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# AQA A2 Chemistry - Equilibria Kc Calculations watch

1. Started doing summary quesions for the equilibria chapter of AQA A2 Unit 4 chemistry
however I am finding it really hard to understand equilibria kc calculation questions where you are giving starting moles and have to deduce information ect.

If someone could give me a step by step working out of these questions would really help.

4.2 Exercise 1 - Kc

1. For each of the following equilibria, write the expression for the equilibrium constant Kc and state its units:

4. The reaction for the formation of hydrogen iodide does not go to completion but reaches an equilibrium: H2(g) + I2(g) == 2HI(g)
A mixture of 1.9 mol of H2 and 1.9 mol of I2 was prepared and allowed to reach equilibrium in a closed vessel on 250 cm3 capacity. The resulting equilibrium mixture was found to contain 3.0 mol of HI. Calculate the value of Kc.

5. Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.

8. A 0.04 sample of SO3 is introduced into a 3.04 litre vessel and allowed to reach
equilibrium. The amount of SO3 present at equilibrium is found to be 0.0284 mole. Calculate the value of Kc for the reaction 2SO3(g) == 2SO2(g) + O2(g).

10. When 1.0 mole each of ethanoic acid and ethanol were allowed to reach equilibrium in a sealed vessel of volume 500 cm3, the amount of ethanoic acid present at equilibrium was found to be 0.33 mole. Calculate the value of Kc for the reaction CH3COOH + CH3CH2OH == CH3COOCH2CH3 + H2O(l)
2. (Original post by PowerPuff)
Started doing summary quesions for the equilibria chapter of AQA A2 Unit 4 chemistry
however I am finding it really hard to understand equilibria kc calculation questions where you are giving starting moles and have to deduce information ect.

If someone could give me a step by step working out of these questions would really help.

4.2 Exercise 1 - Kc

You should be fully familiar with what Kc is before leaping to do questions. Besides, if you have had examples shown in class, you should look at them again and attempt them without looking at the worked solution. That should help with some of these questions.

1. For each of the following equilibria, write the expression for the equilibrium constant Kc and state its units:

4. The reaction for the formation of hydrogen iodide does not go to completion but reaches an equilibrium: H2(g) + I2(g) == 2HI(g)
A mixture of 1.9 mol of H2 and 1.9 mol of I2 was prepared and allowed to reach equilibrium in a closed vessel on 250 cm3 capacity. The resulting equilibrium mixture was found to contain 3.0 mol of HI. Calculate the value of Kc.
Always set up eqm table
H2 I2 2 HI
initial mol 1.9 1.9 0
final mol ? ? 3

You need to fill in ? in order to be able to get Kc.
Reasoning: If you produce 3 mol of HI, then mole of H2 reacted = mole of I2 reacted = 1/2 mole of HI produced (from balanced eqn) = 1/2 (3) = 1.5 mol

if 1.5 mol of H2 has reacted, mole of H2 left at eqm = 1.9 - 1.5 = 0.4

the same goes for I2, mole of I2 left at eqm = 1.9 - 1.5 = 0.4

Hence Kc = [HI]^2/[H2][I2] = (3)(3)/(0.4)(0.4) = ?

5. Consider the equilibrium: N2O4(g) == 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction.

8. A 0.04 sample of SO3 is introduced into a 3.04 litre vessel and allowed to reach
equilibrium. The amount of SO3 present at equilibrium is found to be 0.0284 mole. Calculate the value of Kc for the reaction 2SO3(g) == 2SO2(g) + O2(g).

10. When 1.0 mole each of ethanoic acid and ethanol were allowed to reach equilibrium in a sealed vessel of volume 500 cm3, the amount of ethanoic acid present at equilibrium was found to be 0.33 mole. Calculate the value of Kc for the reaction CH3COOH + CH3CH2OH == CH3COOCH2CH3 + H2O(l)

Work step by step, set up eqm table first, then fill in information
3. No need for an atttitude if I was "fully aware" I wouldnt be asking for assistance
4. (Original post by PowerPuff)
No need for an atttitude if I was "fully aware" I wouldnt be asking for assistance
ermm, if you have a careful look at my post, intercalated in red comments are worked solutions of one particular question.

the rest, i am sure you are "fully aware" how to do based on that and of course, textbooks.
5. Can you please do question 5.
I am ok with the numbers but when it comes to reading it I find it difficult, (probably because I am dyslexic)
I thought that it ment that there were 0.1 mol dm^-3 of N2O4 at the start.... Anyways I could not work out how to do the question, but I got full marks on the rest of the question. It's just really bothering and my teacher is on paternity leave at the moment and won't be back for a couple of weeks
6. (Original post by Katie146)
Can you please do question 5.
I am ok with the numbers but when it comes to reading it I find it difficult, (probably because I am dyslexic)
I thought that it ment that there were 0.1 mol dm^-3 of N2O4 at the start.... Anyways I could not work out how to do the question, but I got full marks on the rest of the question. It's just really bothering and my teacher is on paternity leave at the moment and won't be back for a couple of weeks
5. Consider the equilibrium: N2O4(g) <==> 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction

If 50% have dissociated then this means 0.5 mol

At equilibrium there are 0.5 mol N2O4 remaining

and 1.0 mol NO2 formed.

Vessel is 10dm3 hence to get concentrations divide moles by 10

N2O4 = 0.05 mol dm-3
NO2 = 0.1 mol dm-3

Now plug values into the equilibrium law equation:

kc = (0.1)2/0.05

kc = 0.01/0.05 = 0.2
7. (Original post by charco)
5. Consider the equilibrium: N2O4(g) <==> 2NO2(g).
1 mol of dinitrogen tetroxide, N2O4, was introduced into a vessel of volume 10 dm3. At equilibrium 50% had dissociated. Calculate Kc for the reaction

If 50% have dissociated then this means 0.5 mol

At equilibrium there are 0.5 mol N2O4 remaining

and 1.0 mol NO2 formed.

Vessel is 10dm3 hence to get concentrations divide moles by 10

N2O4 = 0.05 mol dm-3
NO2 = 0.1 mol dm-3

Now plug values into the equilibrium law equation:

kc = (0.1)2/0.05

kc = 0.01/0.05 = 0.2
Thank you so much. It seams so simple now...

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