Question

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)

Answer #1

Apply HEss Laws

inverting equation = negative change

multiplying = multiplying Hrxn

then

2C(s)+O2(g)→2CO(g)ΔH=−221.0kJ

2C(s)+O2(g)+4H2(g)→2CH3OH(g)ΔH=−402.4kJ

then

CO(g)+2H2(g)→CH3OH(g)

invert (1)

2CO(g)→2C(s)+O2(g)ΔH= - - 221.0kJ = +221 kJ

2C(s)+O2(g)+4H2(g)→2CH3OH(g)ΔH=−402.4kJ

divide (1) by 2 and (2) by 2

CO(g)→C(s)+1/2O2(g)ΔH= 221/2 = 110.5 kJ

C(s)+1/2O2(g)+2H2(g)→CH3OH(g)ΔH= - 201.2 kJ

add both equations

CO(g)+C(s)+1/2O2(g)+2H2(g)→C(s)+1/2O2(g) + CH3OH(g) HRxn = 110.5 - 201.2 = -90.7 kJ

cancel common terms

CO(g) +2H2(g)→ CH3OH(g) **HRxn = -90.7 kJ**

**which is what we wanted**

From the enthalpies of reaction
H2(g)+F2(g)→2HF(g)ΔH=−537kJ
C(s)+2F2(g)→CF4(g)ΔH=−680kJ
2C(s)+2H2(g)→C2H4(g)ΔH=+52.3kJ
Calculate ΔH for the reaction of
ethylene with F2: C2H4(g)+6F2(g)→2CF4(g)+4HF(g)

use standard enthalpies of format to calculate Hrxn
for each reaction.
1:2H2S (g) +3O2 (g) 2H2O (l)
2SO2 (g)
2: 3NO2 (g)+H2O
(l) 2HNO3 (aq) +NO (g)
3: N2O4 (g) + 4H2 (g) N2
(g) +4H2O (g)
4: CR2O3 (S) +3CO (g) 2CR
(S)+3CO2 (g)
Calculate Kc for each reaction and predict whether
reactant and product s will be favor at equilibrium
1:
CH3OH(g) CO (g) +
2H2 (g)
2: 2CO2 (g)+2CF2
(g) 4COF2 (g)
3: 6COF2
(g) 3CO2(g)+3CF2
(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)

From the enthalpies of reaction
2H2(g)+O2(g)→2H2O(g)ΔH=−483.6kJ3O2(g)→2O3(g)ΔH=+284.6kJ calculate
the heat of the reaction 3H2(g)+O3(g)→3H2O(g) Express your answer
using four significant figures. I have no idea how to approach this
problem.

Given the standard reaction enthalpies for these two
reactions:
(1) 2C(s) + 2H2(g) = C2H4(g)...... ΔrH° = 52.3 kJ/mol
(2) 2C(s) + 3H2(g) = C2H6(g)......ΔrH° = -84.7 kJ/mol
calculate the standard reaction enthalpy for the reaction: (3)
C2H4(g) + H2(g) = C2H6(g)......ΔrH° = ?

Applying Hess’s Law, from the enthalpies of reactions,
N2(g) + 3H2(g) → 2 NH3(g) ΔH = − 91.8 kJ
O2(g) + 2H2(g) → 2H2O (g) ΔH = − 483.7 kJ
N2(g) + O2(g) → 2NO(g) ΔH = 180.6 kJ
Calculate the enthalpy change (ΔHrxn) for the reaction: 4NH3(g)
+ 5O2(g) → 4 NO (g) + 6H2O(g)

Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g)
given the following reactions and enthalpies of formation:
12N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ
12N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ

Use the standard reaction enthalpies given below to determine
ΔH°rxn for the following reaction:
2 S(s) + 3 O2(g) → 2 SO3(g) ΔH°rxn = ?
Given: SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ 2 SO2(g) + O2(g)
→ 2 SO3(g) ΔH°rxn = -197.8 kJ
Please explain in detail.

Given the following reactions and their enthalpies:
ΔH(kJ/mol)−−−−−−−−−−− H2(g)⟶2H(g) +436 O2(g)⟶2O(g) +495
H2+12O2(g)⟶H2O(g) −242 Part A Devise a way to calculate ΔH for the
reaction H2O(g)⟶2H(g)+O(g)

Use standard enthalpies of formation to calculate ΔH∘rxn for the
following reaction: SO2(g)+12O2(g)→SO3(g) ΔH∘rxn =

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