Question

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)

Answer #1

Given the following data:

2C(s) + 2H2(g) + O2(g) ---- > CH3OCHO(l) ΔH°=-366.0 kJ -----1

CH3OH(l) + O2(g) ---- > HCOOH(l) + H2O(l) ΔH°=-473.0 kJ ------2

C(s) + 2H2(g) + 1/2O2(g) ---- > CH3OH(l) ΔH°=-238.0 kJ ------3

H2(g) + 1/2O2(g) ---- > H2O(l) ΔH°=-286.0 kJ -------4

calculate ΔH° for the reaction:

HCOOH(l) + CH3OH(l) ----> CH3OCHO(l) + H2O(l)

As follows:

First reverse equation number 3 and 2 and multiply equation 3 and 4 by 2 , than add all :

HCOOH(l) + H2O(l) ---- > CH3OH(l) + O2(g) ΔH°=+ 473.0 kJ

2CH3OH(l) ---- >2 C(s) + 4H2(g) + O2(g) ΔH°=+476.0 kJ

2C(s) + 2H2(g) + O2(g) ---- > CH3OCHO(l) ΔH°=-366.0 kJ

2H2(g) + O2(g) ---- > 2 H2O(l) ΔH°=-572.0 kJ

================================================

HCOOH(l)+ CH3OH(l) ----> CH3OCHO(l) + H2O(l) ΔH°=11 kJ

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 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:
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.

Use Hess's Law to calculate the enthalpy of reaction, ΔH rxn,
for the reaction in bold below given the following chemical steps
and their respective enthalpy changes. Show ALL work!
2 C(s) + H2(g) → C2H2(g) ΔH°rxn = ?
1. C2H2(g) + 5/2 O2(g) → 2CO2 (g) + H2O (l) ΔH°rxn = -1299.6
kJ
2. C(s) + O2(g) → CO2 (g) ΔH°rxn = -393.5 kJ
3. H2(g) + ½ O2(g) → H2O (l) ΔH°rxn = -285.8 kJ

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)

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)

Problem 5.64 Part A From the enthalpies of reaction
2C(s)+O2(g)→2CO(g)ΔH=−221.0kJ
2C(s)+O2(g)+4H2(g)→2CH3OH(g)ΔH=−402.4kJ calculate ΔH for the
reaction CO(g)+2H2(g)→CH3OH(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 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

We can use Hess's law to calculate enthalpy changes that cannot
be measured. One such reaction is the conversion of methane to
ethylene: 2CH4(g)⟶C2H4(g)+2H2(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 ΔH∘=−3120.8kJ

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