Chemistry 3/4 2013 Thread

[Alwin]

I had a q, from coursework, but its beyond vce standard (so pretty much @nliu ). It was something like this:
A + B --> C delta H >0

Draw a rate of reaction vs time graph of the forward and reverse reactions when temperature is increased.
Also whether the temperature is suddenly increased (rather infeasible) or slowly increased is not made clear.

I had something like this, assuming instantaneous increased temperature, based on the premise both forward and back reaction rates increase, and the net reaction rate forward increases.

Spoiler

It doesn't look right though . . .

Thanks~!!

EDIT: added smiley cos that makes all the difference

[lzxnl]

You CAN suddenly increase the temperature by reacting the system with some compounds that do not interfere with the system and only react with each other exothermically. It's a bit impractical admittedly.

Yeah what you have is right. Reaction endothermic, so temperature increase means forward reaction rate goes up more. VCAA is hoping students will forget that both reaction rates increase when the temperature goes up.

[Alwin]

You CAN suddenly increase the temperature by reacting the system with some compounds that do not interfere with the system and only react with each other exothermically. It's a bit impractical admittedly.

Yeah what you have is right. Reaction endothermic, so temperature increase means forward reaction rate goes up more. VCAA is hoping students will forget that both reaction rates increase when the temperature goes up.

Okay, cool beans! its just didn't look right to me.

Thanks for the pm too, congrats man!!!

[Homer]

VCAA is hoping students will forget that both reaction rates increase when the temperature goes up.

could you please explain this to me?

[lzxnl]

When you heat the system up, the average kinetic energy of the system increases, so the frequency of collisions increases. This increase in collision frequency is independent on whether the particle is product or reactant, so the number of collisions for the forward and backward reactions increase, leading to a larger number of successful collisions and hence increased reaction rates going forward and backwards. It's just that for a reaction with higher activation energy, this increase in energy will increase the proportion of particles that can break the activation barrier to a greater extent than for the reaction with lower activation energy. I've proved this many times before.

And also, increasing the energy increases the raw proportion of particles that can get over the activation energy barrier too. How could I forget.

[Stick]

Just a pre-warning that Chapters 23 and 24 of the Heinemann textbook are more 'examinable' than you might think. Don't read into it too much, but don't dismiss it either.

[lolipopper]

Just a pre-warning that Chapters 23 and 24 of the Heinemann textbook are more 'examinable' than you might think. Don't read into it too much, but don't dismiss it either.

I thought it was absolutely necessary anyways. But hey, get some sleep Stick, You still gotta do that UMAT 

[EternalSpirit]

When you heat the system up, the average kinetic energy of the system increases, so the frequency of collisions increases. This increase in collision frequency is independent on whether the particle is product or reactant, so the number of collisions for the forward and backward reactions increase, leading to a larger number of successful collisions and hence increased reaction rates going forward and backwards. It's just that for a reaction with higher activation energy, this increase in energy will increase the proportion of particles that can break the activation barrier to a greater extent than for the reaction with lower activation energy. I've proved this many times before.

And also, increasing the energy increases the raw proportion of particles that can get over the activation energy barrier too. How could I forget.

Does this mean, regardless of whether the reaction is endothermic or exothermic, there are more products formed?
And according to LCP, the equilibrium in endothermic reactions will shift to the right, whereas in an exothermic reaction the equilibrium will shift to the left?
Does it shift because the number of particles on the side that the equilibrium is shifting towards, needs to be increased to partially oppose the effect of more products formed?

[Alwin]

Does this mean, regardless of whether the reaction is endothermic or exothermic, there are more products formed?
And according to LCP, the equilibrium in endothermic reactions will shift to the right, whereas in an exothermic reaction the equilibrium will shift to the left?
Does it shift because the number of particles on the side that the equilibrium is shifting towards, needs to be increased to partially oppose the effect of more products formed?

nah. LCP is based on the net rate of reaction. for temperature increase:
exotherimic: forward reaction rate increases, but to a lesser extent compared to the reverse reaction rate. so, the overall rate is back, hence reactants favored. PoE moves to the left
endothermic: forward rate increases more than the reverse reaction rate (which also increased). So, the forward reaction dominates and more products produced. PoE moves to the right

make more sense?

Edit: grammar

[EternalSpirit]

Makes sense to me now.  But for the exothermic reaction, more reactants is formed?

[lzxnl]

Yes

[Scooby]

100 mL - is this to one significant figure or three? I know 100.00 mL would be five but yeah... what about without the decimal places?

[thushan]

100 mL - is this to one significant figure or three? I know 100.00 mL would be five but yeah... what about without the decimal places?

Technically, 100 mL is considered 3 sig figs. To express it to one sig fig, you have to write 1 x 10^3 mL.

[aestheticatar]

Yo,
A quick one for clarification.

How is the rate of the forward reaction affected when I add:

1) more hydrogen gas? Explain.

2) argon gas? Explain.

(Assuming volume and temperature remains constant in both)

Thanks in advance.

[jgoudie]

Yo,
A quick one for clarification.

How is the rate of the forward reaction affected when I add:

1) more hydrogen gas? Explain.

2) argon gas? Explain.

(Assuming volume and temperature remains constant in both)

Thanks in advance.

Interesting question.

1) the rate of the back reaction will increase instantaneously.  The backwards reaction would then slightly decrease then the rate of forward reaction will slowly catch up.  In the end the rate of both reactions will increase as the concentration of ethene and hydrogen would end up larger than they were.

2) the rates would be unchanged as the concentration of substances have not changed, volume stays the same.  The addition of an inert gas does not affect the reaction.

Hope this helps and is clear.