Question

Fe2O3(s) + 3 CO(g) —> 2 Fe(s) + 3 CO2(g)

delta H = ___

a. Calculate delta H for the reaction above using the following know thermochemical reactions

2Fe(s) + 3/2 O2(g) —> Fe2O3(s) delta h=-824.2 kJ

CO(g) + 1/2 O2(g) —> CO2(g) delta h=-282.7.

b. Is this reaction endothermic, exothermic, or both? How do you know?

Answer 1

From the Hess law,

for a chemical reaction, the enthalpy is constant, and irrespective of the reaction happening in one step or many steps

so let us consider for a chemical equation can be written as the sum of other equation. now if we have to calculate the change in enthalpy for initial equation then it will be equal to the sum of the enthalpy changes of the other equations.

so from the above problem, we have to calculate the enthalpy change of the following reaction.

Fe2O3(s) + 3 CO(g) —> 2 Fe(s) + 3 CO2(g)

first equaiton given as

2Fe(s) + 3/2 O2(g) —> Fe2O3(s) delta h=-824.2 kJ

If we will look at the initial equation the reaction is Fe2O3, so the first equaiton we have to flip as

Fe2O3(s) -->2Fe(s) + 3/2 O2(g) now delta H will be +824.2 KJ

for the second equation given, CO(g) + 1/2 O2(g) —> CO2(g) delta H=-282.7

in order to eleminate O2 and balanced equion to form the inital equaiton we have to multiply the equaiton with 3

so, 3CO(g) + 3/2O2(g) ---> 3CO2 (g) so, delta H will be -282.7*3 = -848.1 KJ

so, by adding two equaiton together to get the inital equaiton

so the Delta H of the initial equation will be ((+824.2 ) + (-848.1)) = -23.9 KJ

so for the eqution Fe2O3(s) + 3 CO(g) —> 2 Fe(s) + 3 CO2(g) delta H = -23.9 KJ

as delta H is nagative the reaction is Exothermic