Hard Kc A2 equilibrium question!

When a 0.02mol sample of SO3 is introduced into a vessel of volume 1.52dm3, 0.0142mol of SO2 was found to be present after equilibrium was reached.

Calculate Kc for the reaction.

Kc= [SO2]*2/[SO2]*2 X [O2]

Reply 1

eurasianfeline

2 SO3 --> 2SO2 + O2

start:
mol SO3 = 0.02 / 1.52 = 0.013mol

delta:
mol SO3 = - 9.34 x 10^-3
mol SO2 = 0.0142 / 1.52 = + 9.34 x 10^-3
mol O2 = 9.34 x 10^-3 / 2 = + 4.67 x 10^-3
......................................................................................
equil:
mol SO3 = 0.013 - 9.34 x 10^-3 = 3.66 x 10^-3
mol SO2 = 9.34 x 10^-3
mol O2 = 4.67 x 10^-3

.........(9.34 x 10^-3)^2 x 4.67 x 10^-3
Kc = ................................................
......................(3.66 x 10^-3)^2

Kc = 0.03 moldm^-3

Reply 2

eurasianfeline

good...someone agrees with me. :biggrin:

Reply 3

hitchhiker_13

vinny2256

(Thought this might be easier to understand than my first post)
SO3: inital no. of moles = 0.02
reacting: 0.0142 (equal to no. of moles of SO2 due to 1:1 mole ratio)
at equilibrium: 0.02-0.0142 = 5.8*10^-3
conc: (5.8*10^-3)/1.52 = 3.816*10^-3

SO2: initial no. of moles: 0
reacting: 0.0142
at eqiulibrium: 0.0142
conc: 0.0142/1.52= 9.342*10^-3

O2: initial no. of moles: 0
reacting: 0.0142/2 = 7.1*10^-3 (1:2 mole ratio)
at equilibrium: 7.1*10^-3
conc: (7.1*10^-3)/1.52 = 4.671*10^-3

Kc = [(9.342*10^-3)^2(4.671*10^-3)]/(3.816 *10^-3)^2
= 0.0280moldm^-3

Latest

Trending

Articles for you

Six things you must do on a law school open day visit

Studying for a law degree: how to choose the best university for you

Expert tips for revising A-level and Btec law

What to expect from a fully online law degree: common myths busted

Try out the app

Continue on web