Q.Calculate the standard enthalpy of formation of $CH_3OH(l)$ from the following data: $CH_3OH(l) + \tfrac{3}{2} O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$, $\Delta_r H = -726$ kJ mol$^{-1}$; $C(graphite) + O_2(g) \rightarrow CO_2(g)$, $\Delta_c H = -393$ kJ mol$^{-1}$; $H_2(g) + \tfrac{1}{2} O_2(g) \rightarrow H_2O(l)$, $\Delta_f H = -286$ kJ mol$^{-1}$.

Concept: Bomb Calorimetry Enthalpy — The given combustion enthalpy of methanol is used with standard formation enthalpies of CO₂ and H₂O to find the formation enthalpy of CH₃OH(l) via Hess’s law.

Step 1: Write the target formation reaction:

$C(graphite) + 2H_2(g) + \tfrac{1}{2} O_2(g) \rightarrow CH_3OH(l)$

Step 2: Reverse the combustion of methanol:

$CO_2(g) + 2H_2O(l) \rightarrow CH_3OH(l) + \tfrac{3}{2} O_2(g)$, $\Delta H = +726$ kJ mol$^{-1}$

Step 3: Add the formation reactions of CO₂ and H₂O:

$C + O_2 \rightarrow CO_2$, $\Delta H = -393$ kJ mol$^{-1}$ …

The problem gives you three thermochemical equations and asks for the standard enthalpy of formation of liquid methanol. That’s the $\Delta_f H^\circ$ for the reaction:

You don’t measure this directly — you use Hess’s Law, which says enthalpy change for a reaction is the same whether it happens in one step or many. The combustion of methanol is given, and the formation enthalpies of CO₂ and H₂O are known. So you can treat the formation of methanol as the reverse of its combustion, plus the formation of its products from elements.

Let’s write down what we have:

1. Combustion of methanol:

2. Formation of CO₂:

3. Formation of H₂O(l):

We want the formation reaction of methanol. A clean way: start with the combustion equation reversed (so methanol becomes a product), then add the formation reactions for CO₂ and H₂O in the right amounts.

Step 1: Reverse the combustion equation.

Reversing flips the sign of $\Delta H$:

Step 2: Add the formation of CO₂.

We need CO₂ on the left to cancel it. So take:

Step 3: Add the formation of 2 moles of H₂O.

We need 2H₂O on the left to cancel. So multiply the water formation equation by 2:

Step 4: Add all three equations.

Write them stacked:

$\begin{aligned} &CO_2(g) + 2H_2O(l) \rightarrow CH_3OH(l) + \tfrac32 O_2(g) &&\Delta H = +726 \\ &C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) &&\Delta H = -393 \\ &2H_2(g) + O_2(g) \rightarrow 2H_2O(l) &&\Delta H = -572 \\ \end{aligned}$ …