How Much Heat is Needed to Convert Hydrazine from Solid to Gas?

[Mitchtwitchita]

Hey guys, I was just wondering if anybody could help me out with this problem?

Calculate the heat (in KJ) required to transform 43.90 g of hydrazine from a solid at a temperature of 1.4 degreed C to a gas at 128 degrees C. report your answer to one decimal place.

Data:
molar mass of hydrazine: 32.045 g/mol
melting point = 1.4 degrees C
Boiling point = 114 degrees C
Enthalpy of fusion = 12.6 KJ/mol
Enthalpy of vaporization = 41.8 KJ/mol
Molar heat capacity of the liquid phase = 98.9 J/mol x K
Molar heat capacity of the gas phase = 49.6 J/mol x K

q=ms(delta T)
=(43.90 g)(98.9 J/mol x K)(387 K - 274.4 K)
=488.9 KJ

q=n(molar heat of vaporization)
=(43.90 g)(1 mol/32.045 g/mol)(41.8 KJ/mol)
=57.3 KJ

q=ms(deltaT)
=(43.90 g)(49.6 J/mol x K)(401 K - 387 K)
=30.5 KJ

488.9 KJ + 57.3 KJ + 30.5 KJ
=576.7 KJ

This seems right to me, however, I don't know why the enthalpy of fusion was listed here. Can anybody explain to me if an error has occurred here?

 

[desichick07]

At this step:

q=n(molar heat of vaporization)
=(43.90 g)(1 mol/32.045 g/mol)(41.8 KJ/mol)
=57.3 KJ

you need to use the molar heat of fusion (melting) rather than vaporization to get the compound to the liquid state.

then use this part:

q=ms(delta T)
=(43.90 g)(98.9 J/mol x K)(387 K - 274.4 K)
=488.9 KJ

to get the substance to its boiling point.
at this point you need to do q = n(molar heat of vaporization)
to vaporize the substance.

and then the last part:

q=ms(deltaT)
=(43.90 g)(49.6 J/mol x K)(401 K - 387 K)
=30.5 KJ

gets the substance from its boiling point to the desired temp of 128 C.

also check your sig figs, especially when converting from degrees C --> K. They seem a bit off.