Answer to Question #297134 in General Chemistry for Barbara Cavalcanti

Question #297134

Q8.

Calculate the standard free energy change for the combustion of methane at 25oC. Assume that all reactants and products are gases. Provide an answer correct to 3 significant figures. Do not use scientific notation or include units to report your answer.


1
Expert's answer
2022-02-14T17:04:04-0500

CH4(g)+2O2(g)CO2(g)+2H2O(g)CH_{4(g)}+2O_{2(g)}\to\>CO_{2(g)}+2H_2O_{(g)}


C(s)+2H2(g)CH4(g)C_{(s)}+2H_{2(g)}\to\>CH_{4(g)} ΔHfθ=74.9kJ/mol\Delta\>H_f^\theta=-74.9kJ/mol


C(s)+O2(g)CO2(g)   ΔHfθ=393.5kJ/molC_{(s)}+O_{2(g)}\to\>CO_{2(g)}\>\>\>\Delta\>H_f^\theta=-393.5kJ/mol



2H2(g)+O2(g)2H2O(g)  ΔHfθ=241.8kJ/mol2H_{2(g)}+O_{2(g)}\to\>2H_2O_{(g)}\>\>\Delta\>H_f^\theta=-241.8kJ/mol



ΔHcombθ(CH4)=ΔHfθ(CO2)+2ΔHfθ(H2O)ΔHfθ(CH4)\Delta\>H_{comb}^\theta(CH_4)=\Delta\>H_f^\theta(CO_2)+2\Delta\>H_f^\theta(H_2O)-\Delta\>H_f^\theta(CH_4)


=393.5+2(241.8)(74.9)=-393.5+2(-241.8)-(-74.9)

=802.2kJ/mol=-802.2kJ/mol

=802=-802






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