Help with a chemistry problem in how much heat is produced in this reaction?

Question

The reaction is 3 Mg + Fe2O3 = 3MgO + 2Fe using the following data Mg + 1/2 O2 = MgO = -609.6kj and 2Fe + 3/2 O2 = Fe2O3 = -809.6kj.

I don't know how to calculate this so if possible i need someone to tell me step by step how to calculate it thank you

Answer 1

Doi... is correct.

For any reaction,

DeltaH = sum{DeltaHformation(products)} - sum{DeltaHformation(reactants)}

To see why this must be true, you could imagine doing the reaction on paper (not in reality) by converting the reactants to their elements, and then rearranging the elements to give the products.

So this equation is just an example of conservation of energy (or, for the more jargon minded, the fact that H is a state function independent of path)

Answer 2

Heat produced in the reaction = -1019.2 kJ

Here is the work:

Mg + 1/2 O2 = MgO - 609.6 kJ

==> Mg + 1/2 O2 + 609.6 kJ = MgO  .........(*)

This means that a mole of Mg (Magnesium) will react with 1/2 mole of O2 (Oxygen) when 609.6 kJ of heat is applied to get a mole of MgO (Magnesium Oxide).

Multiplying both sides of equation (*) by 3, yields:

3Mg + 3/2 O2 + 1828.8 kJ = 3MgO .........(1)

On the other hand

2Fe + 3/2 O2 = Fe2O3 - 809.6 kJ

==> Fe2O3 = 2Fe + 3/2 O2 +  809.6 kJ  .......(2)

This means that if a mole of Fe2O3 is converted to 2Fe (Iron) and 3/2 O2 (Oxygen), then an amount of heat (809.6 kJ) will be released.

Now, lets add the two previous equations (eq.(1) & eq.(2)) together:

3Mg + 3/2 O2 + 1828.8 kJ = 3MgO         ..............(1)

Fe2O3 = 2Fe + 3/2 O2 +  809.6 kJ         ..............(2)

--------------------------------------...

3Mg + 3/2 O2 + 1828.8 kJ + Fe2O3 =

3MgO + 2Fe + 3/2 O2 +  809.6 kJ

3/2 O2 will cancel each other in both sides

subtracting 1828.8 kJ from both sides:

3Mg + Fe2O3 = 3MgO + 2Fe +  809.6 kJ - 1828.8 kJ

3Mg + Fe2O3 = 3MgO + 2Fe - 1019.2 kJ

The last was your main equation with the term - 1019.2 kJ at the left side. This quantity is the amount of heat produced by this reaction. Since the the heat produced is negative, it's clear that the reactants should absorb 1019.2 kJ to react.

Hope it's clear!

Answer 3

deltaH=deltaH(products)-deltaH(reactans) in your case

deltaH=deltaH(3xMgO) - deltaH(Fe2O3)

elements like Mg and Fe's enthalpies are 0

deltaH=[3x(-609.6)] - (-809.6)