[sallyhansen]

Okay I have a question on this exam question
1.   The combustion of methanol is shown by the following equation:(4  marks)

You can see it in the attachment it is the very first question

i.   Given the data which follows:
a.   Find the heat of reaction for the equation above. 478 kJ
b.   State the molar heat of combustion of methanol. 239 KJ
c.   State whether the reaction is endothermic or exothermic. Exothermic

 
What I do not get is how 393 kJ and 242 will be come negative if they have not been flipped?

[UG]

What I do not get is how 393 kJ and 242 will be come negative if they have not been flipped?

But they didn't become negative, what were the figures in your calculation?

[sallyhansen]

What I do not get is how 393 kJ and 242 will be come negative if they have not been flipped?

But they didn't become negative, what were the figures in your calculation?

the one highlighted in red are the answer, but for the equation to be flipped would cause it to be negative

[UG]

You don't need to flip the first two equations, the only one you flip is the third

[sallyhansen]

You don't need to flip the first two equations, the only one you flip is the third

Then how come it needs to be negative KJ for the first 2 equations?

[UG]

It seems that there is a bit of misunderstanding here. This is what I gave gathered so far from your question, the correct answer to the first part is 478 kJ. You are not getting this answer correct and you want to know why, am I correct? You keep mentioning that '393 kJ and 242 kJ will be negative', do you have some sort of answer sheet which is telling you this or where are you getting this from? The way I see it, the problem can be solved by multiplying reaction one by 2, reaction two by 4 and flipping reaction three and multiplying by 2. This would give 2 x 393 + 242 x 4 - 638 x 2 which gives 478 kJ

[sallyhansen]

It seems that there is a bit of misunderstanding here. This is what I gave gathered so far from your question, the correct answer to the first part is 478 kJ. You are not getting this answer correct and you want to know why, am I correct? You keep mentioning that '393 kJ and 242 kJ will be negative', do you have some sort of answer sheet which is telling you this or where are you getting this from? The way I see it, the problem can be solved by multiplying reaction one by 2, reaction two by 4 and flipping reaction three and multiplying by 2. This would give 2 x 393 + 242 x 4 - 638 x 2 which gives 478 kJ

   ANS:   
i.   a.   

   =[(2  –393 kJ) + (4  –242 kJ)] – [2  –638 kJ]

   = –478 kJ

[sallyhansen]

It seems that there is a bit of misunderstanding here. This is what I gave gathered so far from your question, the correct answer to the first part is 478 kJ. You are not getting this answer correct and you want to know why, am I correct? You keep mentioning that '393 kJ and 242 kJ will be negative', do you have some sort of answer sheet which is telling you this or where are you getting this from? The way I see it, the problem can be solved by multiplying reaction one by 2, reaction two by 4 and flipping reaction three and multiplying by 2. This would give 2 x 393 + 242 x 4 - 638 x 2 which gives 478 kJ

Okay I understood the approach now, I was plugging it into the calculator incorrectly, but I may I have clarification on one more question

[sallyhansen]

It seems that there is a bit of misunderstanding here. This is what I gave gathered so far from your question, the correct answer to the first part is 478 kJ. You are not getting this answer correct and you want to know why, am I correct? You keep mentioning that '393 kJ and 242 kJ will be negative', do you have some sort of answer sheet which is telling you this or where are you getting this from? The way I see it, the problem can be solved by multiplying reaction one by 2, reaction two by 4 and flipping reaction three and multiplying by 2. This would give 2 x 393 + 242 x 4 - 638 x 2 which gives 478 kJ

Also for number 7 the question is the one posted

[sallyhansen]

and my answer to number 7 is attached

[sallyhansen]

And for 8-10 I am completely lost on how I should solve them.

[sallyhansen]

Another question for number 11 how come they have a charge of 2e- and not 3e- like I did in my answer

[sallyhansen]

This is how I answered number 11. I put 3e- and I am not understanding why it is 2e-

[sallyhansen]

Also can anyone tell me what equation is used for this question

[UG]

Can you show us what you did for 5? Did you manage to set up the general rate law equation? Do you know what n and m represent?
Regarding question 7, you need to put in the equilibrium concentrations when solving for K, i.e. K = x/([0.225-x][0.150-x])
I do believe there is a typo in question 11 as the answer has Pb²⁺ while the question has Pb₂⁺ so either the question is wrong or the answer is wrong, I reckon it is the question. There is nothing much you can do about that, perhaps try the balance again with Pb²⁺ and see if you get the right answer