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Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The ΔHf of C2H4 is 52.26 kJ/mol, CO2 is -393.509 kJ/mol, and H2O is -241.818 kJ.) C2H4 (g) + 3O2(g) 2CO2 (g) + 2H2O(g)
ΔHrxn =
The reaction is .

Respuesta :

daienrode

Answer: -355.642

See picture for explanation
Ver imagen daienrode

Answer: The enthalpy change of the reaction is -1322.91 kJ

Explanation:

The chemical equation for the combustion of propane follows:

[tex]C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_2H_4(g))})+(3\times \Delta H^o_f_{(O_2(g))})][/tex]

We are given:

[tex]\Delta H^o_f_{(H_2O(g))}=-241.818kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.509kJ/mol\\\Delta H^o_f_{(C_2H_4(g))}=52.26kJ/mol[/tex]  

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(2\times (-393.509))+(2\times (-241.818))]-[(1\times (52.26))+(3\times (0))]\\\\\Delta H^o_{rxn}=-1322.91kJ[/tex]

Hence, the enthalpy change of the reaction is -1322.91 kJ