Question

Given the following data:

HNO_{3}(l) → 1/2N_{2}(g) + 3/2O_{2}(g)
+ 1/2H_{2}(g) |
ΔH° = 174.1 kJ |

2N_{2}(g) + 5O_{2}(g) →
2N_{2}O_{5}(g) |
ΔH° = 28.4 kJ |

H_{2}(g) + 1/2O_{2}(g) → H_{2}O(l) |
ΔH° = -285.8 kJ |

Calculate ΔH° for the reaction:

2HNO_{3}(l) → N_{2}O_{5}(g) +
H_{2}O(l)

**Note that you should be able to answer this one without
needing to use any additional information from the thermo
table.**

Answer #1

HNO_{3}(l) → 1/2N_{2}(g) + 3/2O_{2}(g) +
1/2H_{2}(g) : ΔH°_{1} = 174.1
kJ -------(1)

2N_{2}(g) + 5O_{2}(g) →
2N_{2}O_{5}(g) ; ΔH°_{2}
= 28.4
kJ
-------(2)

H_{2}(g) + 1/2O_{2}(g) →
H_{2}O(l) ; ΔH°_{3} =
-285.8
kJ
------(3)

2HNO_{3}(l) → N_{2}O_{5}(g) +
H_{2}O(l) : ΔH° =
?
-------(4)

Eqn (4) can be obtained from remaining equations as follows:

Eqn(1) = [2xEqn(1)] + [(1/2)xEqn(2)] +Eqn(3)

So ΔH° = [2x ΔH°_{1} ] + [ (1/2)x ΔH°_{2} ] +
ΔH°_{3}

= [2x(174.1 kJ)] + [(1/2)x(28.4 kJ)] + (-285.8 kJ)

= +76.6 kJ

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 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)

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)

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)

Given the following reactions and their enthalpies:
ΔH(kJ/mol)−−−−−−−−−−−
H2(g)⟶2H(g) +436
O2(g)⟶2O(g) +495
H2+1/2O2(g)⟶H2O(g) −242
A. Devise a way to calculate ΔH for the reaction
H2O(g)⟶2H(g)+O(g)
B. estimate the H-O bond energy

Use the following calorimetric data to calculate the heat of
combustion (burning) of the carbon compound shown below:
C3H8 (g) + 5O2 (g) →
3CO2(g) + 4H2O(g)
reaction
ΔHo (kJ)
C(s) + O2(g) → CO2(g)
-393.5
C(s) → C(g)
+718.4
H2O(l) → H2O(g)
+44.0
H2(g) + 1/2 O2(g) →
H2O(l)
-285.8
3C(s) + 4H2(g) → C3H8(g)
-103.8
C(s) + 2H2(g) + 1/2O2(g) →
CH3OH(l)
-238.7
6C(s) + 3H2(g) → C6H6(g)
+82.9
C6H6(l) →
C6H6(g)
+33.9
CH3OH(l) → CH3OH(g)
+38.0
ΔHo...

Given tha delta hydrogen = -393.5 kJ for the rxn.: C(s) +O2(g)
-> CO2(g), and
given that delta Hydrogen = -285.8 kj for the rxn.: H2(g)+
1/2O2(g) -> H2O(l), and given
that delta hydrogen = -84.7 kJ for the rxn .: 3H2(g) +2C
+2C(graphite) -> C2H6(g),
calculate the dleta hydrogen for the rxn C2H6(g) + 3 1/2O2(g)
-> 2CO2(g) + 3HO(l)
Delta hydrogen = 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 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).

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)

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