Unit 4 Questions MEGATHREAD!

[thushan]

Yup, HenryP is right.

[luffy]

I see what you're getting at but I think thats just there to illustrate the reaction that takes place. The question tells you that the lactic acid is dissolved in water (4.5g in 500ml), this makes it an aqueous solution and will be included in the Ka calculation

Yes, but doesn't that mean the reactant is aqueous? Hence, the state in the reaction should also be aqueous?

[thushan]

Terminology issues -

reaction occurs in two stages.

Dissolution of acid --> HA (l) --> HA (aq)
Then ionisation of acid: HA (aq) <--> H+ (aq)  + A- (aq).

Given that dissolution step is complete, we consider the equilibrium constant of ionisation of acid.
If you really want to be nitpicky, it'd simply be a badly worded question.

[Wilbo]

Thermochemistry question for you,

A student mixes 50mL of 1.00M HCl with 50mL of 1.00M of NaOH and allows the following reaction to occur

H+ (aq) + OH (aq) ---> H2O (l); enthalpy= - 57.2 kJ mol-1
the student observes the temperature of the mixture rising from 25.0 degrees Celsius to 31.2 degrees Celsius.

Calculate the specific heat capacity of water.

Thanks!

well we know, we have 0.05 mol of each HCl, and NaOH, so the total heat released by this experiment would be 2.86kJ, or 2860J
We also know there is a a total temperature change of 6.2 degrees.
We also know that there is a total of 100ml of water.
So over, we need 2860J to raise the temperature of 100ml of water by 6.2 degrees.
Hence we need 4612J for each degree change for each gram of water, assuming 1ml of water= 1g
hence the specific heat capacity for water in this situation would be, 4.612JC-1g-1

I know its an old question, but yea

[david10d]

i still can't get this around my head...

why is a battery flat when it is in equilibrium?

[vea]

i still can't get this around my head...

why is a battery flat when it is in equilibrium?

The reactions are technically still occurring in both directions, however, the forward reaction rate is equal to the backwards reaction rate.
This means that there is no net forward reaction and so no electricity being produced.

[david10d]

i see, thank you.


another question: do we have to know about partial pressures? this wasn't in the heinemann book

[BoredSatan]

i see, thank you.


another question: do we have to know about partial pressures? this wasn't in the heinemann book

pretty sure this was in unit 2?

[REBORN]

^ yep

[david10d]

sorry i missed out on something

it goes under the notation of Kp?

[mickeymouse]

Okay confused about calorimetry, attempted a question that gave

- Initial Temperature

- Temperature after calibration

- Temperature after combustion

so to work out the energy required for combustion I subtracted AFTER COMBUSTION from INITIAL ... 
but its AFTER COMBUSTION - AFTER CALIBRATION ?

[b^3]

Okay confused about calorimetry, attempted a question that gave

- Initial Temperature

- Temperature after calibration

- Temperature after combustion

so to work out the energy required for combustion I subtracted AFTER COMBUSTION from INITIAL ... 
but its AFTER COMBUSTION - AFTER CALIBRATION ?

In this case the calibration was performed before the combustion. So the temperature before combustion starts is the temperature after the calibration has finished. So when you calibrate first, the energy required for combustion will be after combustion - after calibration.

[HenryP]

sorry i missed out on something

it goes under the notation of Kp?

You're talking about calculating the equilibrium constant with partial pressure right? I remember learning about this from a book but I have never seen VCAA put it on an exam before. I wouldn't worry about it.

[mickeymouse]

Okay confused about calorimetry, attempted a question that gave

- Initial Temperature

- Temperature after calibration

- Temperature after combustion

so to work out the energy required for combustion I subtracted AFTER COMBUSTION from INITIAL ... 
but its AFTER COMBUSTION - AFTER CALIBRATION ?

In this case the calibration was performed before the combustion. So the temperature before combustion starts is the temperature after the calibration has finished. So when you calibrate first, the energy required for combustion will be after combustion - after calibration.

oh okay thanks so with all calorimetry questions the final temp from calibration will ALWAYS be used as the starting for actual combustion reaction?   

[b^3]

Okay confused about calorimetry, attempted a question that gave

- Initial Temperature

- Temperature after calibration

- Temperature after combustion

so to work out the energy required for combustion I subtracted AFTER COMBUSTION from INITIAL ... 
but its AFTER COMBUSTION - AFTER CALIBRATION ?

In this case the calibration was performed before the combustion. So the temperature before combustion starts is the temperature after the calibration has finished. So when you calibrate first, the energy required for combustion will be after combustion - after calibration.

oh okay thanks so with all calorimetry questions the final temp from calibration will ALWAYS be used as the starting for actual combustion reaction?   

Not always, I've across some questions that do the calibration first, and some that do it last. You have to look at each situation and logically apply the right temperature change for the right process.