[Sush_eera]

Can someone please help me with this question?? i tried but couldnt figure out the answersat all. please andd thank you

[Corribus]

You have to show your work to receive help. This is a forum policy.

[mikasaur]

Hello and welcome to the forums Sush_eera. Before we can help you must show your own attempt. It's part of our forum rules.

What do you know about Enthalpy and state functions? Have you learned about Hess's Law? How might those concepts help you here?

[Sush_eera]

Fe2O3 (s) + 3CO (g)  -----> 2Fe (s) + 3CO2 (g) ΔHo = -26.7 kJ ----eq 1
CO (g) + 1/2 O2 (g) ------> CO2 (g) ΔHo = -283.0 kJ--- eq 2
readjusting the above equations to get 2Fe (s) + 3/2 O2 (g)Fe2O3 (s)

reversing eq 1
2Fe (s) + 3CO2 ____> Fe2O3 (s) + 3CO (g)  ΔHo = 26.7 kJ
CO (g) + 1/2 O2 (g) ------> CO2 (g) ΔHo = -283.0 kJ (multiplying eq by 3) t get:
3CO(g) + 3/2 O2(g) ---> 3Co2(g) ΔHo =-849 --- eq (3)
Adding the enthalpies and subtracting the eqn 1 and 3
we get
2Fe (s) + 3CO2 ____> Fe2O3 (s) + 3CO (g)  ΔHo = 26.7 kJ
3CO(g) + 3/2 O2(g) ---> 3Co2(g) ΔHo =-849
we get 2Fe (s)+3/2 O2(g)---> Fe2O3 (s) ΔHo =-822.3 kJ/mol (answer)
This is how i worked it out but i am not too sure

[mikasaur]

That's the answer I get. It's kind of hard to follow your work (you can use the formatting buttons to make subscripts and rxn arrows which makes it much easier for us to follow your work).

It seems like you're going about it the right way though. Enthalpy is a state function which means that no matter how you get there, going from one state to another results in the same change in enthalpy. Equation 1 is basically a combination of equations 2 and 3 if you multiply one and reverse the other.

[Sush_eera]

sorry i will use them next time.. thank you so much