Question

Calculate the Δ*H*∘ 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) |

Answer #1

Multiply first equation with 4,

4CH4(g) + 8O2(g) ---> 4CO2(g) + 8H2O(l) dHo = -3561.2 kJ

Multiply the second equation and reverse it,

2C2H6(g) ----> 2C2H4(g) + 2H2(g) dHo = 272.6 kJ

Add the two and subtract third equation from it,

2C2H6(g) + 4CH4(g) + 8O2(g) ---> 4CO2(g) + 8H2O(l) + 2C2H4(g) + 2H2(g)

- 2H2(g) + O2(g) ----> 2H2O(l)

-------------------------------------------------------------------------------------------------------------

2C2H6(g) + 4CH4(g) + 7O2(g) ----> 4CO2(g) + 6H2O(g) + 2C2H4(g)

pl. note we need another equation to cancel out CH4 and C2H4 on each side.

dHo = -3561.2 + 272.6 + 571.6 = -2717.0 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

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)
Part A
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
Express your answer to four significant figures and include the
appropriate units.

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)

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)

What mass of natural gas (CH4) must you burn to emit 275 kJ of
heat?
CH4(g)+2O2(g)→CO2(g)+2H2O(g)ΔH∘rxn=−802.3kJ
m =
Pentane (C5H12) is a component of gasoline that burns according
to the following balanced equation:
C5H12(l)+8O2(g)→5CO2(g)+6H2O(g)
Part A
Calculate ΔH∘rxn for this reaction using standard
enthalpies of formation. (The standard enthalpy of formation of
liquid pentane is -146.8 kJ/mol.)
Express your answer using five significant figures.
ΔH∘rxn =
kJ

the combustion of methane, CH4, in oxygen.
CH4(g) + 2O2(g) >> CO2(g) + 2H2O (l)
the heat of reaction at 77C and 1.00 atm is -885.5 kJ. what is
the change in volume when 1.00 mol CH4 reacts with 2.00 mol O2.
what is w for this change?
calculate delta U (change in U) for the change indicated by the
chemical equation

Imagine that a chemist puts 8.48 mol each of
C2H6 and O2 in a 1.00-L container
at constant temperature of 284 °C. This reaction occurs:
2C2H6(g) +
7O2(g) ⇄ 4CO2(g) +
6H2O(g)
When equilibrium is reached, 1.06 mol of CO2 is in
the container. Find the value of Keq for the reaction at
284 °C. Find the Equilbrium constant using both concentration
values and pressure values.

Calculate ΔH° for the combustion of methane (products are
CO2(g) and H2O(l)), given the following
thermochemical equations:
CH4(g) + O2(g) →CH2O(g) + H2O(g) ΔH = -284 kJ
CH2O(g) + O2(g) →CO2(g) + H2O(g) ΔH = -518 kJ
H2O(g) →H2O(l) ΔH = +44 kJ

1.) Describe the relationship between∆Esys and
∆Esurr,.
2.) Calculate ΔH° for the following reaction using
standard enthalpies of formation.
2C2H6(g) + 7O2(g)—> 6H2O(g) + 4CO2(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.

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