[wordizlife]

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?

[Borek]

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?

[wordizlife]

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)= ∑n_prod(hf°+Δh)_prod-∑n_reac(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

[Borek]

Can you calculate it for each gas separately?

[wordizlife]

No, it is a mixture. (Syngas)

[Borek]

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).

[wordizlife]

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

[Borek]

They cannot be treated separately.

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

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.

[wordizlife]

Yes I agree with you.
But treating them searatly adds an extra step to calculations.

Thanks!