Chemistry · Thermodynamics · NEET
This is the exact confusion NCERT points at. In the whole, un-broken CH4 molecule, all four C–H bonds ARE identical — same length, same energy. The difference shows up only when you break them one after another. Breaking the first bond turns CH4 into CH3 + H. Now the leftover CH3 is a new species with different electron arrangement. So the second C–H bond you break is no longer in CH4 — it is in CH3, a different chemical environment. Different environment means a different energy. So: the bonds are equal at the start, but the four dissociation STEPS are not equal because each step starts from a different fragment.
NCERT gives the four steps (approximate values, kJ/mol): CH4 → CH3 + H needs about 427; CH3 → CH2 + H needs about 439; CH2 → CH + H needs about 452; CH → C + H needs about 347. They are clearly not the same number. Add them: 427 + 439 + 452 + 347 = 1665 kJ/mol. That total is the enthalpy of atomization of methane (breaking CH4 completely into 1 C atom and 4 H atoms, all in gas phase). You do NOT need to memorize each step value for NEET — you need the idea that they differ, plus the total 1665 and the mean 416.
Because the four step values are all different, we cannot give one true 'C–H bond energy' for methane. So NCERT defines the mean bond enthalpy: divide the total atomization energy by the number of bonds broken. For CH4: mean C–H bond enthalpy = (1/4) × 1665 = 416 kJ/mol. This single average value is what you use in calculations and in tables. Remember: for a diatomic molecule like H2 there is only one bond, so bond dissociation enthalpy = bond enthalpy exactly. For polyatomic molecules like CH4, we must use the MEAN because the individual bonds break with different energies.
Not always — the pattern is not a simple 'first is highest'. From the NCERT step values, the first three steps actually rise (427 → 439 → 452), then the last step drops (347). The reason is the stability of each leftover fragment. If breaking a bond leaves behind a more stable radical, that bond needs less energy; if it leaves a less stable, higher-energy fragment, it needs more. For NEET, do not try to predict the exact order of the four numbers. Just know they DIFFER, and know why: each step leaves a different fragment.
Bond dissociation enthalpy is the energy to break ONE specific bond in one specific step — like breaking the first C–H in CH4 (about 427 kJ/mol). Mean bond enthalpy is the AVERAGE over all the same-type bonds — like (1/4)(1665) = 416 kJ/mol for C–H in CH4. For a diatomic molecule (H2, HCl, O2) there is only one bond, so the two are equal and also equal to the enthalpy of atomization. For polyatomic molecules they differ, and tables list the MEAN value. NEET loves this exact contrast, so keep it clear.
Almost, but not exactly. NCERT says the mean C–H bond enthalpy differs slightly from compound to compound — for example it is a little different in CH3CH2Cl or CH3NO2 compared to CH4 — because the chemical environment around the C–H bond changes. But it does NOT differ by a large amount, so we can still use one average C–H value (about 416 kJ/mol) in bond-enthalpy calculations for reactions. This is why bond-enthalpy answers are approximate, not exact.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the energy to break one mole of CH4(g) completely into one mole of gaseous C atoms and four moles of gaseous H atoms. NCERT value: about 1665 kJ/mol, which is the sum of the four successive C–H bond dissociation energies (427 + 439 + 452 + 347).
Divide the total atomization energy by the number of C–H bonds: (1/4) × 1665 = 416 kJ/mol. This mean value is used because the four individual step energies are all different.
For diatomic molecules such as H2, HCl, O2, N2 — they have only one bond, so bond dissociation enthalpy = bond enthalpy = enthalpy of atomization. For polyatomic molecules like CH4 we must use the mean bond enthalpy instead.
Because the chemical environment (the atoms and groups near the C–H bond) is different, which changes the electron distribution slightly. NCERT notes the change is small, so one average C–H value is still used in calculations.
Yes. NEET tests bond dissociation enthalpy vs mean bond enthalpy, atomization enthalpy, and using bond enthalpies to find reaction enthalpy (ΔrH = bonds broken − bonds formed). Understanding why the four CH4 steps differ makes all of these easy to reason through instead of memorize.
