Question

Consider the reaction:

2Al(s) + Fe^2O^3(s) = Al^2O^3(s) + 2Fe(s)

The Fe^2O^3(s) = -824.3 kJ/mole. The H^f for Al^2O^3(s) = -1675.7 kJ/mole.

Finish the equation.
H^rxn = [(1)( _ kJ/mole) + (2)( _ kJ/mole)] - [(1)( _ kJ/mole) + (2)( _ kJ/mole)]

Answer 1

@Photon336

Answer 2

so you need to find the heat of reaction

Answer 3

this is very simple, if you notice anything that naturally occurs, i.e. any element on the periodic table will have an Hf of zero.

Answer 4

Enthalpy of reaction that is

Answer 5

I am confused who is right??

Answer 6

\[H_{products}-H_{reactants} = H_{rxn}\]

Answer 7

exactly that is the formula they give me but dont know where to plug in the numbers.

Answer 8

@Photon336

Answer 9

can you tell me what the formula I gave you means? this will help you

Answer 10

The molar enthalpy of formation is the amount of heat required to form one mole of the compound from its elements. It is symbolized by Hf

Answer 11

FYI I believe Aluminum and Iron would both be zero do you know why that might be the case?>

Answer 12

to be honest no :( sorry

Answer 13

elements have an enthalpy for formation of 0, because they are naturally occurring well lol at least for Fe and Al so it would be zero for both of those.

Answer 14

so H^rxn = [(1)(1675.7 kJ/mole) + (2)(1675.7 kJ/mole)] - [(1)(-824.3kJ/mole) + (2)(-824.3 kJ/mole)]
is this right @Photon336?

Answer 15

@desmarie this is a molar quantity. remember we have one mole of aluminum oxide being formed and 1 mole of iron oxide.

aluminum oxide is the product, iron oxide is the reactant

so it would be

\[H_{f} Al_{2}O_{3}-H_{f}Fe_{2}O_{3} = H_{products}-H_{reactants} = -1675.7-(-824.3) \frac{ kj }{ mol }\]