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Chemistry Paper 2

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Thanks :D... I didn't know we could use this method

So when you use this formula, it is combustion of REACTANTS minus combustion of PRODUCTS in this one?
(compared to formation of products minus formation of reactants)
Yeah.
 
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T
They had written in bold rate of formation of methanol.It had nothing to do with le chatleirs principle.It was simple:in each case it would increase.

Exactly. And what reasons did you write?
By the way, I did not mean using Le Chatelier's principle directly, but the same concepts applied.

A low temperature favours forward exothermic reaction --> Rate of formation increases
A high temperature favours the side where the number of moles are lesser --> Rate of formation increases
The catalyst lowers the activation energy by providing an alternate route for the reaction to take place quickly --> Rate of formation increases
 
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why does argon not react with chlorine?
Ans: The Ea for reaction is very high
will we get full marks for this?
the same question has come in may june 2010 p2 variant 22

The answer was that argon has a complete octet of electrons/has noble gas configuration.
Did they ask us to find the enthalpy change for the reaction in any previous parts? No.
Did you have any evidence to conclude that the reaction of argon with chlorine is an endothermic reaction? No.
 
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Ahan, Im darn confused about that. What I drew was one -COOH group to ester group and the other remains.
This is what wiki says

" It hydrolyzes, producing maleic acid, cis-HOOC–CH=CH–COOH. With alcohols, the half-ester is generated, e.g., cis-HOOC–CH=CH–COOCH3."

^BIG RELIEF ! By it, they mean maleic anhydride which on hydrolysis produces the acid.
Both the COOH group had to react with the alcohol :( to form an ester, I saw it in a past paper, in which there were 3 OH groups and 1 carboxylic acid, and in the marking scheme, they said that all the OH groups have to be esterified. And the stupidest thing is, I had read this the night before the paper. Still I made that mistake :(
 
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Both the COOH group had to react with the alcohol :( to form an ester, I saw it in a past paper, in which there were 3 OH groups and 1 carboxylic acid, and in the marking scheme, they said that all the OH groups have to be esterified. And the stupidest thing is, I had read this the night before the paper. Still I made that mistake :(
Well, here it's slightly different isnt it? And wiki is with us too, let's keep our fingers crossed and hope for the best :)
 
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:( I don't know why I drew the displayed formula instead of skeletal for cis-trans isomerism. How many marks was that part? :(
2, you get one if you've drawn displayed formula right, cuz 1 mark is for the orientation of atoms or groups of atoms
 
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Exactly. And what reasons did you write?
By the way, I did not mean using Le Chatelier's principle directly, but the same concepts applied.

A low temperature favours forward exothermic reaction --> Rate of formation increases
A high temperature favours the side where the number of moles are lesser --> Rate of formation increases
The catalyst lowers the activation energy by providing an alternate route for the reaction to take place quickly --> Rate of formation increases
You are very right!!!
 
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