Question

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Answer 1

You are asked to calculate the standard enthalpy change of

\[2Al(s) + Fe_2O_3 \rightarrow 2Fe(s) + Al_2O_3(s) \]

In reaction #1 (in which you were given Delta H as -1601 kJ/mol) Al2O3(s) is on the product side and it is also on the product side of the equation you are asked to find Delta H for.

In reaction # 2 (in which Delta H you were given is -821 kJ/mol) Fe2O3(s) is on the product side, however for the equation you are asked to solve Delta H for Fe2O3 is on the reactant side.

So what do you need to do with reaction # 2?

Answer 2

Would you need to combine equations?

Answer 3

Before combining equations. You need to do something to equation #2 so that it will match the equation 2Al(s)+Fe2O3→2Fe(s)+Al2O3(s)

Answer 4

Balance it?

Answer 5

It is balanced.

Answer 6

Then I'm not sure then.

Answer 7

If you move Fe2O3 to the reactant side what do you need to do to the rest of the equation (#2)?

Answer 8

Move it as well? Make it cancel out?

Answer 9

If you move Fe2O3 to the reactant side, that means the Fe2O3 will decompose (come apart) the product then will be 2Fe + 3/2 O2. So if to create the Fe2O3 the enthalpy is -821kJ/mole if we are decomposing it the enthalpy is reversed, so it is +821kJ/mole.

Can you now rewrite both equations #1 keep as it is but #2 reverse?

Answer 10

2Al+Fe2 + O3--> Fe2O3+ 2Al

Answer 11

For now, keep the 2 equations separate, the first equation write as is; the second equation write beneath the first and reverse it (products on reactants side and reactants on product side).

Answer 12

2Al+Fe2O3--> 2Fe+Al2O3
Fe2O3-->2Fe+ (3/2) O2

Answer 13

Like that?

Answer 14

The second one you did correct. The first should be the same way it is in the original problem.

2Al+Fe2O3--> 2Fe+Al2O3 incorrect
Fe2O3-->2Fe+ (3/2) O2 correct

Answer 15

2Al+(3/2) O2 --> Al2O3
Fe2O3 --> 2Fe + (3/2) O2

Answer 16

Good. Now just add after each equation the enthalpy (Delta H values) remember you need to reverse the value for the 2nd equation.

Answer 17

As in it's positive?

Answer 18

Correct

Answer 19

-1601 kJ/mol + 821kJ/mol

Answer 20

=-780 kJ/mol

Answer 21

Write it after the equation. (You can just copy and paste the equation from your previous post.)

Answer 22

2Al+(3/2) O2 --> Al2O3 -1601kJ/mol
Fe2O3 --> 2Fe + (3/2) O2 821kJ/mol

Answer 23

Now add everything together; all reactants on the left and all products on the right. The delta H value should be the sum of the 2 (as you had done 2 posts ago).

Answer 24

2Al+Fe2O3+(3/2) O2--Al2O3+(3/2) O2 -780 kJ/mol

Answer 25

You're missing 2Fe

Answer 26

2Al+Fe2O3+(3/2) O2--2Fe + Al2O3+(3/2) O2 -780 kJ/mol

Answer 27

Good. Now cancel out everything that is the same on both sides.

Answer 28

2Al+Fe2O3-->2Fe+Al2O3 -780kJ/mol

Answer 29

You have your answer, technically you need to put the state of the elements and compounds (after the element or compound). You also need to write your answer in scientific notation.

Answer 30

So 7.80 x 10^2

Answer 31

is it negative or positive?

Answer 32

-7.80 x 10^2

Answer 33

Correct

Answer 34

Thank you!!

Answer 35

You're welcome.