Chemistry · Thermodynamics · NEET
ΔrH = Σ aᵢ H(products) − Σ bᵢ H(reactants). In words: add up the enthalpies of all products, add up the enthalpies of all reactants, then subtract reactants from products. The 'Σ' (sigma) just means 'add them all up'. This is NCERT equation 5.14 and it is the base rule behind every enthalpy calculation you will do in NEET.
aᵢ and bᵢ are the stoichiometric coefficients — the plain numbers written in front of each substance in the balanced equation. For CH4 + 2O2 → CO2 + 2H2O, the coefficient of O2 and H2O is 2, and the coefficient of CH4 and CO2 is 1. You multiply each substance's molar enthalpy by its own coefficient before adding. So the H2O term contributes 2 × H(H2O), not just H(H2O).
Yes, always. Molar enthalpy is 'per one mole'. If the reaction makes 2 moles of water, you get 2 times that enthalpy. Forgetting to multiply by the coefficient is the single most common mistake in these problems and NTA loves to test it. First balance the equation, then attach each coefficient to its substance, then add and subtract.
Because a change (Δ) always means 'final − initial'. Products are the final state, reactants are the initial state. So ΔrH = H(final) − H(initial) = H(products) − H(reactants). If you flip the order you get the right number with the wrong sign, which turns an exothermic reaction into an endothermic one on your answer sheet.
Yes, and that is how most NEET numericals are actually solved. The same structure becomes ΔrH° = Σ aᵢ ΔfH°(products) − Σ bᵢ ΔfH°(reactants) (NCERT eq 5.15). You just replace 'molar enthalpy H' with 'standard enthalpy of formation ΔfH°'. Remember ΔfH° of any element in its most stable form is zero, so those terms drop out.
Treat it exactly by its coefficient in the balanced equation, including fractions like ½. For the formation of XY: ½X2 + ½Y2 → XY, the reactant terms carry ½. Fractions are allowed because ΔrH is 'per mole of reaction' as you have balanced it. If you double the whole equation, ΔrH also doubles.
The standard heat of formation (kcal/mol) of Ba²⁺ is: [ΔfH° of SO4²⁻(aq) = −216, heat of crystallisation of BaSO4(s) = −4.5, ΔfH° of BaSO4(s) = −349]
The bond dissociation energies of X2, Y2 and XY are in the ratio 1 : 0.5 : 1. ΔH for the formation of XY is −200 kJ/mol. The bond dissociation energy of X2 will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. At constant pressure the reaction enthalpy ΔrH equals the heat absorbed or released (qp). A negative ΔrH means the reaction gives out heat (exothermic); a positive ΔrH means it takes in heat (endothermic).
kJ mol⁻¹, meaning 'per mole of reaction' as you have written and balanced it. If you scale the equation up or down, the value of ΔrH scales the same way.
The basic form ΔrH = Σ H(products) − Σ H(reactants) works at any fixed conditions. When every substance is in its standard state (1 bar, specified temperature), it becomes the standard reaction enthalpy ΔrH°, which is what NEET usually asks for.
We cannot measure absolute enthalpy, only changes. That is why in practice we replace each H with a measurable change like the standard enthalpy of formation ΔfH°. The 'products − reactants' structure stays exactly the same.
