Question on equil knost etc (1 Viewer)

[mr EaZy]

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Hydrogen iodide equation:

2HI (---------) H2 + I2
(G) (G) (G)


Q: 4 moles of hydrogen iodide are introduced into a 10L container at 500 deg C. At equil, 70% of the hydrogen iodide remains.

a) Evaluate the equilibrium concentrations for each gas in mole L^-1.

b) Calculate K

c) Describe the equilibrium as being to the left or right.


Tell me how to do this and DO IT! "DO IT!"

 

[mr EaZy]

IS this kinda quest gonna appear in the exam??

 

[04er]

... this option already doesn't look fun... and if you want help, try to be less demanding :mad1:

 

[Heinz]

Im DOING IT (funny film)

Q: 4 moles of hydrogen iodide are introduced into a 10L container at 500 deg C. At equil, 70% of the hydrogen iodide remains.

a) Evaluate the equilibrium concentrations for each gas in mole L^-1.

The initial concentration of HI is 0.4mol/L (4 moles in 10 litres)

initially, H₂ = 0, I₂ = 0 and 2HI = 0.4
at equilibrium, 2HI = 0.28 (70% remaining)

Thus, 0.12 mol/L of HI has reacted to form H₂ + I₂
From the equation, HI reacts with H₂ and I₂ in a 2:1 ratio

Therefore at equilibrium, H₂ and I₂ both have a 0.06mol/L concentration

K = ([H₂][I₂])/[HI]²
=[0.06][0.06]/0.12²
=1/4

Since K < 1, the reaction is favouring the reactants side so to the left of that equation?

I sure hope thats right

 

[TheKing]

more easily explained in if:
aA(aq) + bB(aq) -> cC(aq) + dD(aq)
then
K= [C]^c [D]^d / [A]^a ^b

that is ^= to the power of

 

[kheir]

shouldnt it be
0.12^2/[0.06][0.06]
giving 1/25
therefore k<1 and more product will be made hence equilibrium shifts to the right

 

[Heinz]

Originally posted by kheir
shouldnt it be
0.12^2/[0.06][0.06]
giving 1/25
therefore k<1 and more product will be made hence equilibrium shifts to the right

Yeah, it should be K < 1 still doesnt change much . (0.12^2/.06^2) = 4... It doesnt really matter though since either way, itll favour the 2HI side. Changing the direction of the equation reciprocates the equilibrium constant so you end up with the same answer.

 

[Beats]

Could someone explain to me (sorry if this is really basic) - but since the Q says that at equilibrium 70% HI remains - wouldn't the equilibrium concentration of HI be 70% of 4 = 0.28 mol/L. Where are you getting the 0.12 from? Once again sorry if i totally do not know what i am going on about.

 

[CM_Tutor]

Beats, you are right, [HI] = 0.28 mol/L at equilibrium, not 0.12 mol/L, and so K = 0.046

 

[Heinz]

Originally posted by CM_Tutor
Beats, you are right, [HI] = 0.28 mol/L at equilibrium, not 0.12 mol/L, and so K = 0.046

Okay, now im lost. Umm, if theres 70% concentration remaining, then dont you use the concentration that reacted (30%) to determine the concentration of products thats been created?

 

[CM_Tutor]

Yes, but there is still 70 % HI remaining.

 

[Heinz]

Originally posted by CM_Tutor
Yes, but there is still 70 % HI remaining.

*Smacks head* Oh I see my error now! thanks