Thermochemistry Practice:

1.) Calculate DE when q = +47 kJ and w is -32 kJ

2.) Given the following data:

2A + 3B ---> C + D DHo = -64 kJ
E + A ----> C DHo = 100 kJ
A + 0.5D -----> B DHo = 74 kJ

Calculate DHo for the overall reaction:

3A + B-----> E

3.) Calculate DHo for this reaction:

2CH3OH(l) + 3O2 (g) ----> 2CO2 (g) + 4H2O(l)


Answers:

1.) DE = q + w = 47 kJ - 32 kJ = 15 kJ

2.) First, you must figure out how to use the equations above to get what you want. You can see that E will be a product, so the second rxn will be reversed. You can also see that in order to get 1 B on the left side, the bottom rxn will have to be multiplied by two (because then you will have 2B, and you can subtract 2B from each side, leaving 1 B on the left side). So the three reactions we will use are:

2A + 3B ---> C + D
C ------> E + A
2A + D -----> 2B

The net reaction from these three wil be:

3A + B ----> E (The C and Ds will cancel out)

Now, to calculate the heat of formation....

We did not modify the first rxn, so DHo = -64 kJ. The second rxn was flipped, so we must make the DHo negative, giving us DHo = -100 kJ. The last rxn was doubled, so the heat of formation will be doubled as well, leaving us DHo = 148 kJ.

Adding all of these up....

-64 - 100 + 148 = -16 kJ

3.) To do this problem, you should calculate the heat of formation of all the reactants and subtract it from the heat of formation of the products. Looking at your thermodynamic data, you can see that methanol has a heat of formation of -239 kJ/mol, while oxygen has a heat of formation of 0. Carbon dioxide has a heat of formation of -393.5 kJ/mol and water has a heat of formation of -286 kJ/mol.

So, to find the heat of formation of products....

(-393.5 kJ/mol * 2) + (-286 kJ/mol * 4) = -787 - 1144 = -1932 kJ/mol heat of formation of products.

Reactants: (-239 kJ/mol * 2) = -478 kJ/mole heat of formation of reactants

-1932 kJ/mol - (-478 kJ/mol) = -1453 kJ/mol

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