Calculate the standard enthalpy of formation of CH3OH (l) from the fol

1.	Calculate the standard enthalpy of formation of CH3OH (l) from the following data:(i) CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l);ΔrH° = -726 kJ mol-1(ii) C(g) + O2(g) → CO2(g);ΔcH° = -393 kJ mol-1(iii) H2(g) + 1/2O2(g) → H2O(l);ΔfH° = -286 kJ mol-1
Answer» (i) CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l); Δ_rH° = -726 kJ mol^-1 (ii) C(g) + O₂(g) → CO₂(g); Δ_cH° = -393 kJ mol^-1 (iii) H₂(g) + 1/2O₂(g) → H₂O(l); Δ_fH° = -286 kJ mol^-1 We aim at C + 2H₂ + 1/2 O₂ → CH₃OH In order to get this thermochemical equation, multiply Eq. (ii) by 1 and Eq. (iii) by 2 and subtract Eq. (i) from their sum. C + 2H₂ + 1/2O₂ → CH₃OH ΔH = (-393) + 2(-286) - (-726) = -393 - 572 + 726 = -965 + 726 = -239 kJ mol^-1 Thus, the heat of formation of CH₃OH is ΔH_f = -239 kJ mol^-1

Calculate the standard enthalpy of formation of CH3OH (l) from the following data:(i) CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l);ΔrH° = -726 kJ mol-1(ii) C(g) + O2(g) → CO2(g);ΔcH° = -393 kJ mol-1(iii) H2(g) + 1/2O2(g) → H2O(l);ΔfH° = -286 kJ mol-1

Answer»

(i) CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l);

Δ_rH° = -726 kJ mol^-1

(ii) C(g) + O₂(g) → CO₂(g);

Δ_cH° = -393 kJ mol^-1

(iii) H₂(g) + 1/2O₂(g) → H₂O(l);

Δ_fH° = -286 kJ mol^-1

We aim at

C + 2H₂ + 1/2 O₂ → CH₃OH

In order to get this thermochemical equation, multiply Eq. (ii) by 1 and Eq. (iii) by 2 and subtract Eq. (i) from their sum.

C + 2H₂ + 1/2O₂ → CH₃OH

ΔH = (-393) + 2(-286) - (-726)

= -393 - 572 + 726

= -965 + 726 = -239 kJ mol^-1

Thus, the heat of formation of CH₃OH is

ΔH_f = -239 kJ mol^-1

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Calculate the standard enthalpy of formation of CH3OH (l) from the following data:(i) CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l);ΔrH° = -726 kJ mol-1(ii) C(g) + O2(g) → CO2(g);ΔcH° = -393 kJ mol-1(iii) H2(g) + 1/2O2(g) → H2O(l);ΔfH° = -286 kJ mol-1

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