Relationship between HHV and LHV of gaseous propane

[MickeyBlue]

Homework Statement

The Higher Heating Value (HHV) of gaseous propane is given as 2220.0 kJ/mol. What is the value of the Lower Heating Value (LHV) of gaseous propane?

Data for water (all in kJ/mol):
H(g, 25 deg. C) = 45.85 ;
H(g, 100 deg. C) = 48.17 ;
H(l, 25 deg. C) = 1.886 ;
H(l, 100 deg. C) = 7.544 ;
∆Hvap(25 deg. C) = 43.97 ;
∆Hvap(100 deg. C) = 40.63

A. 2057 kJ/mol

• B. 2387 kJ/mol

• C. 2044 kJ/mol

• D. 2014 kJ/mol

• E. 2035 kJ/mol

Homework Equations

HHV = LHV + nΔH_vap
I know LHV is HHV less heat of vaporisation of water, but the idea of calculating from what I think is the reference state is confusing me. I realize I may be missing something obvious, but we didn't spend long covering this topic.

The Attempt at a Solution

LHV = HHV - nΔH_vap

LHV = (2220) - H_sensible - (ΔH_vap - ΔH_{vap, ref}))
= (2220) - (7.544 - 1.886) - (40.63 - 43.97)
= 2218 kJ/mol

 

[collinsmark]

I'll give this one a try since nobody else has replied yet.

Caveat: This subject is not my area of expertise.

Homework Statement

The Higher Heating Value (HHV) of gaseous propane is given as 2220.0 kJ/mol. What is the value of the Lower Heating Value (LHV) of gaseous propane?

Data for water (all in kJ/mol):
H(g, 25 deg. C) = 45.85 ;
H(g, 100 deg. C) = 48.17 ;
H(l, 25 deg. C) = 1.886 ;
H(l, 100 deg. C) = 7.544 ;
∆Hvap(25 deg. C) = 43.97 ;
∆Hvap(100 deg. C) = 40.63

A. 2057 kJ/mol

• B. 2387 kJ/mol

• C. 2044 kJ/mol

• D. 2014 kJ/mol

• E. 2035 kJ/mol

Homework Equations

HHV = LHV + nΔH_vap
I know LHV is HHV less heat of vaporisation of water, but the idea of calculating from what I think is the reference state is confusing me. I realize I may be missing something obvious, but we didn't spend long covering this topic.

The Attempt at a Solution

LHV = HHV - nΔH_vap

LHV = (2220) - H_sensible - (ΔH_vap - ΔH_{vap, ref}))
= (2220) - (7.544 - 1.886) - (40.63 - 43.97)
= 2218 kJ/mol

Question: Answer C is in boldface in your original post. Is it in boldface because it is the correct answer? (I just want to make sure. It's not clear why it is in boldface in the original post.)

You might find this description of HHV and LHV useful:
https://en.wikipedia.org/wiki/Heat_of_combustion
https://en.wikipedia.org/wiki/Heat_of_combustion
I do not understand what you are doing with H_sensible. What is that?

Nor do I understand why you are subtracting one ΔH_vap from another.

One of the things you will need to determine before arriving at the correct answer is to determine how many moles of water are in the products compared to the number of moles of propane.

In other words, for every mole of propane in the combustion, there will be n moles of water that are produced in the products. What is n?

 

[MickeyBlue]

*Update: I've since come the right answer of 2044 kJ/mol.

Thank you for the input, collinsmark. My confusion here was brought about by other methods I know to calculate heat of a general reaction. You're exactly right. I neglected to write out a balanced equation for the combustion reaction, which is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Sensible heat was meant to represent a change in heat caused by a change in temperature, which I realize now doesn't factor into the LHV.

So LHV = HHV - nΔH^25°Cvap = (2220) - (4×43.97) ≈ 2044 kJ/mol