Question

1. Calculate Δ*H* (in kJ) for the process
Hg_{2}Br_{2}(*s*) → 2 Hg(*l*) +
Br_{2}(*l*) from the following information.

Hg(*l*) + Br_{2}(*l*) →
HgBr_{2}(*s*) Δ*H*^{⁰}_{298} =
−170.7 kJ Hg(*l*) + HgBr_{2}(*s*) →
Hg_{2}Br_{2}(*s*)
Δ*H*^{⁰}_{298} = −36.2 kJ

Answer #1

Hess's Law
Given the following data:
Br2(l) + 5F2(g) →
2BrF5(l)
ΔH°=-918.0 kJ
BrF3(l) + Br2(l) → 3BrF(g)
ΔH°=125.2 kJ
Br2(l) + F2(g) → 2BrF(g)
ΔH°=-117.2 kJ
calculate ΔH° for the reaction:
BrF5(l) → BrF3(l) + F2(g)

Hess's Law Given the following data: 2C(s) + 2H2(g) + O2(g) →
CH3OCHO(l) ΔH°=-366.0 kJ CH3OH(l) + O2(g) → HCOOH(l) + H2O(l)
ΔH°=-473.0 kJ C(s) + 2H2(g) + 1/2O2(g) → CH3OH(l) ΔH°=-238.0 kJ
H2(g) + 1/2O2(g) → H2O(l) ΔH°=-286.0 kJ calculate ΔH° for the
reaction: HCOOH(l) + CH3OH(l) → CH3OCHO(l) + H2O(l)

Given the data 2 S(s) + 3 O2(g) → 2 SO3(g) ΔH = −790 kJ S(s) +
O2(g) → SO2(g) ΔH = −297 kJ SO3(g) + H2O(l) → H2SO4(l) ΔH = −132 kJ
use Hess's law to calculate ΔH for the reaction 2 SO2(g) + O2(g) →
2 SO3(g).

Given the following data:
H2(g) + 1/2O2(g) → H2O(l)
ΔH° = -286.0 kJ
C(s) + O2(g) → CO2(g)
ΔH° = -394.0 kJ
2CO2(g) + H2O(l) →
C2H2(g) + 5/2O2(g)
ΔH° = 1300.0 kJ
Calculate ΔH° for the reaction:
2C(s) + H2(g) → C2H2(g)

Given the following data:
HNO3(l) → 1/2N2(g) + 3/2O2(g)
+ 1/2H2(g)
ΔH° = 174.1 kJ
2N2(g) + 5O2(g) →
2N2O5(g)
ΔH° = 28.4 kJ
H2(g) + 1/2O2(g) → H2O(l)
ΔH° = -285.8 kJ
Calculate ΔH° for the reaction:
2HNO3(l) → N2O5(g) +
H2O(l)
Note that you should be able to answer this one without
needing to use any additional information from the thermo
table.

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)

iii) Calculate ΔH for the reaction 2 Al (s) + 3 Cl2 (g)
→ 2 AlCl3 (s) from the data
2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g) ΔH = -1049.
kJ
HCl (g) → HCl (aq) ΔH = -74.8 kJ
H2 (g) + Cl2 (g) → 2 HCl (g) ΔH = -1845. kJ
AlCl3 (s) → AlCl3 (aq) ΔH = -323. kJ

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f
for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g) ΔH°rxn =
-89 kJ IF7(g) -941 IF5(g) -840

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f
for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g)
ΔH°rxn = -89 kJ
IF7(g) -941
IF5(g) -840
Answers:
(a) -190 KJ/mol
(b) 101 KJ/mol
(c) 24 KJ/mol
(d) -95 KJ/mol
(e) -146 KJ/mol

Consider the following reaction:
2AgCl(s) → 2Ag(s) +
Cl2(g); ΔH° = 127.1 kJ; ΔS° =
115.7 J/K at 298 K
Suppose 59.4 g of silver(I) chloride is placed in a 63.1 L
vessel at 298 K. What is the equilibrium partial pressure of
chlorine gas? (R = 0.0821 L · atm/(K · mol) = 8.31 J/(K ·
mol))
A.5.7 × 10-17 atm
B.0.95 atm
C.5.1 × 10-23 atm
D.0.081 atm
E.0.16 atm

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