Question

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 .

Answer 1

Answer: -355.642

See picture for explanation

Answer 2

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

Explanation:

The chemical equation for the combustion of propane follows:

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

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

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

We are given:

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

Putting values in above equation, we get:

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

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