Tough Equilbrium Questions

[tcg93]

Questions 5 and 6 refer to the following information.

The reaction between water vapour and carbon monoxide can be represented by the chemical equation:

H2O(g)  +  CO(g)     H2(g)  +  CO2(g)    ΔH = -41 kJ mol-1

The value of the concentration ratio of hydrogen gas to water vapour, [H2(g)]/[H2O(g)], at equilibrium for a given set of conditions is 3.5.


6   Compared with the set of conditions used, what effect would doubling the pressure and maintaining the same temperature have on the equilibrium concentration ratio of hydrogen gas to water vapour?

A.    The value of the concentration ratio would double.
B.    The value of the concentration ratio would be higher.
C.    The value of the concentration ratio would be lower.
D.    The value of the concentration ratio would be the same.

(Answer is D - How would one be sure that the ratio does not change - I thought it was K that did not change?)

_______________________________________________________________

9.   A diabetic person sometimes produces species that cause the blood pH to be lowered.   The body automatically attempts to correct the pH.   Part of this correction is achieved by an increase in the breathing rate to enable more carbon dioxide to be expelled from the lungs.   The following reactions are involved:

Reaction 1   CO2(g) ⇌ CO2(aq)
Reaction 2   CO2(aq) + H2O(l) ⇌ H2CO3(aq)
Reaction 3   H2CO3(aq) + H2O(l) ⇌ H3O+(aq) + HCO3-(aq)

For the pH to be raised in this way, the equilibrium position for:
A.    Reaction 3 only must shift to the left.
B.    Reaction 3 only must shift to the right.
C.    all reactions must shift to the left.
D.    all reactions must shift to the right.

(Answer C - How can you be sure that the 2nd equation goes left/right since you are increasing H2O and H2CO3 from the 3rd equation?)

[vexx]

Q6.
The question is basically asking, what affect would the doubling on pressure have on the ratio of products to reactants, yes? The amount of products = the amount of reactants, as there are two compounds on each side, so when the pressure is increased, the change in concentration of both will change evenly, as there is not a dominant side, so the ratio remains the same = D

Q9.
It's asking for the pH to be raised, and according to the reaction to produce H30+ ions
CO2(g)-->CO2(aq)--->H2CO3(aq)---> H30+
(with water along the way)
so if you want to reduce the number of H30+, which causes an increased pH, youd want to do the reverse, so make sure everything goes to the left, to reverse this process of producing H30+ ions.
Hence all to the left, =C

Make sense?

[tcg93]

Q6.
The question is basically asking, what affect would the doubling on pressure have on the ratio of products to reactants, yes? The amount of products = the amount of reactants, as there are two compounds on each side, so when the pressure is increased, the change in concentration of both will change evenly, as there is not a dominant side, so the ratio remains the same = D

Q9.
It's asking for the pH to be raised, and according to the reaction to produce H30+ ions
CO2(g)-->CO2(aq)--->H2CO3(aq)---> H30+
(with water along the way)
so if you want to reduce the number of H30+, which causes an increased pH, youd want to do the reverse, so make sure everything goes to the left, to reverse this process of producing H30+ ions.
Hence all to the left, =C

Make sense?

thanks...

Q6 - didnt pick up it was 2:2 mole ratios

Q9 - so just when u raise the pH => decrease the [H3O+], which increases both [CO2] and [H2O], why isnt the reaction favoured both left and right since a reactant [] has increased and a product [] has increased?

+1 karmaed

[vexx]

^ not quite sure what you mean for q9.
because its asking for the pH to be raised, which all has to go to the left if you follow along, as all is a net back reaction

Reaction 1   CO2(g) ⇌ CO2(aq)
Reaction 2   CO2(aq) + H2O(l) ⇌ H2CO3(aq)
Reaction 3   H2CO3(aq) + H2O(l) ⇌ H3O+(aq) + HCO3-(aq)

A change as occured that has decreased CO2 lets say, so a net back reaction has occurred to account for this, and now since CO2 has decreased, reaction 2 attempts to increase CO2 by driving reaction 2 backwards, which now H2Co3 has decreased, so reaction 3 tries to account for this by a net back reaction.
this results in decreased H30+, hence increased pH.
therefore all are driven to the left.
is that okayy now?

[tcg93]

^ not quite sure what you mean for q9.
because its asking for the pH to be raised, which all has to go to the left if you follow along, as all is a net back reaction

Reaction 1   CO2(g) ⇌ CO2(aq)
Reaction 2   CO2(aq) + H2O(l) ⇌ H2CO3(aq)
Reaction 3   H2CO3(aq) + H2O(l) ⇌ H3O+(aq) + HCO3-(aq)

A change as occured that has decreased CO2 lets say, so a net back reaction has occurred to account for this, and now since CO2 has decreased, reaction 2 attempts to increase CO2 by driving reaction 2 backwards, which now H2Co3 has decreased, so reaction 3 tries to account for this by a net back reaction.
this results in decreased H30+, hence increased pH.
therefore all are driven to the left.
is that okayy now?

okk this is what I'm trying to say

You have raised the pH
Therefore you have decreased [H3O+]
This creates a net back reaction for Reaction 3, and therefore concentrations of both H2CO3(aq) + H2O(l) must increase

1st point - Obviously an increase in concentration of H2CO3(aq) will create a net back reaction for Reaction 2

2nd point - but an increase in concentration of H2O(l) will also create a net forward reaction for Reaction 2

so thats what I was asking.... why is it the 1st point and not the 2nd point?

[vexx]

Oh i think i know what you are saying.

Yes increase in H2Co3 will create a net back reaction which increases CO2 and H20, so this leads to the netback reaction in reaction 1 due to increased CO2,
but your asking why does H20 that increase, not cause a net forward reaction?

Well, in an equilirbium reaction,
A + B --> C
If C is increased, that means A and B will also increase, and this will result in equilibrium being formed. There is no need to further decrease the increased B, because you've aready established equilibrium despite more B.
it's just how it works, so lost some C, to increase the other side to get equilibrium again, and then its done, no more changes need to occur.
this is the same with the increase in water, it doesnt need to change anymore as its already increased to the left to establish equilibrium, anymore changes would only occur if more water was added, not since it was done by the system itself to actually get the equilibrium state.
all good now xD?

[tcg93]

Oh i think i know what you are saying.

Yes increase in H2Co3 will create a net back reaction which increases CO2 and H20, so this leads to the netback reaction in reaction 1 due to increased CO2,
but your asking why does H20 that increase, not cause a net forward reaction?

Well, in an equilirbium reaction,
A + B --> C
If C is increased, that means A and B will also increase, and this will result in equilibrium being formed. There is no need to further decrease the increased B, because you've aready established equilibrium despite more B.
it's just how it works, so lost some C, to increase the other side to get equilibrium again, and then its done, no more changes need to occur.
this is the same with the increase in water, it doesnt need to change anymore as its already increased to the left to establish equilibrium, anymore changes would only occur if more water was added, not since it was done by the system itself to actually get the equilibrium state.
all good now xD?

THX heaps, too bad i cant karma u more