CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 6, Problem 6.108P

(a)

Interpretation Introduction

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 (ΔHrxn°) is as follows:

  ΔHrxn°=mΔHf (products)°nΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(a)

Expert Solution
Check Mark

Answer to Problem 6.108P

1.25×103kJ is released when 25.0 g of methane burns in excess O2.

Explanation of Solution

The expression to calculate the moles of CH4 is:

  MolesofCH4=(mass of CH4Molar mass of CH4)        (1)

Substitute 25g for mass of CH4 and 16.04g/mol for molar mass of CH4 in the equation (1).

  MolesofCH4=(25g16.04g/mol)=1.5586mol

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 (ΔHrxn°) is as follows:

  ΔHrxn°=[{1ΔHf°[CO2(g)]+2ΔHf°[H2O(l)]}{1ΔHf°[CH4(g)]+2ΔHf°[O2(g)]}]        (2)

Substitute 393.5kJ/mol for ΔHf°[CO2(g)], 285.840kJ/mol for ΔHf°[H2O(l)], 74.87kJ/mol for ΔHf°[CH4(g)] and 0 for ΔHf°[O2(g)] in the equation (2).

  ΔHrxn°=[{(1mol)(393.5kJ/mol)+(2mol)(285.840kJ/mol)}{(1mol)(74.87kJ/mol)+0}]=802.282 kJ/mol

The heat released by 1.5586mol of CH4 is calculated as follows:

  Heat relased=(1.5586mol)(802.282 kJ/mol)=1250.4kJ1.25×103kJ.

Conclusion

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 Introduction

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 massof compound(g)(1moleof 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+2B3C

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 1mol of a substance by 1K.

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)

Expert Solution
Check Mark

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 ofCO2=moles ofCH4(1molCO21molCH4)        (3)

Substitute 1.5586mol for moles of CH4 in the equation (3).

  Moles ofCO2=1.5586mol(1molCO21molCH4)=1.5586mol

The formula to calculate the moles of H2O is as follows:

  Moles ofH2O=moles ofCH4(2molH2O1molCH4)        (4)

Substitute 1.5586mol for moles of CH4 in the equation (4).

  Moles ofH2O=1.5586mol(2molH2O1molCH4)=3.1172mol

The formula to calculate the moles of O2 is as follows:

  Moles ofO2=moles ofCH4(2molO21molCH4)        (5)

Substitute 1.5586mol for moles of CH4 in the equation (5).

  Moles ofO2=1.5586mol(2molH2O1molCH4)=3.1172mol

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 ofN2=(moles ofO2)(0.79molN20.21molO2)        (6)

Substitute 3.1172mol for moles of O2 in the equation (6).

  Moles ofN2=(3.1172mol)(0.79molN20.21molO2)=11.72661mol

The formula to calculate the final temperature is as follows:

  q=[(molCO2)(CCO2)(TfinalTinitial)+(molH2O)(CH2O)(TfinalTinitial)+(molN2)(CN2)(TfinalTinitial)]        (7)

Rearrange the equation (7) to calculate Tfinal as follows:

  Tfinal=(q[(molCO2)(CCO2)+(molH2O)(CH2O)+(molN2)(CN2)]+Tinitial)        (8)

Substitute 1.5586mol for molCO2, 57.2 J/mol°C for CCO2, 3.1172mol for molH2O, 36.0 J/mol°C for CH2O, 11.72661mol 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(1000J1kJ)[(1.5586mol)(57.2 J/mol°C)+(3.1172mol)(36.0 J/mol°C)+(11.72661mol)(30.5 J/mol°C)]+0.0°C)=2236.72°C2.24×103 °C.

Conclusion

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 >CUSTOM<

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