partial pressure of N2 &H2 be pN2 +p3H2=P +3P=4P 4P=100P=25 as we know 20% mixture react means α=.2 Acc.to Formula, α=x÷P; x=5 So, N2+3H2=2NH3 Intially. 25 75 0 At Eq. 25-x 75-3x 2x 25-5 75-15 2(5) 20 60 10 P=20+60+10=90
Rinku
13 Points
6 years ago
As 4 moles react to form 2 moles of product therefore 20 atm pressure will react to for just 10 atm pressure thus net pressure will 90 atm. 100-20+10=90
Vinay Suresh
11 Points
6 years ago
The reaction in apparatus :N2+3H2=2NH3 (both directions) Initially assuming we got 1 mole of N2 and 3 moles of H2. Total number of moles (initially) = 4 = [n1] After equilibrium is established we have(alpha.. degree of dissociation = 20/100 =0.2)moles of N2 = 1-0.2 = 0.8moles of H2 = 3-3(0.2) = 2.4moles of NH3 = 2(0.2) = 0.4So... the total no. of moles at equilibrium = 3.6 = [n2]Ideal gas equation initial = P1V1=n1RT1 (eq. 1)final = P2V2=n2RT2 (eq. 2)(since the reaction is being carried out in a closed apparatus.. volume remains constant.. so V1=V2. And the question mentions that the temperature is constant as well so, T1=T2) So dividing eq. 1 by eq.2 P1/P2=n1/n2initial pressure P1=100atmSo, 100/P2=4/3.6P2 = 90atm.
Selim
13 Points
5 years ago
If the pressure of N2 and H2 mixture in closed aperture is hundred atmosphere and 20% of Mi charger and the pressure at the same temperature would be
Divyanshi
13 Points
3 years ago
Answer is 90atm.
 
N2.     +      3H2.         2NH3
1 mole        3 mole.                  2 mole
since D. O. D = 20% = 0.2
(1-0.2)        3-(3*0.2)              (2*0.2)
 
initial moles -- 3+1=4
Moles at equilibrium --(1-0.2)+(3-0.6)+0.4= 3.6
initial pressure -- 100atm for 4 moles ( initial moles) 
Final pressure = x atm for 3.6 moles
 
x = 3.6 *100/4 = 90
So final or total pressure is 90 atm. 
 
 
 
 
 
 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

The reaction in apparatus :
N2+3H2=2NH3 (both directions)
Initially assuming we got 1 mole of N2 and 3 moles of H2.
Total number of moles (initially) = 4 = [n1]
After equilibrium is established we have (alpha.. degree of dissociation = 20/100 =0.2) moles of N2 = 1-0.2 = 0.8moles of H2 = 3-3(0.2) = 2.4
moles of NH3 = 2(0.2) = 0.4
So... the total no. of moles at equilibrium = 3.6 = [n2]
Ideal gas equation initial = P1V1=n1RT1 (eq. 1)
final = P2V2=n2RT2 (eq. 2)
(since the reaction is being carried out in a closed apparatus.. volume remains constant.. so V1=V2.
And the question mentions that the temperature is constant as well so, T1=T2)
So dividing eq. 1 by eq.2 P1/P2=n1/n2
initial pressure P1=100atm
So, 100/P2=4/3.6
P2 = 90atm.

Thanks and Regards

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free