Thermodynamics (Class 11): First Law, Enthalpy and Hess's Law Made Simple

The first law and its sign convention

Here q is heat supplied to the system and w is work done on the system. Both are positive when energy enters. For expansion against a constant external pressure:

The minus sign is not decoration. When a gas expands, ΔV is positive, so w is negative — the system has spent energy pushing the surroundings back.

ΔU versus ΔH

ΔU is the heat change at constant volume; ΔH is the heat change at constant pressure. Since most reactions in a school laboratory happen in open vessels, ΔH is the one usually quoted.

Δn_g counts only gaseous moles: products minus reactants. If a reaction produces and consumes equal moles of gas, Δn_g is zero and ΔH equals ΔU.

Hess's law makes the impossible measurable

Enthalpy is a state function, so the total change depends only on the initial and final states, not the route. That lets you add known equations to reach an unknown one.

The enthalpy of formation of carbon monoxide cannot be measured directly, because burning carbon in limited oxygen always gives some CO₂. But combustion enthalpies of C and CO are both measurable, and subtracting one from the other gives the answer. That is the whole method: arrange known reactions so they sum to the target, reversing signs where you reverse a reaction.

FAQs

What is the difference between ΔU and ΔH?

ΔU is the heat change at constant volume, ΔH at constant pressure. They are related by ΔH = ΔU + ΔnᵍRT.

Why is work negative when a gas expands?

Because the system spends energy pushing against the surroundings. With w = −pΔV, a positive ΔV gives negative w.

Is a spontaneous reaction always fast?

No. Spontaneity is thermodynamic and says nothing about rate. Diamond turning to graphite is spontaneous but immeasurably slow.

What makes enthalpy a state function?

Its value depends only on the current state of the system, not on how that state was reached. This is what Hess's law relies on.