Question

A 0.960mol quantity of Br2 is added to a 1.00 L reaction vessel that contains 1.23mol of H2 gas at 1000 K. What are the partial pressures of H2, Br2, and HBr at equilibrium?
At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ for the reaction H2(g)+Br2(g)⇌2HBr(g).

Dr Bob, you said:
Use PV = nRT to solve for pH2 initial.
Use PV = nRT to solve for p
I obtained approx 100 for pBr2 and 128 for pH2 initially but you need to go through and do it more accurately.

...........H2 + B2 ==> 2HBr
I..........128..100.....0
C...........-p..-p......+2p
E.......128-p..100-p....+2p

How did you get 100 and 128? I got 101 and 78.

Answer 1

pBr2 = ?
n = 0.960
R = 0.08206
T = 273 + 1000 = 1273K

p = nRT/V = (0.960)*0.08206 x 1273/1 = 100.28 BUT I said that was approx and you should go through it with more accuracy. I would round that to 100.

pH2 = (1.28*0.08206 x 1273/1 = 128.49 with the same recommendation. I would round that to 128. I suspect you may have used a slightly different value for R which might account for the slight difference in pBr2 but for pH2 I suspect you just punched in the wrong numbers. OR you may have used a different value for R. Some round the 0.08206 (I think it's 0.082057) off to 0.0821

Answer 2

Hello, this is a few years late, but you should not be adding another 273 to 1000K that is already in kelvin. You should only add 273 when going from Celcius to Kelvin. This in effect alters the rest of your numbers.

Answer 3

Solve for x. 5/8(x+1/2)=100