Question

Calculate the enthalpy of reaction for the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔHfo(Fe2O3(s)) = -824.2 kJ/mol ΔHfo(CO(g)) = -110.5 ΔHfo(Fe(s)) = ? ΔHfo(CO2(g)) = -393.5 kJ/mol

Answer #1

a) C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ/mol

b) 2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔH = -824.2 kJ/mol

c) C(s) + 1/2 O2(g) → CO(g) ΔH = -110.5 kJ/mol

Steps:

1. Multiply (a) by 3

2. Reverse (b) and change sign of ΔH

3. Reverse (c), change sign of ΔH and multiply by 3

3C(s) + 3O2(g) → 3CO2(g) ΔH = (3) -393.5 kJ/mol = -1180.5
kJ
------------ (1)

Fe2O3(s) ========⇒ 2 Fe(s) + 3/2 O2(g) ΔH = +824.2
kJ
------------ (2)

3CO(g) ==========⇒ 3C(s) + 3/2O2 3(+110.5) = +331.5
kJ
------------ (3)

now (2) + (3) - (1)

Fe2O3(s) + 3CO(g) ====⇒ 2Fe(s) + 3CO2(g) **ΔH = -24.8
kJ**

Calculate ΔHrxn for the following reaction:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
Use the following reactions and given ΔH′s.
2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH =
-824.2 kJ
CO(g)+1/2O2(g)→CO2(g), ΔH =
-282.7 kJ

Given the following data:
2Fe(s) + 3CO2(g) → Fe2O3(s) +
3CO(g)
ΔH° = 23.0 kJ
3FeO(s) + CO2(g) → Fe3O4(s) +
CO(g)
ΔH° = -18.0 kJ
3Fe2O3(s) + CO(g) →
2Fe3O4(s) + CO2(g)
ΔH° = -39.0 kJ
Calculate ΔH° for the reaction:
Fe(s) + CO2(g) → FeO(s) + CO(g)

Calculate ΔG° for the reduction of Fe2O3 by CO at 25°C: Fe2O3(s)
+ 3CO(g) → 2Fe(s) + 3CO2(g) (ΔG°f data: CO2(g), –394.4 kJ/mol;
Fe2O3(s), –741.0 kJ/mol; CO(g), –137.3kJ/mol) (Consider: Is the
value that you calculate for ΔG° here the same value that you
calculated for ΔG° using ΔH°f and S° data in the other
question?)
+60.6 kJ
-60.6 kJ
-30.3 kJ
+30.3 kJ
-15.1 kJ

3)From the following data at 25C ∆Hreaction (kJ
mol-1)
Fe2O3(s) + 3C(graphite) → 2Fe(s) + 3CO(g) 492.6
FeO(s) + C(graphite) → Fe(s) + CO(g) 155.8
C(graphite) + O2(g) → CO2(g)
-393.51
CO(g) + 1⁄2 O2(g) → CO2(g) -282.98
Calculate the standard enthalpy of formation of FeO(s) and of
Fe2O3(s).

At 1000 K, Kp = 19.9 for the reaction
Fe2O3(s)+3CO(g)⇌2Fe(s)+3CO2(g).
What are the equilibrium partial pressures of CO and CO2 if CO
is the only gas present initially, at a partial pressure of 0.940
atm ?

At 1000 K, Kp = 19.9 for the reaction
Fe2O3(s)+3CO(g)?2Fe(s)+3CO2(g)
What are the equilibrium partial pressures of CO and CO2 if CO
is the only gas present initially, at a partial pressure of
0.904atm ?
Enter your answers numerically separated by a comma.

Consider the following reaction, which is involved in the
purification of iron from its ores. Some relevant thermodynamic
data is shown below.
Fe2O3 (s) + 3CO(g) → 2Fe(s) + 3 CO2 (g) ΔS° = 15.2 J/K
Substance
∆Hf° (kJ/mol)
S° (J/K/mol)
∆Gf° (kJ/mol)
Fe2O3(s)
–824.2
87.40
??
CO(g)
–110.5
197.6
–137.2
Fe(s)
0
??
0
CO2(g)
–393.5
213.6
–394.4
a) Find ΔH° for the reaction above.
b) Find S° for solid iron.
c) Find ∆Gf° for Fe2O3.
*** I know...

Given the following reactions:
Question (1)
Calculate for the change in enthalpy of the reaction of
(3Fe2O3(s) + CO(g) yields CO2(g) + 2Fe3O4(s))
Fe2O3(s) + 3CO(g) yields 2Fe(s) + 3CO2(s)(Change in enthalpy=
-28.0 kJ)
3Fe(s) + 4CO2(s) yields 4CO(g) + FE3O4(s) (Change in enthalpy=
+12.5 kJ)

Iron(III) oxide reacts with carbon monoxide according to the
equation: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) A reaction mixture
initially contains 22.95 g Fe2O3 and 14.26 g CO.
Once the reaction has occurred as completely as possible, what
mass (in g) of the excess reactant is left?

ch 1-8
22.) The following reaction takes place at a certain elevated
temperature:
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(s)
What is the percent yield of iron if 57.1 g Fe2O3 in excess CO
produces 17.9 g Fe? The M.W. of Fe2O3 is 159.7 g/mol and the M.W.
of CO is 28.01 g/mol.
Recall that the percent yield is the actual yield divided by the
theoretical yield times 100%. Additional help is given in the
feedback.
Procedure: g Fe2O3 -->...

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