Concept explainers
(a)
Interpretation:
The heat that is released when 25.0 g of methane burns in excess O2 is to be calculated.
Concept introduction:
The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactant at the standard conditions.
The formula to calculate the standard enthalpy of reaction (ΔH°rxn) is as follows:
ΔH°rxn=∑mΔH°f (products)−∑nΔH°f (reactants)
Here, m and n are the
(a)
Answer to Problem 6.108P
−1.25×103 kJ is released when 25.0 g of methane burns in excess O2.
Explanation of Solution
The expression to calculate the moles of CH4 is:
Moles of CH4=(mass of CH4Molar mass of CH4) (1)
Substitute 25 g for mass of CH4 and 16.04 g/mol for molar mass of CH4 in the equation (1).
Moles of CH4=(25 g16.04 g/mol)=1.5586 mol
The balanced chemical equation for the reaction of CH4 and O2 is as follows:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
The formula to calculate the standard enthalpy of a given reaction (ΔH°rxn) is as follows:
ΔH°rxn=[{1ΔH°f[CO2(g)]+2ΔH°f[H2O(l)]}−{1ΔH°f[CH4(g)]+2ΔH°f[O2(g)]}] (2)
Substitute −393.5 kJ/mol for ΔH°f[CO2(g)], −285.840 kJ/mol for ΔH°f[H2O(l)], −74.87 kJ/mol for ΔH°f[CH4(g)] and 0 for ΔH°f[O2(g)] in the equation (2).
ΔH°rxn=[{(1 mol)(−393.5 kJ/mol)+(2 mol)(−285.840 kJ/mol)}−{(1 mol)(−74.87 kJ/mol)+0}]=–802.282 kJ/mol
The heat released by 1.5586 mol of CH4 is calculated as follows:
Heat relased=(1.5586 mol)(–802.282 kJ/mol)=–1250.4 kJ≈−1.25×103 kJ.
The standard enthalpy of reaction depends on the standard enthalpy of formation of the product and the standard enthalpy of formation of reactant at the standard conditions.
(b)
Interpretation:
The temperature of the product mixture is to be calculated.
Concept introduction:
The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:
Moles of compound(mol)=[given mass of compound(g)(1mole of compound(mol)molecular mass of compound(g))]
Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.
Consider the general reaction,
A+2B→3C
One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.
Molar heat capacity (Cm) of a substance is the amount of heat needed to raise the temperature of 1 mol of a substance by 1 K.
The formula to calculate heat required is as follows:
q=(mol)(Cm)(ΔT)
Here,
ΔT is the temperature difference.
q is the heat released or absorbed.
Cm is the molar heat capacity of the substance.
(b)
Answer to Problem 6.108P
The temperature of the product mixture is 2.24×103 °C.
Explanation of Solution
The formula to calculate the moles of CO2 is as follows:
Moles of CO2=moles of CH4(1 mol CO21 mol CH4) (3)
Substitute 1.5586 mol for moles of CH4 in the equation (3).
Moles of CO2=1.5586 mol(1 mol CO21 mol CH4)=1.5586 mol
The formula to calculate the moles of H2O is as follows:
Moles of H2O=moles of CH4(2 mol H2O1 mol CH4) (4)
Substitute 1.5586 mol for moles of CH4 in the equation (4).
Moles of H2O=1.5586 mol(2 mol H2O1 mol CH4)=3.1172 mol
The formula to calculate the moles of O2 is as follows:
Moles of O2=moles of CH4(2 mol O21 mol CH4) (5)
Substitute 1.5586 mol for moles of CH4 in the equation (5).
Moles of O2=1.5586 mol(2 mol H2O1 mol CH4)=3.1172 mol
The mole fraction of O2 and N2 is 0.79 and 0.21.
The formula to calculate the moles of N2 is as follows:
Moles of N2=(moles of O2)(0.79 mol N20.21 mol O2) (6)
Substitute 3.1172 mol for moles of O2 in the equation (6).
Moles of N2=(3.1172 mol)(0.79 mol N20.21 mol O2)=11.72661 mol
The formula to calculate the final temperature is as follows:
q=[(molCO2)(CCO2)(Tfinal−Tinitial)+(molH2O)(CH2O)(Tfinal−Tinitial)+(molN2)(CN2)(Tfinal−Tinitial)] (7)
Rearrange the equation (7) to calculate Tfinal as follows:
Tfinal=(q[(molCO2)(CCO2)+(molH2O)(CH2O)+(molN2)(CN2)]+Tinitial) (8)
Substitute 1.5586 mol for molCO2, 57.2 J/mol°C for CCO2, 3.1172 mol for molH2O, 36.0 J/mol°C for CH2O, 11.72661 mol for molN2, 30.5 J/mol°C for CN2, 0.0 °C for Tinitial and 1250.4 kJ for q in the equation (8).
Tfinal=(1250.4 kJ(1000 J1 kJ)[(1.5586 mol)(57.2 J/mol°C)+(3.1172 mol)(36.0 J/mol°C)+(11.72661 mol)(30.5 J/mol°C)]+0.0 °C)=2236.72 °C≈2.24×103 °C.
The amount of species in the reaction depends on the stoichiometry of the reaction by the relationship between the reactants and products.
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Chapter 6 Solutions
Chemistry: The Molecular Nature of Matter and Change
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