ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: m@tty on September 20, 2010, 04:55:45 pm
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A few things to clear up.
Multiple choice question 3.
Which of the following changes will always ensure an increase in the rate of a chemical reaction?
I Adding a catalyst
II Increasing temperature
III Increasing concentration
IV Increasing the surface area of the reactants
The solutions give B(I, II and III only) as the correct answer, professing that IV (which they claim is "adding more reactants") will not increase the rate.
Increasing surface area does increase rate, does it not? So all 4 of the changes will result in an increased rate.
Multiple choice question 20.
Which of the following will cause the calculated enthalpy value for an exothermic reaction carried out in a bomb calorimeter to be lower than the actual value?
I ...
II a thermometer that reads 3oC lower than the actual temperature
III ...
IV ...
The answer which they gave did not include II as they assumed that the same thermometer was used in calibration. This is an unreasonable assumption, isn't it?
Short answer 3.c.
}_{yellow} + SCN^-(aq) \rightleftharpoons \underbrace{FeSCN^{2+}(aq)}_{dark \ red})
They claim that if
is added the solution becomes darker red only.
However, I think that it will become both darker in redness and also more yellow. According to Le Chatelier's principle, there will be higher concentration of both Iron(III) and Iron(III) Thiocyanate at the new equilibrium.
I don't know how the actual colour of the solution will appear with more yellow, but it should be acknowledged, should it not?
Thanks.
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MCQ 3
Increasing surface area only increases the rate for solid and liquid reactants, not ones in gaseous form as they are completely miscible. Therefore, it does not always increase rate.
Short answer 3c)
If u add
, you increase the concentration of
, hence equilibrium will shift to the right in order to partially oppose the increase in concetration, resulting in more production of
and hence, a more darker colour.
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Yes, but you can't exactly increase the surface area of a gas, can you. The way I see it a gaseous substance is already at its maximum surface area. Therefore whenever you increase surface area, rate must also increase.
Again, this is true. But at the new equilibrium there is more Iron(III). Should the solution not, therefore, be more yellow also?
There were four boxes to choose from: No change, Clear, More yellow, Darker red.
By my logic both "more yellow" and "darker red" should be ticked. But the solutions only chose "darker red".
Sorry if this is too overly repetitive...
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Yes, but you can't exactly increase the surface area of a gas, can you. The way I see it a gaseous substance is already at its maximum surface area. Therefore whenever you increase surface area, rate must also increase.
Again, this is true. But at the new equilibrium there is more Iron(III). Should the solution not, therefore, be more yellow also?
There were four boxes to choose from: No change, Clear, More yellow, Darker red.
By my logic both "more yellow" and "darker red" should be ticked. But the solutions only chose "darker red".
Sorry if this is too overly repetitive...
I think it would initially become more yellow and then old Le Chatelier will kick in and then the red would intensify... in doing so the 'more yellow' would fade into red.
If that makes sense
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But it'd still be more yellow than it was initially, yes?
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But it'd still be more yellow than it was initially, yes?
That is correct but I'm guessing that the increase to a more darker colour is greater than the increase to more yellow, therefore overall it has gone more darker.
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Is it too pedantic to say both "more yellow" and "darker red"?
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And for the first question, i think they wrote the question wrong, in the answers, it refers to IV as "Adding additional reactant"
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Yeah. That too. The solutions for this year is pretty terrible...
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Yeah. That too. The solutions for this year is pretty terrible...
Ehh true, still better than a lot of other companies.
Gotta love insight for their MC explanations <33
I made all the same mistakes you did so we have to be doing something right here..
With 3c) I agree, it should be more yellow as well, the reaction will only partially oppose itself
Also, if it's diluted, won't it become less yellow because the concentration decreases?
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Is it too pedantic to say both "more yellow" and "darker red"?
I agree with you, i think overall is would still be darker yellow since its 'partially' oppose so the amount of yellow should be more than what is being reduced by mr chat
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What about short answer 3) b) ?
An increase in temperature should technically increase the pressure in the vessel, because there is a greater frequency of collisions with the sides of the vessel. Right?
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The pressure of the vessel is depended on the volume, not the temperature.
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I think colour should be directly proportional to concentration, but when mixing of colours comes into play it is annyoing
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The pressure of the vessel is depended on the volume, not the temperature.
What about charles' law? That pressure is directly proportional to temperature?
(You are referring to boyle's law here).
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The partial pressures are unaltered
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Doesn't increasing temperature decrease reaction rates in one of the Nitric acid production steps?...
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decrease yield
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Oxidation of nitrogen monoxide
"unlike many reactions, this reaction is unusual in that its rate increases with decreasing temperature." Heinemann Chemistry
???
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Wow that's a first
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I know, its very odd.
Just thought I'd point it out as the question clearly states "always".
I'd be tempted (as I studied Nitric Acid) to not include that, but then I'd be wrong.
being very picky though...
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Chemistry: There's an exception for EVERYTHING
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could someone please clarify for me why one
Q5 C
for the ratio its 4 x 10^3 /6.99 x10^3
i dnt get where the 4X10^3 comes from :S
THANKS :D