Hydrogen cyanide is produced in the following balanced reaction.
2NH3 (g) + 3O2 (g) + 2CH4 (g) → 2HCN (g) + 6H2O (g)
Given the following standard heats of formation, calculate the approximate ΔH°rxn of this reaction.
Substance
ΔHf°(kJ/mol)
NH3
-46
CH4
-75
HCN
135
H2O
-242
Choose one answer.
a. -500 kJ
b. -1000 kJ
c. 500 kJ
d. -1500 kJ
e. -4519
f. not enough information is given to make the calculation
dHrxn = (n*dHf products) - (n*dHf reactants)
When i did that I got -940 kJ but that isn't and option and neither is -1000kJ. So by default would the answer be F.
No, I wouldn't do that because there is enough information to actually calculate the delta H reaction. I ran through the calculation and came up with -940 kJ also so I would pick the closest answer which is -1000 kJ
Okay... but I tried that answer and it was incorrect.
To calculate the approximate ΔH°rxn (standard enthalpy change of reaction) of the given reaction, you need to sum up the standard heats of formation of the products and subtract the sum of the standard heats of formation of the reactants.
The balanced reaction is:
2NH3 (g) + 3O2 (g) + 2CH4 (g) → 2HCN (g) + 6H2O (g)
Using the given standard heats of formation (ΔHf°), we can calculate the approximate ΔH°rxn as follows:
Reactants:
2NH3 (g): 2 × (-46 kJ/mol) = -92 kJ/mol
3O2 (g): 3 × 0 kJ/mol (O2 is in its standard state)
2CH4 (g): 2 × (-75 kJ/mol) = -150 kJ/mol
Products:
2HCN (g): 2 × 135 kJ/mol = 270 kJ/mol
6H2O (g): 6 × (-242 kJ/mol) = -1452 kJ/mol
Now, sum up the products and subtract the sum of the reactants:
(270 kJ/mol + (-1452 kJ/mol)) - ((-92 kJ/mol) + 0 kJ/mol + (-150 kJ/mol))
= -1182 kJ/mol + 242 kJ/mol
= -940 kJ/mol
Therefore, the approximate ΔH°rxn of this reaction is -940 kJ.
The correct answer is option b. -1000 kJ.