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October 20, 2025, 12:46:36 am

Author Topic: question help required  (Read 2669 times)  Share 

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tarquin008

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question help required
« on: August 22, 2008, 10:05:12 pm »
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i found this question in my textbook and i couldnt figure this. 

upon dissolving in water, a sugar a-d-glucose undergoes conversion into an isomer called b-d-glucose. this process is called mutarotation and reaches equilibrium when 63.6% of the original a-d-glucose has been converted. calculate the value of the equilibrium constant for this process.

by the way, the a and b shown in front of the glucose ( i.e. a-d-glucose, b-d-glucose ) are actually the greek letters alpha and beta.

any help much appreciated

Mao

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Re: question help required
« Reply #1 on: August 22, 2008, 10:13:07 pm »
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if we assume a 100M initial concentration in a 1L vessle: (i know this is not realistic :P )

(no unit)
« Last Edit: August 22, 2008, 10:16:27 pm by Mao »
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tarquin008

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Re: question help required
« Reply #2 on: August 22, 2008, 10:18:58 pm »
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yeah, the answer given was 1.74

thanks dude

lanvins

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Re: question help required
« Reply #3 on: August 23, 2008, 10:19:45 am »
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another question,

A temperature rise of 1.78oC was observed when 1.00*10^-3 mol of propane gas was burnt in a calorimeter. The calibration factor of the calorimeter was previouly determined to be 1250J oC^-1

1. calculate the heat of combustion of propane in kJ mol-1

I did this 1250 J / oC  x   1.78 oC =   2225 J

convert to kJ   2.225 kJ

2.225 kJ / 0.001 mol= 2225kj/mol

.......but the back of the book says its -2.23MJ mol-1, i'm i wrong?

cara.mel

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Re: question help required
« Reply #4 on: August 23, 2008, 10:35:26 am »
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negative sign is there because the reaction is releasing energy (as temperature goes up)

Other than that, your answer is equivilant :)

lanvins

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Re: question help required
« Reply #5 on: August 23, 2008, 10:44:39 am »
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how do you go from -2225kj/mol to -2.23MJ mol-1?

lanvins

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Re: question help required
« Reply #6 on: August 23, 2008, 10:56:43 am »
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oic...i think
« Last Edit: August 23, 2008, 11:00:01 am by lanvins »

cara.mel

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Re: question help required
« Reply #7 on: August 23, 2008, 11:00:37 am »
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1000kJ = 1MJ (k = 10^3, M = 10^6)
And then kJ/mol is the same as kJ mol^-1

lanvins

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Re: question help required
« Reply #8 on: August 23, 2008, 11:01:44 am »
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lol...yea i get it...thanks

lanvins

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Re: question help required
« Reply #9 on: August 23, 2008, 01:43:28 pm »
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more questions,

An oxy-acetylene welding torch uses ethene (acetylene) at the rate of 500ml per minute, measured at STP. The reaction is 2C2H2 + 502---> 4CO2+ 2H2O ; H= -2599kj mol-1

1. What is the heat of combustion of ethyne in kj mol-1 and kj g-1

2. Calculate the rate of enery production by the torch in kj per minute.

lanvins

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Re: question help required
« Reply #10 on: August 23, 2008, 04:14:32 pm »
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no... that's how the book has it

lanvins

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Re: question help required
« Reply #11 on: August 23, 2008, 04:19:56 pm »
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...the back of the book says the answers are; q1. -49.97kj g-1 and q2. 29kj min-1

Mao

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Re: question help required
« Reply #12 on: August 23, 2008, 04:24:01 pm »
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argh, i wasnt thinking in the previous post...


do you mean ethene or ethyne?

if you meant ethyne:

Mr=26








at STP, 1 mol = 22.4 L



hence the heat released woud be:
« Last Edit: August 23, 2008, 05:27:41 pm by Mao »
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Glockmeister

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Re: question help required
« Reply #13 on: August 23, 2008, 04:27:05 pm »
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"this post is more confusing than actual chemistry.... =S" - Mao

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lanvins

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Re: question help required
« Reply #14 on: August 23, 2008, 04:44:33 pm »
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