Question

Using the following thermochemical data, calculate ΔHfˆ of Tm_2O_3(s) . 2TmCl_3(s) + 3H_2O(l) \rightarrow Tm_2O_3(s) + 6HCl(g) ΔHˆ=388.1kJ/mol 2Tm(s) + 3Cl_2(g) \rightarrow 2TmCl_3(s) ΔHˆ=−1973.2kJ/mol 4HCl(g) + O_2(g) \rightarrow 2Cl_2(g) + 2H_2O(l) ΔHˆ=−202.4kJ/mol

Answer #1

Using the following thermochemical data,

calculate ΔHfˆ of Tm_{2}O_{3}(s)

The equation for formation of Tm2O3(s) will be

2Tm(s) + 3/2 O_{2}(g) -->
Tm_{2}O_{3}(s)
...............................................................(1)

The other equation given are

2TmCl_{3}(s) + 3H_{2}O(l) --->
Tm_{2}O_{3}(s) + 6HCl(g) ΔHˆ=388.1kJ/mol
......................(2)

2Tm(s) + 3Cl_{2}(g) ---> 2TmCl_{3}(s)
ΔHˆ=−1973.2kJ/mol .....................................(3)

4HCl(g) + O2(g) ---> 2Cl_{2}(g) + 2H_{2}O(l)
ΔHˆ=−202.4kJ/mol ..............................(4)

Now we will combine the equation 2,3 and 4 will be combined to form equation (1)

Equation (1) = Equation (3) + Equation (2) + 1.4 X equation (4)

ΔH (1) = ΔH (3) +ΔH (2) + 1.5 ΔH (4) = 388.1 - 1973.2 - 1.5(202.4) = -1888.7 KJ / mole

Using the following thermochemical data, calculate ΔHf° of
Cr2O3(s).
2CrCl3(s) + 3H2O(l) → Cr2O3(s) + 6HCl(g) ΔH° = 276.9 kJ/mol
2Cr(s) + 3Cl2(g) → 2CrCl3(s) ΔH° = -1113.0 kJ/mol
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(l) ΔH° = -202.4 kJ/mol

Determine the lattice energy for the formation of MgO(s) from
Mg(s) and O_22(g) using the following equations:
Mg(s) \rightarrow→ Mg(g) , + 147 kJ/mol
Mg(g) \rightarrow→ Mg^{2+}2+(g) + 2e^-− , + 2189
kJ/mol
O_22(g) \rightarrow→ 2O(g) , + 498 kJ/mol
O(g) + 2e^-− \rightarrow→ O^{2-}2−(g) , +702 kJ/mol
Mg(s) + 1/2O_22(g) \rightarrow→ MgO(s) , -602 kJ/mol

Calculate the ΔH∘ for this reaction using the following
thermochemical data:
CH4(g)+2O2(g)⟶CO2(g)+2H2O(l)
ΔH∘=−890.3kJ
C2H4(g)+H2(g)⟶C2H6(g)
ΔH∘=−136.3kJ
2H2(g)+O2(g)⟶2H2O(l)
ΔH∘=−571.6kJ
2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(l)

Calculate the standard enthalpy of formation of liquid water
(H2O) using the following thermochemical information:
4 B(s) + 3 O2(g) 2 B2O3(s) H = -2509.1 kJ
B2H6(g) + 3 O2(g) B2O3(s) + 3 H2O(l) H = -2147.5 kJ
B2H6(g) 2 B(s) + 3 H2(g) H = -35.4 kJ

Octane (C8H18) undergoes combustion according to the following
thermochemical equation:
2C8H18(l) + 25O2(g) 16CO2(g) +
18H2O(l) ∆H°rxn = −10,800
kJ/mol
Given that ∆H°f[CO2(g)] = −394 kJ/mol and ∆H°f[H2O(l)] = −286
kJ/mol, calculate the standard enthalpy of formation of
octane.
a. −326 kJ/mol
b.326 kJ/mol
c.210 kJ/mol
d.-218 kJ/mol

Calculate the deltaH rxn for the production of CO2 and H20 via
propane combustion, using thermochemical equations below. Show all
work! 3 C (s) + 4 H2 (g) = C3H8 (g) Delta H = (-) 103.9 kJ/mol C
(s) + O2 (g) = CO2 (g) Delta H = (-) 393.5 kJ/mol H2 (g) + 1/2 O2
(g) = H2O (g) Delta H = (-)241. 8 kJ/mol Delta H rxn : ?

Given the following thermochemical data:
½H2(g)+AgNO3(aq) → Ag(s)+HNO3(aq) ΔH = -105.0 kJ
2AgNO3(aq)+H2O(l) → 2HNO3(aq)+Ag2O(s) ΔH = 44.8 kJ
H2O(l) → H2(g)+½O2(g) ΔH = 285.8 kJ
Use Hess’s Law to determine ΔH for the reaction:
Ag2O(s) → 2Ag(s)+½O2(g)

Calculate the standard enthalpy of formation of dinitrogen
pentoxide (N2O5) using the following thermochemical information: 2
H2O(l) 2 H2(g) + O2(g) H = +571.7 kJ 2 HNO3(l) N2O5(g) + H2O(l) H =
+92.0 kJ N2(g) + 3 O2(g) + H2(g) 2 HNO3(l) H = -348.2 kJ H = kJ

Calculate ΔH° for the combustion of methane (products are
CO2(g) and H2O(l)), given the following
thermochemical equations:
CH4(g) + O2(g) →CH2O(g) + H2O(g) ΔH = -284 kJ
CH2O(g) + O2(g) →CO2(g) + H2O(g) ΔH = -518 kJ
H2O(g) →H2O(l) ΔH = +44 kJ

4. Use data from the table of thermochemical data (Appendix C)
to calculate Delta G Standard for the reactions shown.
A. 2Fe2O3(s) >> 4Fe(s) + 3O2(g)
B. 2PCl3(g) + O2(g) >>2POCl3(g)

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