[Korokian]

1)  Based on the following equations:

2CO₂(g) ----> O₂(g) + 2CO(g)                change in enthalpy = +566.0kJ

2C(g) + O₂(g) --------> 2CO(g)                               change in enthalpy = -221.0 kJ

Determine the enthalpy change for the folowing:

C(graphite) + O₂(g) ---------> CO₂(g)
-----------------------------------------------------------------------------------------------------------------------------------------------

so i think i can't use hess's law becuase it doesnt have C(s) in any of those equations so i used
Change enthalpy rxn = sum (mol)(change enthalpy of products) - sum(mol)(change enthalpy of reactants)

since C-graphite and O₂gas are the standard condition they have 0 kJ and so the final answer is -393.5kJ which is the
enthalpy of CO₂(g)

is that right or did i do it wrong?




2) Calculate the change in enthalpy for the reaction of ammonia gas and oxygen to form ONE MOLE of nitrogen and steam

so i did 2NH₃(g)+ O₂(g) ----------> N₂(g) + 2H₂O(g)

i am not sure on how to calculate one mole of the change in enthalpy
i tried to use Change enthalpy rxn = sum (mol)(change enthalpy of products) - sum(mol)(change enthalpy of reactants)
but instead of putting the 2 for the mol i put 1 for both of them and ended up with 286.9 kJ/mol
i think thats wrong i have no idea what to do

[Korokian]

help please

[Donaldson Tan]

1) Based on the following equations:

2CO₂(g) ----> O₂(g) + 2CO(g) change in enthalpy = +566.0kJ

2C(g) + O₂(g) --------> 2CO(g) change in enthalpy = -221.0 kJ

Determine the enthalpy change for the folowing:

C(graphite) + O₂(g) ---------> CO₂(g)

Hess' Law is definitely valid. In fact, your pathway is wrong. It is C (s) -> CO (g) -> CO₂ (g), so the enthalpy change is given as " [ -221.0kJ - (+556.0 kJ) ] / 2"

The plus or minus sign of the parenthess depends whether I am going against the direction of the arrow sign in the given reaction steps above. I divide by 2 because the addition in the numerator is for 2 moles of carbon dioxide gas produced.

[Donaldson Tan]

2) Calculate the change in enthalpy for the reaction of ammonia gas and oxygen to form ONE MOLE of nitrogen and steam

so i did 2NH₃(g)+ O₂(g) ----------> N₂(g) + 2H₂O(g)

What data are you supplemented to solve this problem? eg. standard heat of formation, standard heat of combustion, etc

[Korokian]

2) Calculate the change in enthalpy for the reaction of ammonia gas and oxygen to form ONE MOLE of nitrogen and steam

so i did 2NH₃(g)+ O₂(g) ----------> N₂(g) + 2H₂O(g)

What data are you supplemented to solve this problem? eg. standard heat of formation, standard heat of combustion, etc

standard heat of formation
H2O(g)  -241.8kJ/mol
H2O(l)    - 285.8kJ/mol
NH3(g)    -46.1kj/mol


is this correct? 2(-285. -  2(-46.1) = -391.4 kJ

[Korokian]

Hess' Law is definitely valid. In fact, your pathway is wrong. It is C (s) -> CO (g) -> CO₂ (g)

good catch geo!!! i got the problem now doing hess's law and i verified its correctness.. i got -393.5 kJ thanks

[Donaldson Tan]

standard heat of formation
H2O(g)  -241.8kJ/mol
H2O(l)    - 285.8kJ/mol
NH3(g)    -46.1kj/mol

is this correct? 2(-285. -  2(-46.1) = -391.4 kJ

You have the useful data, but your reaction equation is not balanced. There is 6H on the LHS but 4H on the RHS. Balance it correctly and you will get the right answer.

[Korokian]

i got the answer to be -633 kJ/mol thanks have free scooby snacks