Can someone help me with these chemistry questions (1 Viewer)

[sunshine_c22]

Hi,
I am stuck with these questions and i'm not getting the answer can someone please provide me with the answer + explanation

 

[carrotsss]

5. It will not shift as there are equal moles of gas on each side. This means that favouring either side would be pointless, as neither shift would decrease the pressure.

6. Try to do this question yourself. Remember that Keq=[products]/[reactants], and if there’s a two/three in front of something then you have to square/cube its concentration for Keq

7. Same as question 6, give it a go yourself, it should be pretty simple. The equation for Qc is identical to the equation for Keq, and you’ve been given all of the concentrations so you just have to sub into equations

8. For this question, set up your equation for Kc e.g. Kc=[CH3OH][CO]/[CH3COOH], then sub in your values for the concentrations you’ve been given and for Kc. If you rearrange the equation, you’ll be able to find [CH3COOH]

9. Draw a RICE table, where cis-but-2-ene has an initial concentration of 1 and trans-but-2-ene has an initial concentration of 0. Then, there’s a change in concentration of s for both into equilibrium (1:1 ratio). If you sub these equations (1-s and s) into Kc you should be able to solve for s, which is your answer

 

[sunshine_c22]

5. It will not shift as there are equal moles of gas on each side. This means that favouring either side would be pointless, as neither shift would decrease the pressure.

6. Try to do this question yourself. Remember that Keq=[products]/[reactants], and if there’s a two/three in front of something then you have to square/cube its concentration for Keq

7. Same as question 6, give it a go yourself, it should be pretty simple. The equation for Qc is identical to the equation for Keq, and you’ve been given all of the concentrations so you just have to sub into equations

8. For this question, set up your equation for Kc e.g. Kc=[CH3OH][CO]/[CH3COOH], then sub in your values for the concentrations you’ve been given and for Kc. If you rearrange the equation, you’ll be able to find [CH3COOH]

9. Draw a RICE table, where cis-but-2-ene has an initial concentration of 1 and trans-but-2-ene has an initial concentration of 0. Then, there’s a change in concentration of s for both into equilibrium (1:1 ratio). If you sub these equations (1-s and s) into Kc you should be able to solve for s, which is your answer

I've tried! But I'm literally not getting the answers

 

[Eagle Mum]

I've tried! But I'm literally not getting the answers

Do they provide the answers to questions 6 & 7? @carrotsss has provided you with the correct approach, so if you provide their answers, we can determine if the textbook might be wrong (happens not uncommonly) and possibly reverse engineer from the answer.

 

[sunshine_c22]

Do they provide the answers to questions 6 & 7? @carrotsss has provided you with the correct approach, so if you provide their answers, we can determine if the textbook might be wrong (happens not uncommonly) and possibly reverse engineer from the answer.

It's a quiz, and it doesn't tell me the answers, it only says its wrong

 

[Eagle Mum]

I’m not sure what they are trying to test with question 5.
carrotsss is correct about pressure per se not affecting the equilibrium when there are equal moles of gases on each side, but then the question mentions that the (forward) reaction is exothermic, so I don’t know if they are trying to test knowledge that pressure is proportional to temperature and therefore by heating the system, you’d favour the endothermic direction (ie. to the left).

 

[carrotsss]

It's a quiz, and it doesn't tell me the answers, it only says its wrong

I can’t do them right now but if you send your answers I can check them tomorrow

 

[Eagle Mum]

It's a quiz, and it doesn't tell me the answers, it only says its wrong

For Q6, I used an online calculator to get 0.39. What did you get?

 

[sunshine_c22]

5. It will not shift as there are equal moles of gas on each side. This means that favouring either side would be pointless, as neither shift would decrease the pressure.

6. Try to do this question yourself. Remember that Keq=[products]/[reactants], and if there’s a two/three in front of something then you have to square/cube its concentration for Keq

7. Same as question 6, give it a go yourself, it should be pretty simple. The equation for Qc is identical to the equation for Keq, and you’ve been given all of the concentrations so you just have to sub into equations

8. For this question, set up your equation for Kc e.g. Kc=[CH3OH][CO]/[CH3COOH], then sub in your values for the concentrations you’ve been given and for Kc. If you rearrange the equation, you’ll be able to find [CH3COOH]

9. Draw a RICE table, where cis-but-2-ene has an initial concentration of 1 and trans-but-2-ene has an initial concentration of 0. Then, there’s a change in concentration of s for both into equilibrium (1:1 ratio). If you sub these equations (1-s and s) into Kc you should be able to solve for s, which is your answer

Is the answer for question 6 = 0.39
Question 7 = 0.91
Question 8= 0.94
I still can't figure out question 9

 

[sunshine_c22]

For Q6, I used an online calculator to get 0.39. What did you get?

Is the answer for question 6 = 0.39
Question 7 = 0.91
Question 8= 0.94
I still can't figure out question 9

 

[sunshine_c22]

I can’t do them right now but if you send your answers I can check them tomorrow

It's due tonight

 

[Eagle Mum]

Is the answer for question 6 = 0.39
Question 7 = 0.91
Question 8= 0.94
I still can't figure out question 9

We agree on the answers to Q6 and Q8.

I get 1.52 as the answer to Q7.

 

[Eagle Mum]

For Q9,

K = [product] / [reactant]
= [trans] / [cis]
0.17 = [trans] / 1
[trans] = 0.17 M

 

[carrotsss]

It's due tonight

Sorry too late now, you should try not to do it on the night before next time

 

[sunshine_c22]

Sorry too late now, you should try not to do it on the night before next time

Haha I know, thank you tho!