calculate delta Hrxn for the following reaction Fe2O33CO2Fe

calculate delta Hrxn for the following reaction:

Fe2O3+3CO=2Fe +3CO2

Use the following reactions and given delta H\'s

2Fe +3/2O2(g)=Fe2O3 delta H =-824.2 kJ

CO +1/2 O2 = CO2 delta H=-282.7 k/J

Solution

1st equation because you need the Fe2O3 on the reactant side so you now have

Fe2O3>>>>2Fe + 3/2 O2

Remember to change the sign if the delta H so that it is positive because you flipped the equation.

Next you multiply the second equation by 3 so now you have:

3CO + 3/2 O2>>>> 3CO2

Cancel out the 3/2O2

so now the equations are:

Fe2O3 >>> 2Fe /\\H=824.2kJ
3CO >>> 3CO2 /\\H=-282kJ

/\\Hrxn= 824 + 3(-282.7) = -23.9