Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

page 38 q10 in chemicaleq module i did not get

page 38 q10  in chemicaleq module i did not get

Grade:11

1 Answers

Naveen Kumar
askIITians Faculty 60 Points
7 years ago
The reaction is: 2NH3(g).................= N2(g).........+3H2(g)
the total pressure of the system=10atm
Temperature=700K
and the equilibrium mol% of NH3=Ne=10%
We have to find Kp(?)

Solution: As all the species present here is gas and we know that for same temperature and pressure,the pressure is directely proportional to the mole of the particular gas.
So we can say that the pressure % of NH3=Ne=10%
Now as the total pressure =10 atm, so individual prtessure of NH3=Ne=10atm*10%=1atm
So the pressure due to N2 and H2 at equilibrium=total pressure-(pressure due to NH3+pressure due to Ne)
=10-1-1=8atm
But the moles of N2 and H2 formed would be in the ratio of 1:3 (see the above equation). So the pressure due to them would be also in the same ratio.
Now let the pressure due to N2=p, hence due to H2=3p
and so applyng the Dalton’s law of partial pressure again, we would have:
1atm(of NH3)+1atm(of Ne) +p (of N2)+ 3p(of H2)=10 atm
4p=8atm
p=2atm
hence, 3p=6atm
Now Kp= p*(3p)3/(1atm)2
(as inert gas d not participate in the reaction and we only need the partial pressures of reactant and product at equilibrium to calcul;ate the Kp)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free