Question

What is the standard enthalpy of formation of liquid methylamine (CH3NH2) ?C(s)+O2(g) -> CO2(g); ?H=-393.5 kJ 2H2O(l) -> 2H2(g)+O2(g); ?H=571.6 kJ N2(g)+O2(g) -> NO2(g); ?H=33.10 kJ 4CH3NH2(l)+13O2(g) -> 4CO2(g)+4NO2(g)+10H2O(l); ?H=-4110.4 kJ The calculated answer is -47.3 kJ/mol. Show the work to confirm or deny the answer

Answer 1

ΔH =-47.3kJ

Step-by-step explanation:

You must know standard enthalpy is defined as change of enthalpy during the formation of 1 mole of the substance from its constituent elements

You can find the standard enthalpy of any reaction from the sum of another similar reactions (Hess's law) as follows:

For methylamine, CH₃NH₂ is:

C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂ (l)

1. C(s) + O₂(g) → CO₂(g); ΔH=-393.5 kJ

2. 2H₂O(l) → 2H₂(g) + O₂(g); ΔH=571.6 kJ

3. 1/2N₂(g) + O₂(g) → NO₂(g); ΔH=33.10kJ

4. 4CH₃NH₂(l) + 13O₂(g) → 4CO₂(g) + 4NO₂(g) + 10H₂O(l); ΔH=-4110.4 kJ

The sum of (1)+(3) produce:

C(s) + 2O₂(g) + 1/2N₂(g) → CO₂(g) + NO₂(g) ΔH=-393.5kJ + 33.10kJ = -360.4kJ

-5/4 (2):

C(s) + 13/4O₂(g) + 1/2N₂(g) + 5/2H₂(g) → CO₂(g) + NO₂(g) + 5/2 H₂O(l)

ΔH= -360.4kJ -5/4 (571.6kJ) = -1074.9kJ

And this reaction -1/4 (4):

C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂(l)

ΔH= -1074.9kJ -1/4(-4110.4kJ)

ΔH =-47.3kJ

Now, you can confirm the calculated answer!