Question

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w07_qp_1.pdf
Q33

Answer 1

can anyone explain option 3 please

Answer 2

is this because the compound will decompose only under certian pressure and temperature

Answer 3

The equilibrium constant (Keq) gives you information about where your reaction will be at equilibrium (i.e. the relative amounts of reactants vs. products). The larger Keq is, the more strongly the forward reaction is favoured (i.e. formation of products). It is calculated FROM the rate of the forward and reverse reactions. This means the forward rate alone can't depend on Keq, since Keq depends on the forward rate!

I would say the activation energy is the answer to this question, because this represents the main barrier to the forward reaction...well...going forward. The enthalpy change can tell you whether the reaction is exothermic or endothermic, but on it's own it doesn't really affect forward reaction rate (you would have to consider free energy as well to make any speculations).

Answer 4

Actually just to add to the bit about enthalpy, even with the free energy, that only tells you how spontaneous the reaction is - it gives no information about reaction rates (a spontaneous reaction could be very fast or very slow depending on the specific reaction).

Answer 5

reaction is endothermic like any decomposition but what you are saying is quite not related to the question what i think is that beacause the internal pressure inside the pressure and the atmospheric pressure are equal

Answer 6

internal pressure=atmospheric pressure

Answer 7

Oops! I was looking at Q33 from the other PDF you posted - didn't realize this was a different one! Funny how both talk about equilibrium...

Answer 8

coincidences everywhere

Answer 9

Anyways for this Q33, #1 is wrong because in this case, at equilibrium the gas pressure will be EQUAL to atmospheric pressure.

#2 is correct because that's just one of your Le Chatelier principles.

#3 is correct as well, you just need to think about what happens. If you shift towards the reactants, you have fewer moles of gas particles. Therefore, if volume didn't change, the pressure inside would decrease. However, we know that at equilibrium, the gas pressure will be equal to the atmospheric pressure (this is what determines the movement of the plunger). Therefore, with fewer moles of gas, the volume will need to DECREASE in order to bring the gas pressure back up to balance atmospheric pressure.

Answer 10

if the volume was constant then the only variable controlling equilibrium will be both temperature and pressure, therefore i should decrease volume to increase pressure and equilbrium would shift towrds Pcl5 but if o dissociation had occur how would the volume be greater if thats true than pressure in the internal gas would decrease and would be pulled out by atmoshpheric pressure

Answer 11

Temperature we don't have to worry about because we assume it doesn't change (the question talks about a specific temperature). Option #3 is talking about what happens when your moles of gas is lower to begin with (in other words, we aren't actively changing volume to shift the reaction toward reactants - we're just saying what if PCl5 didn't dissociate). Since we have fewer moles of gas, they'll need to occupy a smaller volume in order to achieve the same pressure.

Answer 12

thank you so much for your help