Q.Calculate the standard enthalpy of formation of CH3OH(l) from the following data i. CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l), ΔH0 = -726 kJ mol-1 ii. C(Graphite) + O2(g) → CO2(g), ΔcH0 = -393 kJ mol-1 iii. H2(g) + 1/2 O2(g) → H2O(l), ΔfH0 = -286 kJ mol-1 Ans. : (- 239 kJ mol-1)

Hess's Law: The Chemistry Shortcut

Imagine you're climbing a mountain. You can take a direct, steep trail straight to the summit, or you can take a longer, winding path that goes through a valley first. Either way, you start at the base camp and end at the same peak. The net change in your altitude is exactly the same, no matter which route you take. The altitude difference between base and summit is a property of those two points alone — it doesn't care about the path.

That's the core idea behind Hess's Law.

In chemistry, enthalpy ($H$) is like altitude. It's a state function — its value depends only on the current state of the system (temperature, pressure, composition), not on how you got there. When a reaction happens, the change in enthalpy ($\Delta H$) is simply the difference between the enthalpy of the products and the enthalpy of the reactants:

$\Delta H = H_{\text{products}} - H_{\text{reactants}}$

Since enthalpy is a state function, $\Delta H$ for a given reaction is fixed. It doesn't matter if the reaction occurs in one step or in a dozen steps — the total enthalpy change will be identical.

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The Precise Statement …