Today was a normal friday in chem class. After we all asked matt for a cupcake, Mr. Lieberman began to explain the lab we were about to do.
Mass of Candle: 78.28g
Mass of Index Card: 2.24g
Volume of Water: 99.2 mL
Temp of Water: 21.22 degrees C
Highest temp of Water: 35.2 degrees C
Final mass of candle: 78.07g
Final mass of index card: 2.28g
There are 12 post lab questions that are due on Tuesday. They are not too difficult if you simply think about them and apply your data.
After we had completed the lab, we took some notes. We learned how to calculate the change in H for the reaction. The equation that you use is ... ΔH (reaction) = ΣΔHfinal(products) - ΣΔHfinal(reactants).
****Remember that H(final) for an element in its standard state is 0****
Here is an example of a problem using this equation.
CALCULATE THE ΔH FOR THE REACTION
4NH(3) + 7O(2) --> 4NO(2) + 6H(2)O
First you have to look in your textbook or online and find the given values for Σ.
NH(4) --> -46KJ/mol (multiply by 4)
O(2) --> 0KJ/mol (multiply by 7)
NO(2) --> 34KJ/mol (multiply by 4)
H(2)O --> -286 KJ/mol (multiply by 6)
Multiply all of the values of Σ by however many moles of it is given in the problem. Then take the multiplied values and put them in the equation. You should get...
(-1716 + 136) - (0 + -184) which equals -1396.
Your answer is is that the ΔH(reaction) = -1396
Have a nice weekend!! The next scribe is... wait. All the name are crossed off. Ummm ill ask Mr. Lieberman on Monday.
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