Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane:

CH4+2O2→2H2O+CO2

ΔHrxn from a given table:

CH4 = 1656 kJ/mol

O2 = 498 kJ/mol

H2O = 928 kJ/mol

CO2 = 1598 kJ/mol

ΔHrxn = [ (1656) + (2498) ] - [ (2928) + (1598) ] = - 802 kJ/mol?

Question

Chemistry

Kevin Wiley

27 April, 11:17

Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane:

CH4+2O2→2H2O+CO2

ΔHrxn from a given table:

CH4 = 1656 kJ/mol

O2 = 498 kJ/mol

H2O = 928 kJ/mol

CO2 = 1598 kJ/mol

ΔHrxn = [ (1656) + (2498) ] - [ (2928) + (1598) ] = - 802 kJ/mol?

+5

Answers (2)

Know the Answer?

Gerald Figueroa · Answer 1

The heat of combustion is obtained by getting the difference between the summation of enthalpies of the products and the enthalpies of the reactants. Hence the equation is enthalpy change = 1598 kJ+2*928 kJ-1656 kJ-498 kJ. The answer is 1300 kJ.

Timothy Mcneil · Answer 2

We subtract the enthalpies of the products by the enthalpies of the reactants, taking into account their stoichiometric coefficients:

2ΔH (H2O) + ΔH (CO2) - ΔH (CH4) - ΔH (2O2) = 2 (928) + 1598 - 1656 - 2 (498) = 802 kJ / 1 mol methane

Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane:CH4+2O2→2H2O+CO2ΔHrxn from a given table:CH4 = 1656 kJ/molO2 = 498 kJ/molH2O = 928 kJ/molCO2 = 1598 kJ/molΔHrxn = [ (1656) + (2*498) ] - [ (2*928) + (1598) ] = - 802 kJ/mol?

Calculate the approximate enthalpy change, ΔHrxn, for the combustion of methane:

CH4+2O2→2H2O+CO2

ΔHrxn from a given table:

CH4 = 1656 kJ/mol

O2 = 498 kJ/mol

H2O = 928 kJ/mol

CO2 = 1598 kJ/mol

ΔHrxn = [ (1656) + (2498) ] - [ (2928) + (1598) ] = - 802 kJ/mol?