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# ocr a f325 revision thread watch

1. Revision anyone, I'm in some need of some motivation and I'm hoping this'll help give me that!

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2. (Original post by MathsNerd1)
Revision anyone, I'm in some need of some motivation and I'm hoping this'll help give me that!

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Bring it on brother
3. (Original post by otrivine)
Bring it on brother
Okay, what chapter do you want to go through? I've done all except from the majority of transition metals and redox titrations.

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4. (Original post by MathsNerd1)
Okay, what chapter do you want to go through? I've done all except from the majority of transition metals and redox titrations.

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lets do chapter 1
5. (Original post by otrivine)
lets do chapter 1
Okay,

Define acid- base pair (2)
Explain the effect of temperature on the rate constant (3)

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6. (Original post by MathsNerd1)
Okay,

Define acid- base pair (2)
Explain the effect of temperature on the rate constant (3)

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1) acid base pair is where it differs by H+ and is where the two species transform into each other by the gain or loss of a proton.

2) increase temperature will result in an increase in the rate constant this is because the molecules or particles attain more energy for more collision due to more kinetic energy and so exceeds the activation energy and as a result more frequent effective successive collisions between molecules.
7. (Original post by otrivine)
1) acid base pair is where it differs by H+ and is where the two species transform into each other by the gain or loss of a proton.

2) increase temperature will result in an increase in the rate constant this is because the molecules or particles attain more energy for more collision due to more kinetic energy and so exceeds the activation energy and as a result more frequent effective successive collisions between molecules.
That's correct, my turn.

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8. Man y'all know how much I love titration based questions right? If anyone needs help give me a heads up
9. (Original post by MathsNerd1)
That's correct, my turn.

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Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

i) Write down an expression for Kc for this reaction
ii) Calculate the concentrations of methanol and carbon monoxide present at equilibrium. Show your working.

iii) Hence calculate Kc (to 3 significant figures) including its units
10. (Original post by rival_)
Man y'all know how much I love titration based questions right? If anyone needs help give me a heads up

I have a question for you
11. (Original post by otrivine)
Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

i)Write down an expression for Kc for this reaction
ii)Calculate the concentrations of methanol and carbon monoxide present at equilibrium. Show your working.

iii)Hence calculate Kc (to 3 significant figures) including its units
My god, what a question!

Just let me get my notepad of paper to work it out and try not to laugh if I get it wrong as this'll probably be the first question I have done of this in quite some time.

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12. (Original post by MathsNerd1)
My god, what a question!

Just let me get my notepad of paper to work it out and try not to laugh if I get it wrong as this'll probably be the first question I have done of this in quite some time.

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I am going to give you 2 minutes more or you would get 0!
13. (Original post by otrivine)
Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

i)Write down an expression for Kc for this reaction
ii)Calculate the concentrations of methanol and carbon monoxide present at equilibrium. Show your working.

iii)Hence calculate Kc (to 3 significant figures) including its units
Okay let's see if this is correct.

i) Kc = [CH3COOH]/{[CH3OH][CO]}

ii) If there is 12.2 moles made up in equilibrium of ethanoic acid then there is only 5.4 moles and 7.4 moles left of methanol and carbon monoxide respectively. Then using n= CxV with the volume being 5dm3 you'd get the concentrations of 1.08 and 1.48 mol-1dm3 of methanol and carbon monoxide respectively.

iii) Then once subbing in all the values worked out and the conc of ethanoic acid being 2.44 mol-1dm3 then you'll get Kc = 1.53 (3sf) with units mol dm-3

How did I do?

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14. (Original post by MathsNerd1)
Okay let's see if this is correct.

i) Kc = [CH3COOH]/{[CH3OH][CO]}

ii) If there is 12.2 moles made up in equilibrium of ethanoic acid then there is only 5.4 moles and 7.4 moles left of methanol and carbon monoxide respectively. Then using n= CxV with the volume being 5dm3 you'd get the concentrations of 1.08 and 1.48 mol-1dm3 of methanol and carbon monoxide respectively.

iii) Then once subbing in all the values worked out and the conc of ethanoic acid being 2.44 mol-1dm3 then you'll get Kc = 1.53 (3sf) with units mol dm-3

How did I do?

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Everything is perfect. However, your units are not correct, it should be dm3 mol-1. you did correct then you put a contradiction.
15. (Original post by otrivine)
Everything is perfect. However, your units are not correct, it should be dm3 mol-1. you did correct then you put a contradiction.
Darn I first put that on my notepad but thought I remembered the units for concentration wrong so rubbed it all out, maybe next time

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16. (Original post by MathsNerd1)
Darn I first put that on my notepad but thought I remembered the units for concentration wrong so rubbed it all out, maybe next time

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No worries

My turn
17. (Original post by otrivine)
No worries

My turn
Okay, here we go.

Nitrogen monoxide and oxygen react together as in the overall equation below:
2NO(g) + O2(g) -------> 2NO2(g)
The rate equation for this reaction is rate =k[NO]^2

a) Explain what is meant by rate-determining step
b) What does the rate equation tell you about the rate determining step?
c) Suggest a possible two step mechanism for this reaction? (6)

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18. (Original post by MathsNerd1)
Okay, here we go.

Nitrogen monoxide and oxygen react together as in the overall equation below:
2NO(g) + O2(g) -------> 2NO2(g)
The rate equation for this reaction is rate =k[NO]^2

a) Explain what is meant by rate-determining step
b) What does the rate equation tell you about the rate determining step?
c) Suggest a possible two step mechanism for this reaction? (6)

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1) rate- determining step is the slowest step reaction mechanism in a multi-step reaction.

2) the rate equation gives the species that are present in the reactant side of the slowest step and the rate also indicates the number of moles of the species in the reactants in the slowest step.

3) NO + NO ---> N2O4 slowest step
N2O4 ---> NO2+ NO2 fastest

overall 2NO+O2 --> 2NO2 ?
19. (Original post by otrivine)
1) rate- determining step is the slowest step reaction mechanism in a multi-step reaction.

2) the rate equation gives the species that are present in the reactant side of the slowest step and the rate also indicates the number of moles of the species in the reactants in the slowest step.

3) NO + NO ---> N2O4 slowest step
N2O4 ---> NO2+ NO2 fastest

overall 2NO+O2 --> 2NO2 ?
Yeah that's all correct, my turn

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20. (Original post by MathsNerd1)
Yeah that's all correct, my turn

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How does buffer work using example of methanoic acid and NaOH (4)

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