6.70
A sample of benzene, C6H6, weighing 3.51g was burned in an excess of oxygen in bomb calorimeter. The temperature of the calorimeter rose from 25.00C to 37.18C. If the heat capacity of the calorimeter and contents was 12.05kJ/C, what is the value of q for burning 1.25 mol of benzene at constant volume and 25C? The reaction is
C6H6(l)+15/2O2(g)6CO2(g)+3H2O(l)
Here is my work:
q=c x delta T
12.05x12.18=146.769kg = q calorimeter
So the value of q for burning 1.25 mol of benzene at constant volume and 25C would be
-146.769
I then converted C6H6 to moles 3.51x78.118=.045mol of C6H6.
Then I multiplied that by 1.25=.05625 moles
Then -146.769kj/.05625=-2609.2266 or -2.61e3kj
is this correct?
Your tone in your message seems a little mean spirited, I am trying the best I can. Have you ever tried to do something your not very good at?
I have really been struggling with this class and it does feel like I am just plugging numbers into a calculator. The odds of me passing are slim but I am not a quieter. I heard about this site to help me out and you guys have more then you know, if you guys help me with an answer then I try to look at it and figure out how it makes sense. But Bobs answer was very offensive, I know I am not good at this but I do not need him to make me feel like an idiot, You at least used understandable sentences to express what I am doing wrong. I will try to label my units more and thank you for the constructive criticism.
There are two question in this problem:
a) Find "q" for the sample burned.
b) find "q" for 1.25 moles of C6H6.
Your numerical answer for the first part is correct but you are not paying attention to unit labels. It is mislabeled.
For the second part, to find kilojoules/mole, you DIVIDE kilojoules by moles. If you LABEL each measurment, that would suggest to you how to use those measurements.
146.77 kJ / 0.0449345 mol = 3270 kJ/mole
(That means 3270 kJ given off by 1 mole)
Now, how do you find the heat released by 1.25 moles of benzene???
BobPursley has helped you and observed you longer than I have. He is trying to tell you something that could benefit you: Pay more attention to the strategy and logic of a solution. That includes unit labels. None of us are good at something, or anything, unless we are completely focused on it.
Your approach to solving the problem is correct, but there are some mistakes in your calculations. Let's go through the steps again:
1. Calculate the heat absorbed by the calorimeter (q calorimeter):
q calorimeter = C x ΔT
q calorimeter = 12.05 kJ/C x (37.18 C - 25.00 C)
q calorimeter = 12.05 kJ/C x 12.18 C
q calorimeter ≈ 146.469 kJ (not 146.769 kg)
2. Next, you need to calculate the moles of benzene burned from the given mass (3.51 g):
Molar mass of C6H6 = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
Moles of C6H6 = mass of C6H6 / molar mass
Moles of C6H6 = 3.51 g / 78.11 g/mol ≈ 0.045 mol (not 0.045 x 78.118 = 3.517 mol)
3. The next step is to find the moles of benzene burned for the given volume of 1.25 mol:
Moles of C6H6 burned = moles of C6H6 per mol of benzene x volume of benzene burned
Moles of C6H6 burned = 0.045 mol/mol x 1.25 mol ≈ 0.05625 mol (not 0.045 mol x 1.25)
4. To find the value of q for burning 1.25 mol of benzene, you need to use the following equation:
q = -q calorimeter (negative because the reaction is exothermic)
q = -146.469 kJ (not -146.769 kJ)
So the final calculation would be:
q = -146.469 kJ / 0.05625 mol ≈ -2606.293 kJ/mol
Therefore, the correct value of q for burning 1.25 mol of benzene at constant volume and 25 °C is approximately -2606.293 kJ/mol.
I am truly amazed at your ability to punch numbers in your calculator while you have no idea what you are doing.
If .045moles gave 146.7kj
then by proportion...
1.25moles/.045moles= ??/146.7kJ
I assume you can find ?? the heat per 1.25moles.
Now go back an look at your work, and verify it is nonsense. You need to understand what you are doing before you use the calculator.