A mixture containing 19.8 moles of H2 and 7.2 mole of I2 was allowed to reach equilibrium in a 5 L closed vessel at ToC according to the equation: H2 (g) + I2 (g) 2HI (g) At equilibrium, 14 moles of H2 was present. The equilibrium constant for this reaction is denoted as Kc(1).

Question

(a) How many grams of hydrogen iodide (HI) will be produced if a mixture of 45 g I2 and 0.5 g H2 was allowed to reach equilibrium in a 100 L closed vessel at T oC? (2 marks)

(b) If chlorine (Cl2) gas is added to the above reaction system, a new equilibrium will be established with an equilibrium constant denoted as Kc(2).

2HI (g) + Cl2 (g) ↔ 2HCl (g) + I2 (g)

(i) Express the equilibrium constant Kc(2) in term of Kc(1). (1 mark)

(ii) An unknown amount of Cl2 gas is added to the system in Question 3(a). so that there is 40.0 g Cl2 at equilibrium. Calculate the number of moles of hydrogen chloride (HCl) gas produced if Kc(2) is twice than that of Kc(1). (2 marks)

(iii) What will be the position of equilibrium for the overall reaction if more HI is added to the reaction mixture at constant temperature and pressure? Explain your answers.

Please, Help me
Thank you Drbob

Answer 1

6.865 = x^2 / ((45/253.8/100)-0.5x)((0.5/2/100)-0.5x)
This equation is OK?because the mole ratio of H2 to Hi is 1:2
So [HI]=x=5.33*10-4M
HI =6.813g?

Answer 2

I will get you started. Is this an exam?
.............H2 + I2 ==> 2HI
I..........19.8...7.2....0
C...........-x....-x.....2x
E........19.8-x..7.2-x...2x
The problem tells you that 19.8-x = 14 which allows you to determine mols reactants and products at equilibrium.
Convert to concentrations, substitute into Kc and solve.

a)mols H2 = 0.5/2 = 0.25 and 0.25/100 = about 0.0025M but you need to check that.
mols I2 = 45/approx 254 = approx 0.18; again, you should check that. It's just an estimate. Then 0.18/100 = approx 0.0018 M
...........H2 + I2 ==> 2HI
I......0.0025.0.0018....0
C.........-x.....-x.....2x
E....0.0025-x.0.0018-x..2x

Substitute the E line into Kc1 and solve for x and the other values.
mols HI = M HI x L HI = ?
grams = mols x molar mass

Answer 3

Thank you for your help
Yes it is past exam paper and I have a exam after tomorrow

Kc1 I got 6.865,so
6.865=(2x)^2/((1.77*10^-3)-x)((2.5*10^-3)-x)
6.865=4x^2/x^2-(4.273*10^-3)+(4.433*10-6)
(2.865)x^2-(0.0293)x+(3.043*10^-5)=0
Are the steps correct above?
And [HI]=x=1.173*10^-3 M
no of mol of HI = 0.1173mol
mass of HI = 0.1173*(126.9+1)=15g
Correct ?

Answer 4

The steps look ok to me but I didn't check the math down through the quadratic. By the way, isn't (HI) = 2x and not x?

Answer 5

Oh...yes.I was careless,(HI) should be equal to 2x

Answer 6

3bi
Kc2 = [HCl(g)]^2/Kc(1) [H2(g)] [Cl2(g)]

3bii
Combine two reaction together, I have
H2 + Cl2 ⇌ 2HCl

Kc=[HCl]^2/[H2][Cl2] = Kc1 X Kc2 = 94.2647

Let y be the change of no. of mole of H2(g)

H2 + Cl2 ⇌ 2HCl
Final no. of mol (0.13287-y) 0.56338 2y

Is it OK?

Answer 7

3bi is ok
3bii
ok for the equation.
ok for k1*k2 (but too many significant figures)
For the last part I'm unclear about the problem Is that 40 g Cl2 at equilibrium and 0.5 g H2 gas before equilibrium is established?

Answer 8

May be the problem is unclear
Anyway, Thank you for your help sir!