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iogann1982 [59]
2 years ago
7

Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhou

se gas, so over-production can have negative effects on the environment. Use enthalpy of formation data to calculate the number of moles of CO2(g) produced per megajoule of heat released from the combustion of each fuel under standard conditions (1 atm and 25 �C).Each must be in mol * MJ^-1(a) coal, C(s, graphite)(b) natural gas, CH4(g);(c) propane, C3H8(g);(d) octane, C8H18(l) (?Hf� = �250.1 kJ �mol^�1).
Chemistry
1 answer:
RUDIKE [14]2 years ago
4 0

Answer:

a) 2.541 mol/MJ;

b) 1.124 mol/MJ;

c) 0.4354 mol/MJ;

d) 0.1835 mol/MJ

Explanation:

The enthalpy of formation (ΔH°f) is the enthalpy of a reaction to form a compound by its constituents. For CO₂, ΔH°f = - 393.5 kJ/mol.

The enthalpy of a reaction is the sum of the enthalpy of the products (each one multiplied by the number of moles) less the sum of the enthalpy of the reactants (each one multiplied by the number of moles). The ΔH°f for simple substances (with one atom) is 0. The combustion is the reaction between the fuel and the oxygen.

a) The combution reaction is:

C(s) + O₂(g) → CO₂(g)

ΔH°rxn = -393.5 kJ/mol = -393.5x10⁻³ MJ/mol

Number of moles per MJ released: 1/|ΔH°rxn|

n = 1/(393.5x10⁻³) = 2.541 mol/MJ

b) The combustion reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

H₂O is in the liquid state because it's at 1 atm and 25ºC.

ΔH°f, H₂O(l) = -285.3 kJ/mol

ΔH°f, O₂(g) = 0

ΔH°f, CH₄(g) = -74.8 kJ/mol

ΔH°rxn = [2*(-285.3 ) + 1*(-393.5)] - [1*(-74.8)]

ΔH°rxn = -889.3 kJ/mol = -889.3x10⁻³ MJ/mol

n = 1/889.3x10⁻³ = 1.124 mol/MJ

c) C₃H₈(g) + 10O₂(g) → 3CO₂(g) + 4H₂O(l)

ΔH°f,C₃H₈(g) = -25.2 kJ/mol

ΔH°rxn = [4*(-285.3) + 3*(-393.5)] - [1*(-25.2)]

ΔH°rxn = -2,296.5 kJ/mol = -2.2965 MJ/mol

n = 1/2.2965 = 0.4354 mol/MJ

d) C₈H₁₈(l) + (25/2)O₂(g) → 8CO₂(g) + 9H₂O(l)

ΔH°f, C₈H₁₈(l) = -250.1 kJ/mol

ΔH°rxn = [9*(-283.5) + 8*(-393.5)] - [1*(-250.1)]

ΔH°rxn = -5,449.4 kJ/mol = -5.4494 MJ/mol

n = 1/5.4494 = 0.1835 mol/MJ

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Problem PageQuestion Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume
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Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 3.16 L of chlorine were consumed at STP.

Be sure your answer has the correct number of significant digits.

Answer: Thus volume of carbon tetrachloride that would be produced is 0.788 L

Explanation:

According to ideal gas equation:

PV=nRT

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n = number of moles = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =273K=

n=\frac{PV}{RT}

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According to stoichiometry:

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Thus 0.141 moles of methane produces = \frac{1}{4}\times 0.141=0.0352 moles of carbon tetrachloride

volume of carbon tetrachloride =moles\times {\text {Molar volume}}=0.0352mol\times 22.4L/mol=0.788L

Thus volume of carbon tetrachloride that would be produced is 0.788 L

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