UA Final Review - Kate, Riya, Matthew

General non-science questions and class announcements.

Chem_Mod
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UA Final Review - Kate, Riya, Matthew

Hi everyone!

As promised, here is our Final Review worksheet covering Equilibrium through Thermodynamics. We hope that you will find it useful for your studying. Some of these problems are meant to be difficult; do not worry. Approach them as you would any test question: write out what you know, what you don't know, and work step-by-step to try and reach the final answer. Of course, due to the coronavirus situation there will be no in-person review session on Thursday. However, we will be posting complete worked-out solutions. It's been a great quarter with you all and we wish you the best!

- Kate, Riya, Matthew

Worksheet
Final Review - W20.pdf

Errors
Q9. $\Delta H_{b}(O=O)=496kJ/mol$; $\Delta H_{b}(O-H)$ was repeated

Rebecca Remple 1C
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Re: UA Final Review - Kate, Riya, Matthew

Hi Kate, Riya, and Matthew,

Thank you all for your hard work this quarter! Your review sessions have been incredibly helpful. I'll definitely miss seeing you all at the final review session, but I'm so grateful for all of your help. Your review worksheet looks great and I look forward to working through it in preparation for the exam. Thank you again for your help and I wish you all the best of luck in your future endeavors! :)

-Rebecca

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Re: UA Final Review - Kate, Riya, Matthew

what a wild way to end winter quarter! thank you all for working so diligently in order to support our continued academic success. Stay safe everyone.

Rebecca Remple 1C
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Re: UA Final Review - Kate, Riya, Matthew

Chem_Mod wrote:Hi everyone!

As promised, here is our Final Review worksheet covering Equilibrium through Thermodynamics. We hope that you will find it useful for your studying. Some of these problems are meant to be difficult; do not worry. Approach them as you would any test question: write out what you know, what you don't know, and work step-by-step to try and reach the final answer. Of course, due to the coronavirus situation there will be no in-person review session on Thursday. However, we will be posting complete worked-out solutions. It's been a great quarter with you all and we wish you the best!

- Kate, Riya, Matthew

Worksheet
Final Review - W20.pdf

To be posted

Hi all,

Thank you so much for sharing your worksheet! I am currently working through it in preparation for the final. I was wondering if you plan to upload the solutions to it soon? I want to make sure that I fully understand how to answer each question. Thank you so much and I hope you have a wonderful day!

-Rebecca

Daniela Shatzki 2E
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Re: UA Final Review - Kate, Riya, Matthew

hi, I was just wondering when the solutions were going to be posted? thank you! :)

Chem_Mod
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Re: UA Final Review - Kate, Riya, Matthew

Answer Key is posted! Thank you all for your patience; it's been a hectic few days. If you notice any discrepancies or have questions feel free to ask on this thread! Good luck on finals and stay safe.

-Matthew, Kate, Riya

Labiba Sardar 2A
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Re: UA Final Review - Kate, Riya, Matthew

For Q4. a), how do you know that 5.00 x 10^-3 mol H2C2O4 = 1.00 x 10^-2 mol H+?

Chem_Mod
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Re: UA Final Review - Kate, Riya, Matthew

Labiba Sardar 2A wrote:For Q4. a), how do you know that 5.00 x 10^-3 mol H2C2O4 = 1.00 x 10^-2 mol H+?

Since oxalic acid is a diprotic acid, each molecule of oxalic acid has 2 protons that can be neutralized by a strong base. Therefore 5 x 10^-3 mol of oxalic acid has 2*(5 x 10^-3 mol) = 1 x 10^-2 mol of acidic protons that must be completely neutralized by NaOH.

AlyssaYeh_1B
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Re: UA Final Review - Kate, Riya, Matthew

For question 11, the van't hoff equation is written like ln(K2/K1) = -(deltaH/R)((1/T2) - (1/T1)). However, in the constants and equations sheet, the T2 and T1 are swapped, which would result in K2=0.0169 instead of 0.0186. I was wondering which form of the equation is correct and which we should use to solve these problems?

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Re: UA Final Review - Kate, Riya, Matthew

AlyssaYeh_1B wrote:For question 11, the van't hoff equation is written like ln(K2/K1) = -(deltaH/R)((1/T2) - (1/T1)). However, in the constants and equations sheet, the T2 and T1 are swapped, which would result in K2=0.0169 instead of 0.0186. I was wondering which form of the equation is correct and which we should use to solve these problems?

Keep in mind the negative sign. The equation used is in the answer key is correct. When you swap T2 and T1, you distribute the negative sign inside the parentheses and the deltaH/R term is positive ie. no negative sign. You most likely forgot about it. Try deriving the van't Hoff equation to verify that the equation is correct.

You also should be able to tell that your answer is incorrect because your K2 is less than K1; however we know it should increase when the temperature is increased because the reaction is endothermic.

Christineg1G
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Re: UA Final Review - Kate, Riya, Matthew

For question 4 part b, how do you get the two equations for k1 and k2?

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Re: UA Final Review - Kate, Riya, Matthew

Christineg1G wrote:For question 4 part b, how do you get the two equations for k1 and k2?

Those are the equations for the first and second dissocation of oxalic acid, respectively. Since oxalic acid is a diprotic acid, it can dissociate 2 protons into solution. The first dissociation is when it gives off its first proton and the second dissociation is when it gives off its second proton. We often write the dissociation of polyprotic acids this way because each successive dissocation has a lower K value.

claudia_1h
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Re: UA Final Review - Kate, Riya, Matthew

Regarding question 11: This is not representative of what would actually happen right? Because when temperature increased, K went up from 0.0177 to 0.0186, but isn't K supposed to decrease with increasing temperature?

Chem_Mod
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Re: UA Final Review - Kate, Riya, Matthew

claudia_1h wrote:Regarding question 11: This is not representative of what would actually happen right? Because when temperature increased, K went up from 0.0177 to 0.0186, but isn't K supposed to decrease with increasing temperature?

No, this is what would actually happen. When a reaction is endothermic, increasing the temperature increases K. When a reaction is exothermic, increasing the temperature decreases K. The reaction has a positive change in enthalpy (endothermic), thus increasing the temperature leads to an increase in K. You learned this conceptually when learning about equilibrium and can now make quantitative assessments using thermodynamics (van't Hoff) and kinetics (Arrhenius).