[mrmedic]

Hi,

I have to do this numeracy assignment in chemistry. Everything has gone well (I hope) apart from two questions which I am not sure about.

Gaseous ferric oxide (Fe2O3) reacts with gaseous ammonia (NH3) to form solid ferrous oxide (FeO), liquid water (H2O) and nitrogen gas (N2).
i)   Write a balanced chemical equation for this reaction, and in conjunction with the data given below calculate the change in standard Gibbs Energy (Δ Go) at 25o C.
ii)   Is this reaction spontaneous?
Δ Gof FeO = -234.1 kJ mol-1;  Δ Gof Fe2O3 = -732.2 kJ mol-1; Δ Gof H2O = -237.6 kJ mol-1; Δ Gof NH3= -16.7 kJ mol-1; Δ Gof N2= 0 kJ mol-1.

So wrote out the equation,
3Fe2O3­­­ + 2NH3 --> 6FeO + 3H2O + N2

∆G = ∆Gproducts - ∆Greactants
 
Then did: (1404.6 + 712.8 - 0) - (2196.6 + 33.4) = -112.6kJmol-1

∆G<0 is  feasible at this temperature as ∆G is greater than 0.

But in context of the question I do not believe that this could be right?

My other question was working out the enthalpy of combustion.

Calculate the enthalpy change for this conversion using Hess’s law given the following enthalpies of combustion:

1.    C6H12O6 (s) + 6 O2 (g)   -> 6 CO2 (g) + 6 H2O (l)  ΔH1 = -2821 kJmol-1


2.    2C3H4O3 (l) + 2H2O+ 5 O2 (g) -> 6CO2 (g) + 6H2O (l)     ΔH2 = -1170 kJmol-1

I got ΔH=-481kJmol-1


This next bit was confusing: from the information given above, calculate the amount of heat energy released in Kcal by burning 1g of glucose. You should assume that complete combustion occurs.

Here's what I did:
Q = mcm∆T =  1mol x 4.18 x -481kJmol- = -2010.58kJmol-1

m = 1/180 mol

cm = 75.312

 ∆T = K

Cs = 4.18 JK-1g-1

Water RMM = 18gmol-1

Water from Cs to Cm = (4.18 J x 18 mol-1)  = 75.24 Jmol-1K-1

∆H3 = ∆H1 - ∆H2

-481kJmol-1 = -2821 – (2 x 1170)

-481kJmol-1 converted in to T◦c = -253.278˚C

∆TK = 19.872

q = mcm∆T

q = 1/180 x 75.24 x 19.872

q = 8.306496kJmol-1

1kJmol-1 = 0.239001kcalmol-1

q = 8.306496kJmol-1 x 0.239001kcalmol-1

q = 1.98526085kcalmol-1

q = 2.0kcalmol-1

 
But this whole thing assumes that the enthalpy of combustion can be used a temperature change.

[mrmedic]

Anyone?

[cakaro13]

For number 2, \( \Delta{H}_{f,comb} \) should be the heat energy released per mole, upon combustion of glucose.

[mrmedic]

didmt i do that

[cakaro13]

You are on the right track, but you don't need to use q=mc\(\Delta T\). \(\Delta H \) is the "heat released" per mole. You only need to do the conversion from joules to kcal.

[mrmedic]

CHEEEEEERS

I get 0.638kcal.
/mol?