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#1
July 9th, 2005, 16:26
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Hess's Law
Hi,
Do you guys know of a place where I can find the equations to find Hess's laws and how to calculate- heat of neutralization per mole of water formed? Thanks in advance if anyone can help. You guys have been very helpful. |
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#2
July 10th, 2005, 07:54
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Hess's Law
Hess's law doesn't have specific equations, it merely states that the difference in energy between two states is immaterial of the route taken to arrive at the states.
i.e going from A to B will involve the same energy change as going from A to C to D to B It is a statement of the law of conservation of energy. Its utility is that if a particular energy change cannot be directly found by experimentation it can be deduced using another route. This is done by adding and subtracting, multiplying and dividing equations (doing the same to the energy) until the required equation is produced. see: http://www.ibchem.com/IB/ibc/energy/energy(hl).htm While I'm here does anyone know the official definition of Coles Law? (mini punto for the correct answer) |
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#3
July 10th, 2005, 13:48
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Re: Hess's Law
Quote:
[Old King Cole was very fond of the new cabbage salad his chef invented. In fact, he wanted the whole kingdom to share in his pleasure. Therefore, he sent out a decree that whenever anyone ate cabbage, it must be shredded and mixed with mayonnaise and bits of carrots. This is known today as Cole's Law.] |
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#4
July 10th, 2005, 15:39
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Another silly question
How does the Heat neutralization of Hcl and base and CH3COOH and base differ? I got very similar results for both- 656 Cal/mole and -650 Cal/mole. What does this mean? I thought they would be different!??
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#6
July 10th, 2005, 19:12
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Quote:
The value for strong acid strong base = 57kJmol-1 water formed (your values are very low suggesting either experimental or calculations errors - in calories it is 57000 / 4.186 = 13619 Calories) In reality there is very little observable difference between weak acids and strong bases as the usual experimental errors mask the difference in enthalpy. It is not surprising that you found no difference when your margin of error was approx 95% !!! Experimental errors arise from 1. heat lost to the environment by radiation, conduction and convection 2. heat used to change the temp of the thermometer and container 3. difference between shc of water and solutions used 4. volume and mass measurement inaccuracy 5. thermometer inaccuracy but I fear that with such a large error you have made a more fundamental mistake in the calculations (correcting for 1 mole of water formed perhaps?) |
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#7
July 10th, 2005, 20:07
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Hi,
I calculated neutralization by using density of 0.5 m NaCl as 1.02 g/mL, heat capacity as 0.96 cal/g and total heat of reaction. Is this incorrect? Is this right?.... 102g x 0.96 cal/g x 6.4 (change in temp) x 18.02 mole = 11292cal/mole Should I multiply 1.02 x 100 mL (total solution)? |
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#8
July 11th, 2005, 03:57
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Your answer to the first post was significantly different to the 11292C/mol quoted here.
The equations are correct although your use of density seems a little extravagent - it would be far easier to actually weigh the solution! E = mass x shc x temp rise Then calculate the number of moles of water formed in your reaction (or the number of moles of H+ used) an divide your answer in part 1 by this number. Using your value above 11292C/mol you are now in the ball park (17% error) and this remaining error can be explained by the heat losses I previouly outlined. You should now be able to see some variation between strong and weak acids with the weak having perhaps a decrease of 5-10% with respect to the strong acids. |
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