Please help guys...Kp question

The colourless gas dinitrogen tetroxide, N

2O4, dissociates into the brown gas
nitrogen dioxide, NO2.
N2O4(g) &#56256;&#56405; 2NO2(g) ∆H = +58.1 kJ mol–1

(b) At 400 K and a pressure of 6.75 atm, dinitrogen tetroxide is 80% dissociated.
Calculate the value of

Kp at this temperature, stating its units.

can some please explain to me fully how to do this? i have the markscheme, but i dont understand how to work it out!

thanku!

Reply 1

alkaeda

3

is the answer = 48atm? just worked it out

Reply 2

littleone

OP

alkaeda

is the answer = 48atm? just worked it out

yes..!!!!!!!!!!!!!!but how?why?

Reply 3

grapesugar

1

goodness..how on earth do you do that? don't u need to know some mole figures??

Reply 4

littleone

OP

ANyone? please unit 4 is tomorow :-(

Reply 5

forkwise

1

N2O4 ------------> 2NO2
initial mole ratio: 1,0
given ratio: 0.2,1.6
(it's 80% dissociated so from 1 to 0.2, and NO2 has double number of moles so add 1.6 to 0)

Mole fractions = 0.2/1.8 and 1.6/1.8 respectively
Multiply this by pressure (6.75 atm) to get partial pressures:
0.75 and 6

Kp = p(NO2)^2/p(N2O4)
= 6^2/0.75
= 36/0.75 = 48 atm

Reply 6

littleone

OP

forkwise

N2O4 ------------> 2NO2
initial mole ratio: 1,0
given ratio: 0.2,1.6
(it's 80% dissociated so from 1 to 0.2, and NO2 has double number of moles so add 1.6 to 0)

Mole fractions = 0.2/1.8 and 1.6/1.8 respectively
Multiply this by pressure (6.75 atm) to get partial pressures:
0.75 and 6

Kp = p(NO2)^2/p(N2O4)
= 6^2/0.75
= 36/0.75 = 48 atm