Homework Problem 8.53

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elizabethrojas1G
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Joined: Fri Sep 25, 2015 3:00 am

Homework Problem 8.53

Postby elizabethrojas1G » Sun Jan 17, 2016 7:26 pm

The reaction of 1.40g of carbon monoxide with excess water vapor to produce carbon dioxide and hydrogen gases in a bmb calorimeter causes the temp of the calorimeter assembly to rise from 22.113C to 22.799C. The calorimeter assembly is known to have a total heat capacit of 3.00 kJ/C. Calculate the internal energy change for the rxn of 1.00 mol CO.


I don't understand what the solutions manual is showing, can someone explain it step by step to me please?

Divya Prajapati 1E
Posts: 40
Joined: Fri Sep 25, 2015 3:00 am

Re: Homework Problem 8.53

Postby Divya Prajapati 1E » Sun Jan 17, 2016 7:39 pm

Based on the given information, we will be using the equation "q rxn = - q cal = - C x delta T". This equation tells us that the heat released in the reaction is the negative of the total heat capacity of the calorimeter times the change in temperature.

The problem gives us C = 3.00 kJ/C and we can calculate delta T = 22.799 - 22.113 = 0.686 C. When we multiply these values, we get q = - 2.058 kJ, meaning the rxn released 2.258 kJ of heat.

To find delta H we use the equation "delta H = q/n" where q is heat of reaction and n is the number of moles. We can find n by converting the given 1.40 grams CO into moles of CO. Therefore, delta H = -2.058 kJ/0.04998 mol CO = -41.2 kJ/mole CO. Because every reactant has a coefficient of 1, the delta H of the whole reaction is still -41.2 kJ/mol.


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