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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: wordizlife on May 30, 2016, 09:33:50 PM

Title: Lower Heating Value Problem
Post by: wordizlife on May 30, 2016, 09:33:50 PM
Problem I can't solve:

Calculate the lower heating value of this gas which is composed of 55%H2 and 45%CO (molar composition) knowing that the gas, the combustion air and the products of combustion are all at 298 degree K.

Anyone have an Idea how to solve this?

Title: Re: Lower Heating Value Problem
Post by: Borek on May 31, 2016, 02:44:33 AM
You have to show your attempts at solving the problem to receive help.

First things first: what is the definition of the lower heating value?
Title: Re: Lower Heating Value Problem
Post by: wordizlife on May 31, 2016, 09:51:16 AM
The lower heating value is the amount of heat extracted during a full combustion of a gaz.

and I know that LHV is calculated like this:
LHV=-(∆H)= ∑nprod(hf°+Δh)prod-∑nreac(hf°+Δh)react

I also know that Δh is = 0 in this case since the temperature is at 298k. I also know the enthalpie of formation (hf°) for CO2, H2O & CO.

What I am having trouble with is the chemical equation. I am not sure how to write it out...

Normaly for a stoechiometrique mixture eq:

CxHyOz+(x+y/4-z/2)(O2+3.76N2) = xCO2 + y/2H2O+3.76(x+y/4-z/2)N2

In this problems case I am tempted to simply write:
H2+CO=H2O+C.... But the %Molar composition is confusing me.

 gas which is composed of 55%H2 and 45%CO
Title: Re: Lower Heating Value Problem
Post by: Borek on May 31, 2016, 01:53:53 PM
Can you calculate it for each gas separately?
Title: Re: Lower Heating Value Problem
Post by: wordizlife on May 31, 2016, 02:12:22 PM
No, it is a mixture. (Syngas)
Title: Re: Lower Heating Value Problem
Post by: Borek on May 31, 2016, 03:27:08 PM
No, I am asking if you know how to calculate it for each gas separately?

Once you know two separate values you can combine them to account for the fact it is a mixture. These are additive things (well, perhaps in some specific cases they are not).
Title: Re: Lower Heating Value Problem
Post by: wordizlife on May 31, 2016, 08:17:07 PM
They cannot be treated separately.

Problem solved:

Equation for stoechiometric mixture:

CxHyOz+(x+y/4-z/2)(O2+3.76N2) = xCO2 + y/2H2O+3.76(x+y/4-z/2)N2

Solution:

x =0.45
y = 0.55 * 2 = 1.1
z = 0.45

0.45CO + 0.55H2 + (0.45 + 1.1/4 - 0.45/2) (O2+3.76N2) = 0.45CO2 + 1.1/2H2O +3.76 (0.45 + 1.1/4-0.45/2)N2

LHV CO2 = -393.52
LHV H2O = -241.83
LHV CO = -110.54

Δh = (0.45 * -393.52 + 0.55 * -241.83) - (0.45*-110.54) = -260.34 MJ / kmol
Title: Re: Lower Heating Value Problem
Post by: Borek on June 01, 2016, 03:21:10 AM
Quote from: wordizlife on May 31, 2016, 08:17:07 PM
They cannot be treated separately.


If they can't why you did so? Your "stoichiometric equation"

Quote
0.45CO + 0.55H2 + (0.45 + 1.1/4 - 0.45/2) (O2+3.76N2) = 0.45CO2 + 1.1/2H2O +3.76 (0.45 + 1.1/4-0.45/2)N2

is in fact a sum of two separate equations, each describing the burning of the CO and H2 respectively:

0.45CO + 0.45/2 (O2+3.76N2) = 0.45CO2 +3.76 (0.45/2)N2

0.55H2 + (1.1/4) (O2+3.76N2) = 1.1/2H2O +3.76 (1.1/4)N2

And that's exactly what I suggested.
Title: Re: Lower Heating Value Problem
Post by: wordizlife on June 01, 2016, 10:50:40 AM
Yes I agree with you.
But treating them searatly adds an extra step to calculations.

Thanks!