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).
(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
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?
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
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 ?
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?
Oh...yes.I was careless,(HI) should be equal to 2x
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?
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?
May be the problem is unclear
Anyway, Thank you for your help sir!
(a) To find the grams of hydrogen iodide (HI) produced, we need to calculate the amount of I2 and H2 that is consumed based on the stoichiometry of the balanced equation.
Given:
Moles of H2 = 7.2 moles
Moles of I2 = 45 g / molar mass of I2
Molar mass of I2 = 2 * atomic mass of I = 2 * 126.9 g/mol
From the balanced equation: H2 (g) + I2 (g) -> 2HI (g)
1 mole of H2 reacts with 1 mole of I2 to produce 2 moles of HI.
Moles of I2 consumed = 1/2 * moles of H2 reacted
Moles of I2 consumed = 1/2 * 7.2 moles
Moles of I2 consumed = 3.6 moles
Moles of I2 remaining = Moles of I2 initial - Moles of I2 consumed
Moles of I2 remaining = (45 g / molar mass of I2) - 3.6 moles
Now we can calculate the moles of HI produced using the stoichiometry of the balanced equation.
Moles of HI produced = Moles of I2 consumed * 2
Moles of HI produced = 3.6 moles * 2
Now we can find the mass of HI produced using the molar mass of HI.
Mass of HI produced = Moles of HI produced * molar mass of HI
(b) (i) To express the equilibrium constant Kc(2) in terms of Kc(1), we need to examine the balanced equation for the reaction in question.
2HI (g) + Cl2 (g) ↔ 2HCl (g) + I2 (g)
The equilibrium constant Kc(2) can be expressed as the product of the equilibrium constants of the individual reactions:
Kc(2) = [HCl]^2 * [I2] / [HI]^2 * [Cl2]
And Kc(1) = [HI]^2 / [H2] * [I2]
Therefore, Kc(2) = Kc(1)^2 * [HCl]^2 / [H2] * [Cl2]
(ii) To calculate the number of moles of hydrogen chloride (HCl) gas produced, we need to find the moles of Cl2 consumed based on the given mass of Cl2 and the molar mass of Cl2.
Given:
Mass of Cl2 = 40.0 g
Molar mass of Cl2 = 2 * atomic mass of Cl = 2 * 35.45 g/mol
Moles of Cl2 consumed = Mass of Cl2 / Molar mass of Cl2
Now we can use the stoichiometry of the balanced equation to find the moles of HCl produced.
From the balanced equation: 2HI (g) + Cl2 (g) ↔ 2HCl (g) + I2 (g)
2 moles of HI react with 1 mole of Cl2 to produce 2 moles of HCl.
Moles of HCl produced = Moles of Cl2 consumed * 2
(iii) If more HI is added to the reaction mixture at constant temperature and pressure, it will shift the equilibrium to the left. According to Le Chatelier's principle, an increase in the concentration of reactants will drive the equilibrium in the direction that consumes these reactants. In this case, increasing the concentration of HI will favor the reverse reaction, resulting in more H2 and I2 being formed and less HI being produced. Therefore, the position of equilibrium will shift towards the reactants, favoring the formation of H2 and I2.