Calculate the standard enthalpy of formation of C2H4(g) from the follo

1.	Calculate the standard enthalpy of formation of C2H4(g) from the following:Thermo-chemical equationC2H4(g) + 3O2 (g) → 2CO2 (g) + 2H2O(g);∆rHΘ = −1323 kJ mol−1Given that ∆fHΘ of CO2 (g) and H2O (g) as -393.5 and -249 kJ mol-1 Respectively.
Answer» ∆_fH^Θof O₂(g) = 0 by convention ∆_rH^Θ = \(\sum\)∆_fH^Θ_(products) - \(\sum\)∆_fH^Θ _(reactants) Substituting the given values: −1323 = [2 × (−393.5) + 2 × (−249)] − [∆_fH^ΘC₂H₄] −1323 = [−787 − 498] − [∆_fH^ΘC₂H₄] −1323 = −1285 − ∆_fH^ΘC₂H₄ \(\therefore\) ∆_fH^ΘC₂H₄ = 1323 − 1285 = 38 kJ mol^-1

Calculate the standard enthalpy of formation of C2H4(g) from the following:Thermo-chemical equationC2H4(g) + 3O2 (g) → 2CO2 (g) + 2H2O(g);∆rHΘ = −1323 kJ mol−1Given that ∆fHΘ of CO2 (g) and H2O (g) as -393.5 and -249 kJ mol-1 Respectively.

Answer»

∆_fH^Θof O₂(g) = 0 by convention

∆_rH^Θ = \(\sum\)∆_fH^Θ_(products) - \(\sum\)∆_fH^Θ _(reactants)

Substituting the given values:

−1323 = [2 × (−393.5) + 2 × (−249)] − [∆_fH^ΘC₂H₄]

−1323 = [−787 − 498] − [∆_fH^ΘC₂H₄]

−1323 = −1285 − ∆_fH^ΘC₂H₄

\(\therefore\) ∆_fH^ΘC₂H₄ = 1323 − 1285

= 38 kJ mol^-1

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Calculate the standard enthalpy of formation of C2H4(g) from the following:Thermo-chemical equationC2H4(g) + 3O2 (g) → 2CO2 (g) + 2H2O(g);∆rHΘ = −1323 kJ mol−1Given that ∆fHΘ of CO2 (g) and H2O (g) as -393.5 and -249 kJ mol-1 Respectively.

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