Question

25.

Calculate the standard enthalpy change for the following chemical equation.

4FeO (s) + O2 (g) → 2Fe2O3 (s)

Use the following thermochemical equations to solve for the change in enthalpy.

Fe (s) + ½ O2 (g) → FeO (s) ΔH = -269 kJ/mol

2Fe (s) + 3/2 O2 (g) → Fe2O3 (s) ΔH = -825 kJ/mol

-2726 kJ/mol |

556 kJ/mol |

-556 kJ/mol |

574 kJ/mol |

-574 kJ/mol |

Answer #1

4FeO (s) + O2 (g) → 2Fe2O3 (s)

Use the following thermochemical equations to solve for the change in enthalpy.

Fe (s) + ½ O2 (g) → FeO (s) ΔH = -269 kJ/mol -------------(1)

2Fe (s) + 3/2 O2 (g) → Fe2O3 (s) ΔH = -825 kJ/mol ---------------(2)

Multiply equation (2) with 2

4Fe (s) + 3 O2 (g) → 2Fe2O3 (s) ΔH = -1650 kJ ---------------(3)

Multiply equation (1) with 4

4Fe (s) + 2O2 (g) →4 FeO (s) ΔH = -1076 kJ -------------(4)

Subtract equation 4 from equation 3

4FeO (s) + O2 (g) → 2Fe2O3 (s) ΔH = -1650 kJ-(-1076 kJ) = -1650+1076 = -574 kJ

the standard enthalpy change for the above chemical equation is -574 kJ/mol.

Calculate the standard enthalpy change, ΔH°rxn, in kJ for the
following chemical equation, using only the thermochemical
equations below: 4KO2(s) + 2H2O(l) → 4KOH(aq) + 3O2(g) Report your
answer to three significant figures in scientific notation.
Equations: ΔH°rxn
(kJ)
4K(s) + O2(g) → 2K2O(s)
-726.4
K(s) + O2(g) → KO2(s) -284.5
K2O(s) + H2O(l) → 2KOH(aq) -318

Calculate the standard enthalpy change for the reaction
2 Al(s) +
Fe2O3(s)
2
Fe(s) + Al2O3(s)
Given that
2 Al(s) + 3/2 O2
(g)
Al2O3(s) ΔH rxn =
-1669.8 kJ/mol
2 Fe (s) +
3/2 O2
(g)
Fe2O3(s) ΔH rxn = -822.2
kJ/mol

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).

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

What is the enthalpy change for the first reaction?
P4(s) + 6Cl2(g) → 4PCl3(l) ΔH
=
P4(s) + 10Cl2(g) → 4PCl5(s) ΔH
= -1,779.8
PCl3(l) + Cl2 → PCl5(s) ΔH =
-123.3
question 2
What is the enthalpy change for the first reaction?
Fe2O3(s) → 2Fe(s) + 3/2O2(g) ΔH
=
4Fe(s) + 3O2(g) → 2Fe2O3 (s) ΔH
= -1,645 kJ
Help me understand, please

1.
Refer to this equation:
2CaCO3 (s) —> 2CaO (s) + 2CO2 (g)
Enthalpy change = -178.1 kJ/ mol reaction
How many grams of CaCo3 must react in order to liberate 545 kJ
of heat?
2. Refer to this equation:
2Al (s) + Fe2O3 (s) —> 2Fe (s) + Al2O3 (s)
Enthalpy change = -847.6 kJ/ mol reaction
How much heat is released if 35.0 g of Al (s) reacts to
completion?

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)

You are given the following thermodynamic data. 2 Fe(s) + 3/2
O2(g) → Fe2O3(s) ΔH° = -823 kJ 3 Fe(s) + 2 O2(g) → Fe3O4(s) ΔH° =
-1120. kJ Calculate the ΔH° for the following reaction. 3 Fe2O3(s)
→ 2 Fe3O4(s) + ½ O2(g)

1.) Using enthalpies of formation, calculate the standard change
in enthalpy for the thermite reaction. The enthalpy of formation of
Fe3O4 is −1117 kJ/mol.
8 Al(s) + 3 Fe3O4(s) → 4 Al2O3(s) + 9 Fe(s)
2. a) Nitroglycerin is a powerful explosive, giving four
different gases when detonated.
2 C3H5(NO3)3(l) → 3 N2(g) + 1/2 O2 (g) + 6 CO2(g) + 5 H2O(g)
Given that the enthalpy of formation of nitroglycerin, ΔHf°, is
−364 kJ/mol, calculate the energy (heat at...

A chemist measures the enthalpy change ΔH during the following
reaction: 2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s) =ΔH−852.kJ
Use this information to calculate ΔH in kJ for the following
reactions:
3Al2O3s + 6Fes → 6Als +
3Fe2O3s
Al2O3s + 2Fes → 2Als +
Fe2O3s
4Als + 2Fe2O3s → 2Al2O3s +
4Fes

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