Unit 4 Questions MEGATHREAD!

[ruchika5]

This is probably really simple, but when using energy profile diagrams, how do you tell the activation energy for a reverse reaction?? thanks.

[Andiio]

This is probably really simple, but when using energy profile diagrams, how do you tell the activation energy for a reverse reaction?? thanks.

On a graph, it's the vert. distance between the reactants and the top of the curve, if that makes sense!

[b^3]

^That would be the reactants of the reverse reaction, i.e. the products of the forward reaction.

[Andiio]

^That would be the reactants of the reverse reaction, i.e. the products of the forward reaction.

Orient the graph either way (i.e. exo and then 'flip' it, reversing it into an endo) and then the Ea would be the distance between the reactants and the top curve? If you get what I mean

[ruchika5]

Hmm okay... I've attached a question from the 2005 VCAA exam and I'm confused the way they've worked it out.

The question was " Give the activation energy for the reverse reaction in kJ mol" and looking at the examiner's report the answer it was 360.

[b^3]

the activation energy would be thee energy of the peak minus the energy of the reactants (as this is the energy needed to be gained) here which would be -40-(-400) which is 360kJmol^-1

[meli001]

2SO2 + O2 <===> 2SO3 (exothermic)

An equilibrium mixture of these gases was made by
mixing sulfur trioxide and carbon dioxide. It consisted of
0.028 mol of CS2, 0.022 mol of SO3 and 0.014 mol of CO2
in a 20 L vessel. Calculate the value of the equilibrium
constant at that temperature.

So... how do I calculate the concentration of oxygen?

[Vincezor]

An equilibrium mixture of these gases was made by
mixing sulfur trioxide and carbon dioxide. It consisted of
0.028 mol of CS2, 0.022 mol of SO3 and 0.014 mol of CO2
in a 20 L vessel. Calculate the value of the equilibrium
constant at that temperature.

So... how do I calculate the concentration of oxygen?

Okay it's probably a good idea to make a table for this question.

It's a bit hard to do it this forum, so I'll just explain maybe what you could do.

I hope this is the equation (to be honest, i've never seen this reaction before )



So anyway, here's my attempt. Do note that I am kinda confused at the whole question, and this is how I interpreted it? :S

I realise there are many unnecessary steps, and you'll probably notice a lot of mistakes (all that erasing haha), so please correct me if I'm wrong!




EDIT: Sorry, quite tired, and sorry for the messy handwriting in advance

[meli001]

Thanks! It's right, the table made more sense than the worked solutions.

[Graphite]

why is it that the conjugate of a strong acid is a weak base HOWEVER, the conjugate of a weak acid is not always a strong base?

[Mao]

why is it that the conjugate of a strong acid is a weak base HOWEVER, the conjugate of a weak acid is not always a strong base?

Conjugate of strong is neutral. Conjugate of weak is weak.

e.g. conjugate of strong acid HNO3 is the neutral nitrate ion. etc.

[nacho]

i have a question in regards to lechat and his principles
i understand that a system will try to oppose and external change imposed upon it, in order to reach equilibrium
say we have the following in a closed environment: 5A + 2B <----> 3C
what i am confused about is :
By decreasing the volume, we increase pressure. In the adjustment period, it will be seen that the concentration of C will gradually increase because "the system will favour the reaction with fewer particles if the pressure is increased/volume is decreased".
But what i don't get it, if we have the same amount of everything in that closed environment, how is it that the pressure will change at all? Why does a (forward reaction in this case) make any difference to the pressure? I hope i am clear about my question, if not ill rephrase it

[vea]

i have a question in regards to lechat and his principles
i understand that a system will try to oppose and external change imposed upon it, in order to reach equilibrium
say we have the following in a closed environment: 5A + 2B <----> 3C
what i am confused about is :
By decreasing the volume, we increase pressure. In the adjustment period, it will be seen that the concentration of C will gradually increase because "the system will favour the reaction with fewer particles if the pressure is increased/volume is decreased".
But what i don't get it, if we have the same amount of everything in that closed environment, how is it that the pressure will change at all? Why does a (forward reaction in this case) make any difference to the pressure? I hope i am clear about my question, if not ill rephrase it

If it was say 5A+2B<--->7C then there will be no change if the pressure is changed.

[kenhung123]

You need to understand that it is the particles themselves who create the pressure in the first place. Less particles=release some pressure.

[nacho]

You need to understand that it is the particles themselves who create the pressure in the first place. Less particles=release some pressure.

but doesnt the size of the particle have any part to play?