Thermochemical question about the heat of combustion of a hydride

AI Thread Summary
The discussion revolves around calculating the heat released from the complete combustion of 22.4x10^3 liters of FeTiH2, a fuel used in hydrogen cars. The user correctly determines that 22.4 liters at STP corresponds to 1000 moles of FeTiH2 and proposes a combustion equation. They apply Hess's Law to find the reaction enthalpy but struggle to locate the heat of formation for FeTiH2 at 1 atm pressure. The user expresses confusion regarding the state of FeTiH2, questioning its gaseous form and the relevance of volume without density. The thread highlights the challenges in finding specific thermochemical data for less common compounds.
pollycampos
Messages
8
Reaction score
0

Homework Statement


What is the heat released by the complete combustion of 22.4x10^3 litres of FeTiH2 - a fuel used in hydrogen cars -, producing liquid water?

Homework Equations


ΔH°(reaction) = ΔH°f(products) - ΔH°f(reactants)

The Attempt at a Solution


1 mol FeTiH2 - 22,4 L (STP)
x - 22.4x10^3 litres
x = 1000 moles of FeTiH2

I guess the equation for the combustion is:
1000 FeTiH2(g) + 2250 O2(g) = 1000 H2O(l) + 500 Fe2O3(s) + 1000 TiO2(s)

Using the Hess's Law:
ΔH°(reaction) = 1000ΔH°f(H2O,l) + 500ΔH°f(Fe2O3,s) + 1000ΔH°f(TiO2,s)
- 1000ΔH°f(FeTiH2,g) - 2250ΔH°f(O2,g)
ΔH°(reaction) = - 285800 - 824200 - 945000 - 1000ΔH°f(FeTiH2,g) - 0

I'm having trouble finding the heat of formation of FeTiH2. There're some heat of formation around Google, but I haven't found one at 1 atm pressure. So I guess there must be another way to resolve this problem :-p

The answer is - 2.86x10^5 kJ/mol
 
Physics news on Phys.org
I don't see how FeTiH2 can be gaseous. I expect solid, so the amount given as a volume (without density) doesn't make much sense.
 
Oh yes, my bad :-p But anyway, I can't find the heat of formation :frown:
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top