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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: sallyhansen on May 10, 2013, 07:47:14 PM

Title: Chemistry Exam Review
Post by: sallyhansen on May 10, 2013, 07:47:14 PM
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?
Title: Re: Chemistry Exam Review
Post by: UG on May 10, 2013, 08:09:27 PM
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?
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 10, 2013, 09:17:17 PM
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
Title: Re: Chemistry Exam Review
Post by: UG on May 10, 2013, 09:27:09 PM
You don't need to flip the first two equations, the only one you flip is the third
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 10, 2013, 10:14:54 PM
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?
Title: Re: Chemistry Exam Review
Post by: UG on May 10, 2013, 10:56:31 PM
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
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 10:34:10 AM
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
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 01:56:23 PM
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
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 02:17:51 PM
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
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 02:21:47 PM
and my answer to number 7 is attached
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 02:24:37 PM
And for 8-10 I am completely lost on how I should solve them.
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 02:54:58 PM
Another question for number 11 how come they have a charge of 2e- and not 3e- like I did in my answer
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 02:58:33 PM
This is how I answered number 11. I put 3e- and I am not understanding why it is 2e-
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 03:30:47 PM
Also can anyone tell me what equation is used for this question
Title: Re: Chemistry Exam Review
Post by: UG on May 11, 2013, 08:07:54 PM
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
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 11, 2013, 09:33:49 PM
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

For number 5 I actually don't know what n and m represent
Title: Re: Chemistry Exam Review
Post by: iamback on May 11, 2013, 11:56:31 PM
^^Do you remember the general rate law ?

Rate = k[X]m[y]n

where is k is rate constant.

Here n is the coefficient of [OH], and m is coefficient of [ClO2] in rate law.

Can you now solve it on your own ?
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 12, 2013, 03:13:23 PM
^^Do you remember the general rate law ?

Rate = k[X]m[y]n

where is k is rate constant.

Here n is the coefficient of [OH], and m is coefficient of [ClO2] in rate law.

Can you now solve it on your own ?

For a question like this how would I calculate for X
1.6 × 10-48 = 4× 3

The answer is × = 7.4 × 10-17 mol/L but how do I get rid of an exponent with a 3É
Title: Re: Chemistry Exam Review
Post by: iamback on May 12, 2013, 10:18:47 PM
You don't have calculate of for X. You are given the values of X and Y here in the question.

X and Y are the concentrations of [ClO2] and [OH] respectively.

Form equations with the data given and solve.

I am forming one for you.

Rate = k[X]m[Y]n

0.18 = k[0.10]m[0.10]n

Now make equations like this and solve for m, n and k.
Title: Re: Chemistry Exam Review
Post by: sallyhansen on May 13, 2013, 03:02:06 PM
You don't have calculate of for X. You are given the values of X and Y here in the question.

X and Y are the concentrations of [ClO2] and [OH] respectively.

Form equations with the data given and solve.

I am forming one for you.

Rate = k[X]m[Y]n

0.18 = k[0.10]m[0.10]n

Now make equations like this and solve for m, n and k.

No the above question was a different question entirely
Title: Re: Chemistry Exam Review
Post by: iamback on May 14, 2013, 12:16:33 AM
I solved it, and got the right answer, which you have mentioned in red letters.