# Determining the heat of a reaction

Determine the heat of reaction for the oxidation of iron,

4 Fe 9 (s) + 3 O_{2 }---> 2 Fe_{2}O_{3 }(s)

given the enthalpies of the reactions below.

2 Fe (s) + 6 H_{2}0 (l) ---> 2 Fe(OH)_{3} (s) + 3 H_{2} (g) change in enthalpy = 321.8 kJ

2 H_{2} (g) + O_{2} (g) ----> 2 H_{2}O (l) change in enthalpy = -571.7 kJ

Fe_{2}O_{3} (s) + 3 H_{2}O (l) ----> 2 Fe(OH)_{3 }(s) change in enthalpy = 288.6 kJ

The answer is -1648.7 kJ, but i don't understand what to do to solve this problem. I need to know what to do starting from the beginning. Thank you for any help.

### You will use Hess's law to

You will use Hess's law to solve this question.Review this topic.

2 Fe (s) + 6 H20 (l) ---> 2 Fe(OH)3 (s) + 3 H2 (g) change in enthalpy = 321.8 kJ---1

2 H2 (g) + O2 (g) ----> 2 H2O (l) change in enthalpy = -571.7 kJ-----2

Fe2O3 (s) + 3 H2O (l) ----> 2 Fe(OH)3 (s) change in enthalpy = 288.6 kJ------3

To get the desired equation

4 Fe (s) + 3 O2 ---> 2 Fe2O3 (s)

Multiply eq -1 with 2 ,eq -2 with 3 and reverse eq 3 and multiply it with 2 ...

you will get new sets of equations

4 Fe (s) + 12 H20 (l) ---> 4 Fe(OH)3 (s) + 6 H2 (g) change in enthalpy = 2*321.8 kJ---1

6 H2 (g) + 3O2 (g) ----> 6 H2O (l) change in enthalpy =3* -571.7 kJ-----2

4 Fe(OH)3 (s) ---->2Fe2O3 (s) + 6 H2O (l) change in enthalpy =2* - 288.6 kJ------3

add all three equations and cancel common substances on different sides of the equations and check that you will get your desired reaction.

Add all the enthalpies with applied changes and you will get your answer.

Delta H (reaction) = 2*321.8 kJ+3* (-571.7 )kJ +2* (- 288.6) kJ

## Ignore theĀ 9 in the very

Ignore the 9 in the very first equation after the 4 Fe. it was a typo