Why a new function: enthalpy
The first law tells us that the heat absorbed at constant volume equals the change in internal energy, $\Delta U = q_V$. That is a tidy result, but it describes a sealed, rigid vessel. The overwhelming majority of laboratory reactions are run not in sealed bombs but in flasks and test tubes open to the atmosphere, where the pressure stays constant and the volume is free to change. Under those everyday conditions, part of the energy budget is spent doing expansion work against the atmosphere, so the heat exchanged is no longer equal to $\Delta U$.
To handle constant-pressure processes cleanly we need a second state function, one whose change directly equals the heat measured at constant pressure. Starting from the first law written as $\Delta U = q_p - p\Delta V$, NCERT rearranges the constant-pressure heat into a form that depends only on state variables. The result is the function we call enthalpy, from the Greek enthalpien, meaning "to warm" or heat content.
| Condition | Heat equals | Why |
|---|---|---|
| Constant volume ($\Delta V = 0$) | $q_V = \Delta U$ | No expansion work, so all heat changes internal energy |
| Constant pressure | $q_p = \Delta H$ | System also does work $p\Delta V$; enthalpy absorbs that term |
Defining H = U + pV
Begin with the first law at constant pressure, where the only work is expansion work, and label the initial and final states 1 and 2:
$$U_2 - U_1 = q_p - p(V_2 - V_1)$$
Rearranging to collect each state on its own side gives $q_p = (U_2 + pV_2) - (U_1 + pV_1)$. The grouping $U + pV$ recurs for both states, so we name it the enthalpy:
$$H = U + pV$$
With this definition the constant-pressure heat becomes $q_p = H_2 - H_1 = \Delta H$. Although $q$ is in general a path-dependent quantity, $H$ is a state function because it is built entirely from the state functions $U$, $p$ and $V$. Therefore $\Delta H$ is independent of path, and so is $q_p$. For a finite change at constant pressure we may write $\Delta H = \Delta U + p\Delta V$.
One immediate consequence is the sign convention for chemical change. Because $\Delta H = q_p$, an exothermic reaction loses heat to the surroundings, so $q_p$ and hence $\Delta H$ are negative; an endothermic reaction absorbs heat, so $\Delta H$ is positive. The combustion-style formation of methane, $\ce{C(s) + 2H2(g) -> CH4(g)}$ with $\Delta H = -74.8\ \text{kJ mol}^{-1}$, is exothermic — its products sit below the reactants on an energy diagram.
ΔH = ΔU + Δn_g RT
For systems made only of solids and liquids the difference between $\Delta H$ and $\Delta U$ is negligible, because such phases barely change volume on heating, so $p\Delta V \approx 0$. The gap becomes significant only when gases are produced or consumed. To make the $p\Delta V$ term explicit for gaseous reactions, treat the gases as ideal. If $n_A$ moles of gaseous reactants occupy volume $V_A$ and $n_B$ moles of gaseous products occupy $V_B$, all at the same constant $p$ and $T$, the ideal-gas law gives $pV_A = n_A RT$ and $pV_B = n_B RT$.
Subtracting one from the other yields $p(V_B - V_A) = (n_B - n_A)RT$, that is $p\Delta V = \Delta n_g RT$. Substituting this into $\Delta H = \Delta U + p\Delta V$ gives the relation that NCERT flags as the workhorse for converting between the two energies:
$$\Delta H = \Delta U + \Delta n_g RT$$
Here $\Delta n_g$ is the number of moles of gaseous products minus the number of moles of gaseous reactants. The equation is symmetric: rearranged as $\Delta U = \Delta H - \Delta n_g RT$, it converts an enthalpy of reaction into an internal-energy change just as readily.
The $p\Delta V$ term and the $q + w$ bookkeeping come straight from the First Law of Thermodynamics. Revisit it if the derivation above moved quickly.
Counting Δn_g correctly
Most errors in $\Delta H \leftrightarrow \Delta U$ problems trace back to a miscounted $\Delta n_g$, not to the formula. The rule is narrow and strict: count only species written in the gaseous state, and ignore everything labelled solid (s), liquid (l) or aqueous (aq). The sign of $\Delta n_g$ then decides whether $\Delta H$ runs above, below, or equal to $\Delta U$.
| Sign of Δn_g | Representative reaction | Δn_g | Relation between ΔH and ΔU |
|---|---|---|---|
| $\Delta n_g > 0$ | $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ | $1 - 0 = +1$ | $\Delta H > \Delta U$ (gas produced) |
| $\Delta n_g = 0$ | $\ce{C(s) + O2(g) -> CO2(g)}$ | $1 - 1 = 0$ | $\Delta H = \Delta U$ |
| $\Delta n_g < 0$ | $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ | $2 - 4 = -2$ | $\Delta H < \Delta U$ (gas consumed) |
When ΔH equals ΔU — and how Δn_g misleads
Two slips recur. First, candidates assume $\Delta H$ always exceeds $\Delta U$; in fact $\Delta H = \Delta U$ exactly when $\Delta n_g = 0$, and $\Delta H < \Delta U$ when gas moles fall (as in ammonia synthesis). Second, aqueous and condensed species get miscounted as gases. In $\ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)}$, only $\ce{H2}$ is gaseous, so $\Delta n_g = +1$, not the total mole change.
Count gas moles only. $\Delta n_g = n_{\text{gas, products}} - n_{\text{gas, reactants}}$, ignoring s, l and aq. Sign of $\Delta n_g$ fixes whether $\Delta H$ is above, equal to, or below $\Delta U$.
Worked example: ΔH ↔ ΔU
Assuming water vapour behaves as a perfect gas, the molar enthalpy change for the vaporisation of 1 mol of water at 1 bar and 100 °C is $41.00\ \text{kJ mol}^{-1}$. Calculate the internal energy change $\Delta U$. (Take $R = 8.3\ \text{J K}^{-1}\,\text{mol}^{-1}$, $T = 373\ \text{K}$.)
The process is $\ce{H2O(l) -> H2O(g)}$. One mole of gas is produced and none is consumed, so $\Delta n_g = 1 - 0 = +1$.
Rearrange the working relation: $\Delta U = \Delta H - \Delta n_g RT$.
$\Delta U = 41.00\ \text{kJ mol}^{-1} - (1)(8.3\ \text{J K}^{-1}\text{mol}^{-1})(373\ \text{K})$
$= 41.00\ \text{kJ mol}^{-1} - 3.096\ \text{kJ mol}^{-1} = \mathbf{37.90\ \text{kJ mol}^{-1}}$.
Because gas is produced ($\Delta n_g > 0$), $\Delta H$ exceeds $\Delta U$ by exactly the work term $\Delta n_g RT$, here about $3.1\ \text{kJ}$.
For the combustion $\ce{C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)}$ at 300 K, the internal energy change is $\Delta U = -1406\ \text{kJ mol}^{-1}$. Find $\Delta H$. (Use $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$.)
Count gaseous moles only. Gaseous products: $\ce{CO2}$ contributes 2; the water is liquid, so it contributes 0. Gaseous reactants: $\ce{C2H4}$ (1) plus $\ce{O2}$ (3) gives 4.
$\Delta n_g = 2 - 4 = -2$.
$\Delta H = \Delta U + \Delta n_g RT = -1406\ \text{kJ} + (-2)(8.314\times10^{-3}\ \text{kJ K}^{-1}\text{mol}^{-1})(300\ \text{K})$
$= -1406\ \text{kJ} - 4.99\ \text{kJ} = \mathbf{-1410.99\ \text{kJ mol}^{-1}}$.
With $\Delta n_g < 0$, gas is consumed, so $\Delta H$ lies slightly below $\Delta U$ — the opposite direction to Example 1.
Heat capacity and q = CΔT
Enthalpy and internal energy are abstractions; what a calorimeter actually records is a temperature rise. Heat capacity is the bridge between them. When heat is supplied to a system the temperature rises in proportion to the heat transferred, and the proportionality coefficient is the heat capacity $C$:
$$q = C\,\Delta T$$
Provided we know $C$, we can deduce the heat by monitoring the temperature change alone. The magnitude of $C$ depends on the size, composition and nature of the system. A large heat capacity means a given amount of heat produces only a small temperature rise: water has a notably large heat capacity, which is why a great deal of energy is needed to warm it.
Specific vs molar heat capacity
Heat capacity $C$ is an extensive property — it scales with the amount of matter, just as mass, volume, internal energy and enthalpy do. To compare substances independent of sample size we divide by the amount, producing two intensive forms. The molar heat capacity $C_m = C/n$ is the heat needed to raise one mole by one degree. The specific heat (specific heat capacity) $c$ is the heat needed to raise one unit mass by one degree.
For a sample of mass $m$, the heat required to change its temperature by $\Delta T$ follows directly:
$$q = c\,m\,\Delta T = C\,\Delta T$$
| Quantity | Symbol | Defining relation | Nature |
|---|---|---|---|
| Heat capacity | $C$ | $q = C\,\Delta T$ | Extensive |
| Molar heat capacity | $C_m = C/n$ | per mole, per degree | Intensive |
| Specific heat capacity | $c$ | $q = c\,m\,\Delta T$ | Intensive |
The relation Cp − Cv = R
A gas can be heated under two reference conditions, and they are not equivalent. At constant volume the heat is $q_V = C_V\,\Delta T = \Delta U$ — every joule goes into raising the internal energy, because no work is done. At constant pressure the heat is $q_p = C_p\,\Delta T = \Delta H$ — but now the gas also expands and does work on the surroundings, so extra heat is required for the same temperature rise. That is why $C_p$ is always larger than $C_V$ for a gas.
The exact gap follows from the enthalpy definition applied to one mole of an ideal gas. Writing $\Delta H = \Delta U + \Delta(pV)$ and using $pV = RT$ for one mole gives $\Delta(pV) = R\,\Delta T$, so $\Delta H = \Delta U + R\,\Delta T$. Now substitute $\Delta H = C_p\,\Delta T$ and $\Delta U = C_V\,\Delta T$:
$$C_p\,\Delta T = C_V\,\Delta T + R\,\Delta T$$
Dividing through by $\Delta T$ leaves the result that NEET tests almost verbatim:
$$C_p - C_V = R$$
The relation holds per mole for an ideal gas, with $C_p$ and $C_V$ here understood as molar heat capacities. It cleanly summarises the whole chapter logic: the difference between the two heat capacities is precisely the work the gas does during constant-pressure heating, packaged into the gas constant $R$.
Enthalpy and heat capacity at a glance
- Enthalpy is the state function $H = U + pV$; its change equals the constant-pressure heat, $\Delta H = q_p$.
- $\Delta H = q_p$ is negative for exothermic and positive for endothermic reactions.
- For gaseous reactions, $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ counts gaseous product moles minus gaseous reactant moles.
- $\Delta H = \Delta U$ when $\Delta n_g = 0$, and effectively for solid/liquid-only systems.
- Heat capacity links temperature to heat: $q = C\,\Delta T$; specific heat gives $q = c\,m\,\Delta T$; molar heat capacity is $C_m = C/n$.
- For one mole of an ideal gas, $C_p - C_V = R$, because constant-pressure heating also pays the expansion-work term.