Define standard heat of formation. Calculate the heat of formation of acetic acid from the following data : [5] C(s) + O2(g) CO2(g) ; H = 99 kcal/mol H2(g) + 12O2(g) H2O(l) ; H = 68 kcal/mol CH3COOH…

Standard Heat of Formation

The standard heat of formation ($\Delta H_f^{\circ }$) is the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states (most stable form at 1 atm and 298 K). For elements in their standard state, $\Delta H_f^{\circ }=0$.

CH₃COOHCH₃COOH

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Calculation of Heat of Formation of Acetic Acid ($\mathrm{CH}{}_3\mathrm{COOH}$)

The formation reaction of acetic acid is: $$2\mathrm{C}(\mathrm{s})+2\mathrm{H}{}_2(\mathrm{g})+\mathrm{O}{}_2(\mathrm{g})\to \mathrm{CH}{}_3\mathrm{COOH}(\mathrm{l})$$ We use Hess’s Law to calculate $\Delta H_f^{\circ }$ of $\mathrm{CH}{}_3\mathrm{COOH}$ from the given data.

2C(s)+2H₂(g)+O₂(g)Hess’s Law calculationacetic acidacetic acid

Step 1: Write the target reaction and given reactions

1. Formation of $\mathrm{CO}{}_2$: $$\mathrm{C}(\mathrm{s})+\mathrm{O}{}_2(\mathrm{g})\to \mathrm{CO}{}_2(\mathrm{g})\Delta H=-99\text{ kcal/mol}$$
2. Formation of $\mathrm{H}{}_2\mathrm{O}$: $$\mathrm{H}{}_2(\mathrm{g})+\frac{1}{2}\mathrm{O}{}_2(\mathrm{g})\to \mathrm{H}{}_2\mathrm{O}(\mathrm{l})\Delta H=-68\text{ kcal/mol}$$
3. Combustion of acetic acid: $$\mathrm{CH}{}_3\mathrm{COOH}(\mathrm{l})+2\mathrm{O}{}_2(\mathrm{g})\to 2\mathrm{CO}{}_2(\mathrm{g})+2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})\Delta H=-208\text{ kcal/mol}$$

C(s)+O₂(g)ΔH = −99 kcal/molCO₂(g)

Step 2: Reverse the combustion reaction and adjust coefficients

To isolate $\mathrm{CH}{}_3\mathrm{COOH}$, reverse the combustion reaction and multiply by $-1$: $$2\mathrm{CO}{}_2(\mathrm{g})+2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})\to \mathrm{CH}{}_3\mathrm{COOH}(\mathrm{l})+2\mathrm{O}{}_2(\mathrm{g})\Delta H=+208\text{ kcal/mol}$$

Step 3: Combine with formation reactions

Multiply the formation reactions to match the reversed combustion reaction:

• For 2 moles of $\mathrm{CO}{}_2$: $$2\mathrm{C}(\mathrm{s})+2\mathrm{O}{}_2(\mathrm{g})\to 2\mathrm{CO}{}_2(\mathrm{g})\Delta H=2(-99)=-198\text{ kcal/mol}$$
• For 2 moles of $\mathrm{H}{}_2\mathrm{O}$: $$2\mathrm{H}{}_2(\mathrm{g})+\mathrm{O}{}_2(\mathrm{g})\to 2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})\Delta H=2(-68)=-136\text{ kcal/mol}$$

Step 4: Sum the reactions

Add the reversed combustion reaction with the adjusted formation reactions: $$\begin{array}{rlr}2\mathrm{C}(\mathrm{s})+2\mathrm{O}{}_2(\mathrm{g})& \to 2\mathrm{CO}{}_2(\mathrm{g})& \Delta H=-198\text{ kcal/mol}\\ 2\mathrm{H}{}_2(\mathrm{g})+\mathrm{O}{}_2(\mathrm{g})& \to 2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})& \Delta H=-136\text{ kcal/mol}\\ 2\mathrm{CO}{}_2(\mathrm{g})+2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})& \to \mathrm{CH}{}_3\mathrm{COOH}(\mathrm{l})+2\mathrm{O}{}_2(\mathrm{g})& \Delta H=+208\text{ kcal/mol}\\ 2\mathrm{C}(\mathrm{s})+2\mathrm{H}{}_2(\mathrm{g})+\mathrm{O}{}_2(\mathrm{g})& \to \mathrm{CH}{}_3\mathrm{COOH}(\mathrm{l})& \Delta H=-198-136+208=-126\text{ kcal/mol}\end{array}$$

Step 5: Calculate $\Delta H_f^{\circ }$ per mole

Since the reaction forms 1 mole of $\mathrm{CH}{}_3\mathrm{COOH}$, the standard heat of formation is: $$\boxed{\Delta H_f^{\circ }=-126\text{ kcal/mol}}$$

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OR

a) Standard Electrode Potential

The standard electrode potential ($E^{\circ }$) is the potential difference measured under standard conditions (1 M concentration, 1 atm pressure, 298 K) between an electrode and the standard hydrogen electrode (SHE). It indicates the tendency of an electrode to gain or lose electrons.

b) Galvanic Cell Notation

Given:

• $E_{\mathrm{Zn}{}^{2+}/\mathrm{Zn}}^{\circ }=-0.76\text{ V}$ (anode, oxidation)
• $E_{\mathrm{Mg}{}^{2+}/\mathrm{Mg}}^{\circ }=-2.38\text{ V}$ (cathode, reduction)

Since $\mathrm{Mg}$ has a more negative potential, it acts as the anode (oxidation), and $\mathrm{Zn}$ acts as the cathode (reduction).

Cell notation: $$\mathrm{Mg}(\mathrm{s})|\mathrm{Mg}{}^{2+}(\mathrm{aq})||\mathrm{Zn}{}^{2+}(\mathrm{aq})|\mathrm{Zn}(\mathrm{s})$$

c) Complete Cell Reaction

• Anode (oxidation): $\mathrm{Mg}(\mathrm{s})\to \mathrm{Mg}{}^{2+}(\mathrm{aq})+2\mathrm{e}{}^{-}$
• Cathode (reduction): $\mathrm{Zn}{}^{2+}(\mathrm{aq})+2\mathrm{e}{}^{-}\to \mathrm{Zn}(\mathrm{s})$

Overall reaction: $$\mathrm{Mg}(\mathrm{s})+\mathrm{Zn}{}^{2+}(\mathrm{aq})\to \mathrm{Mg}{}^{2+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})$$

d) Cell Potential Calculation

The cell potential ($E_{\text{cell}}^{\circ }$) is calculated as: $$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$$ $$E_{\text{cell}}^{\circ }=(-0.76)-(-2.38)=1.62\text{ V}$$