Question

We can use Hess's law to calculate enthalpy changes that cannot
be measured. One such reaction is the conversion of methane to
ethylene: |
H∘ for this reaction using the following
thermochemical data:
Express your answer to four significant figures and include the appropriate units. |

Answer #1

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

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)

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

Use Hess's Law to calculate the enthalpy change for the
reaction:
GdO3 + 3H2(g) -> Gd(s) + H2O
(g)
from the following data:
2Gd(s) + 3O2 --> 2GdO3(s)
^H=-685.4kJ
2H2 (g) + O2 (g) -->
2H2O(g) ^H= -47.8kJ

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)

Use Hess's Law to calculate the enthalpy change for recovering
tungsten from its oxide using the reaction:WO3(s) +
3H2(g) -->
W(s)+3H2O(g)
from the following data :
2W(s)+
3O2(g)-->2WO3(s) ,
change in H = -1685.4 kJ
2H2(g)+O2(g)--->
2H2O(g) , change in H = -477.84 kJ

I am having difficulty with understanding this particular types
of problems.
1) Calculate ΔHrxn for the following reaction:
C(s)+H2O(g)→CO(g)+H2(g)
Use the following reactions and given ΔH values:
C(s)+O2(g)→CO2(g), ΔH= -393.5 kJ
2CO(g)+O2(g)→2CO2(g), ΔH= -566.0 kJ
2H2(g)+O2(g)→2H2O(g), ΔH= -483.6 kJ
2) Calculate ΔHrxn for the following reaction:
5C(s)+6H2(g)→C5H12(l)
Use the following reactions and given ΔH values.
C5H12(l)+8O2(g)→5CO2(g)+6H2O(g),
ΔH=−3244.8 kJ
C(s)+O2(g)→CO2(g), ΔH=−393.5
kJ
2H2(g)+O2(g)→2H2O(g), ΔH=−483.5
kJ

1)
Calculate the change in enthalpy (in kJ) for the reaction using
the Enthalpy tables in
the back of your book.
CaCO3(s) CaO(s) + CO2(g)
2)
5. The combustion of ethane, C2H4, is an exothermic
reaction.
C2H4(g) + 3 O2(g) 2 CO2(g) + 2 H2O(l) ∆H = -1.39 x 103
J
Calculate the amount of heat liberated when 4.79 g of C2H4 reacts
with excess oxygen.

Using the Information below determine the change in enthalpy for
the following reaction:
2NO (g) + 5H2
(g)!2NH3
(g) + 2H2O
(l)
H2 (g) + 1⁄2O2 (g)!H2O (l) ΔH° = -285.8 kJ
N2 (g) + O2 (g)!2NO (l) ΔH° = +180.5 kJ
2NH3 (g)!N2 (g) + 3H2 (g)ΔH° = +92.22 kJ
a)-197.52 kJ
b)-241.7 kJ
c)-483.3 kJ
d)-659.88 kJ
e)-844.3 kJ
please show me which equation is first second and thrid and
reason why? i may be taking the wrong...

A.) Express the equilibrium constant for the combustion of
ethane in the balanced chemical equation.
2C2H6(g)+7O2(g)⇌4CO2(g)+6H2O(g)
K=[C2H6]2[O2]7 / [CO2]4[H2O]6
K=[CO2]4 / [C2H6]2[O2]7
K=K=[CO2]4[H2O]6 / [C2H6]2[O2]7
K=[CO2][H2O] / [C2H6]2[O2]
B.)Consider the chemical equation and equilibrium constant at
25∘C:
H2(g)+I2(g)⇌2HI(g) , K=6.2×102
Calculate the equilibrium constant for the following reaction at
25∘C:
HI(g)⇌12H2(g)+12I2(g)
Express the equilibrium constant to two significant figures.
C.) Consider the following reaction and corresponding value of
Kc:
H2(g)+Br2(g)⇌2HBr(g) , Kc=1.9×1019 at 25∘C
What is the value of Kp...

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