Q.Calculate the enthalpy change for the process CCl4(g)→C(g)+4Cl(g) and calculate the bond enthalpy of C–Cl in CCl4(g). ΔvapH(CCl4)=30.5 kJ mol−1; ΔfH(CCl4)=−135.5 kJ mol−1; ΔaH(C)=715.0 kJ mol−1 (enthalpy of atomisation); ΔaH(Cl2)=242 kJ mol−1.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Bond Enthalpy
Bond Enthalpy: The Energy Cost of Breaking a Bond
Think of a chemical bond as a spring holding two atoms together. To pull the atoms apart, you have to do work — you have to put energy in. That energy, per mole of bonds broken, is the bond enthalpy. The stronger the bond, the more energy you need to supply, and the larger the bond enthalpy.
The intuition is simple: breaking bonds costs energy; forming bonds releases energy. A reaction is exothermic if the energy released in forming new bonds is greater than the energy consumed in breaking old ones.
The Precise Definition
Bond enthalpy (symbol: ΔHbond or B.E.) is defined under a specific set of conditions:
The average enthalpy change when one mole of a particular covalent bond is broken, with all species in the gaseous state.
Three key points are packed into that sentence:
- One mole of bonds — we measure the energy for Avogadro's number of bonds, not just one.
- Gaseous state — the atoms or molecules must be in the gas phase. This removes complications from intermolecular forces (like hydrogen bonding in liquid water) and lets us isolate the bond energy itself.
- Average — in a molecule like water (H2O), the two O–H bonds are not identical in energy. The first O–H bond in H2O requires about 502 kJ/mol to break, but the second (in the remaining OH radical) requires about 427 kJ/mol. So we report the average bond enthalpy for O–H: roughly 464 kJ/mol.
Bond enthalpy is always positive — it is the energy absorbed to break a bond. Bond formation releases the same amount of energy (negative enthalpy change).
How It's Used: Estimating Reaction Enthalpies
You can estimate the enthalpy change of a reaction (ΔHrxn) using bond enthalpies:
ΔHrxn=∑(bond enthalpies of bonds broken)−∑(bond enthalpies of bonds formed)
The logic: you put energy in to break bonds (positive), and you get energy out when bonds form (negative). So:
ΔHrxn=Energy in−Energy out
This method gives an estimate, not an exact value. Bond enthalpies are averages taken from many different molecules, so they don't account for the exact molecular environment. For precise work, use standard enthalpies of formation.
Example: Combustion of Methane
Consider: CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Bonds broken (energy in):
- 4 C–H bonds: 4×413=1652 kJ/mol
- 2 O=O bonds: 2×498=996 kJ/mol
- Total in: 1652+996=2648 kJ/mol
Bonds formed (energy out):
- 2 C=O bonds: 2×799=1598 kJ/mol
- 4 O–H bonds: 4×464=1856 kJ/mol
- Total out: 1598+1856=3454 kJ/mol
ΔHrxn=2648−3454=−806 kJ/mol
The negative sign tells you the reaction is exothermic — more energy is released in forming bonds than was consumed in breaking them. …
Concept: Bond dissociation enthalpy from atomisation data and Hess's law.
The process CClX4(g)C(g)+4Cl(g) breaks all four C–Cl bonds in gaseous CClX4. We construct a thermochemical cycle using the given data.
Step 1: Form CClX4(g) from elements in standard states:
C(s)+2ClX2(g)CClX4(g)ΔH=−135.5 kJ mol−1
Step 2: Atomise the elements:
- C(s)C(g): ΔaH=+715.0 kJ mol−1
- 2ClX2(g)4Cl(g): 2×242=+484 kJ mol−1
Step 3: Apply Hess's law. The enthalpy change for CClX4(g)C(g)+4Cl(g) equals the reverse of formation plus atomisation:
ΔH=−(−135.5)+715.0+484=135.5+715.0+484=1334.5 kJ mol−1
Step 4: Since this breaks four equivalent C–Cl bonds: …
To break CCl4(g) into gaseous atoms, we reverse its formation and add atomisation energies for the elements. The total enthalpy change is 1304 kJ mol−1, giving a C–Cl bond enthalpy of 326 kJ mol−1.
Why this approach works
Bond enthalpy measures the energy needed to break one mole of a particular bond in the gas phase, producing gaseous atoms. For CCl4, we have four C–Cl bonds, so the process CCl4(g)→C(g)+4Cl(g) requires breaking all four. The challenge is that we're not given this dissociation energy directly—instead, we have formation data and atomisation energies.
The key insight: we can construct any thermochemical process by combining others through Hess's Law. We'll build a cycle that takes CCl4(g) back to its elements in their standard states, then atomises those elements into gaseous atoms.
Step-by-step construction
1. Reverse the formation of CCl4(l)
The standard enthalpy of formation tells us:
C(s)+2Cl2(g)→CCl4(l)ΔH=−135.5 kJ mol−1
Reversing this:
CCl4(l)→C(s)+2Cl2(g)ΔH=+135.5 kJ mol−1
2. Vaporise CCl4 to get the gaseous molecule
We need CCl4(g), not the liquid. The vaporisation enthalpy is given:
CCl4(l)→CCl4(g)ΔH=+30.5 kJ mol−1
Combining steps 1 and 2, we can write:
CCl4(g)→C(s)+2Cl2(g)ΔH=+135.5−30.5=+105.0 kJ mol−1
A common mistake is forgetting to account for the phase of CCl4. The formation enthalpy given is for the liquid, so we must subtract the vaporisation energy to work with the gas.
3. Atomise carbon
Now we break the solid carbon into gaseous atoms:
C(s)→C(g)ΔH=+715.0 kJ mol−1
This is the enthalpy of atomisation (or sublimation energy) of carbon.
4. Atomise chlorine
We need four moles of Cl(g), which means breaking two moles of Cl2 bonds:
2Cl2(g)→4Cl(g)ΔH=2×242=+484 kJ mol−1
Each Cl2 molecule requires 242 kJ mol−1 to dissociate.
5. Sum the entire cycle …
- AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQQ.The enthalpy of atomization of CH3NH2(g) is 2313kJmol−1. If ΔC−HH⊖ and ΔN−HH⊖ are 414 and 389 kJmol−1 respectively, then ΔC−NH⊖ (in kJmol−1) will be (A) 293 (B) 1510 (C) 682 (D) 778
›Reveal solutionSolution
Atomization enthalpy is the sum of all bond enthalpies broken. CH3NH2 has 3 C–H, 2 N–H and 1 C–N bond; solving for the unknown C–N bond enthalpy gives 293 kJ/mol.
Concept and Intuition
Enthalpy of atomization is the energy needed to break all bonds in one mole of a gaseous molecule into free gaseous atoms — numerically, the sum of individual bond enthalpies (assuming additivity, as is standard at this level).
Step-by-Step Solution
- Structure of CH3NH2: a carbon bonded to 3 H atoms and 1 N atom; the nitrogen bonded to 2 H atoms and that same carbon. So it has 3 (C–H) + 1 (C–N) + 2 (N–H) bonds.
- Total atomization enthalpy: 3ΔC−HH+ΔC−NH+2ΔN−HH=2313.
- Substitute given values: 3(414)+ΔC−NH+2(389)=2313. …
- AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQQ.Consider the following reaction C2H5Cl(g)→2C(g)+5H(g)+Cl(g) ΔaH=3047 kJmol−1 The ΔbondH of C-Cl and C-C is 431 and 414 kJmol−1 respectively. What is the average ΔbondH of C-H in kJmol−1? (A) 436 (B) 431 (C) 440.4 (D) 446
›Reveal solutionSolution
Atomization enthalpy of C2H5Cl is the sum of all bond enthalpies broken (1 C–C + 1 C–Cl + 5 C–H); solving for the unknown gives the average C–H bond enthalpy = 440.4 kJ mol⁻¹.
Concept and Intuition
C2H5Cl is CH3-CH2-Cl. Complete atomization breaks EVERY bond in the molecule into gaseous atoms: one C–C bond, one C–Cl bond, and five C–H bonds (three on the terminal CH₃, two on the CH₂). The enthalpy of atomization is simply the sum of the enthalpies needed to break each of these bonds (bond enthalpies are always endothermic to break).
Step-by-Step Solution
- Write the atomization enthalpy as a sum of bond enthalpies: ΔaH=ΔH(C-C)+ΔH(C-Cl)+5×ΔH(C-H)
- Substitute known values: 3047=414+431+5ΔH(C-H)
- Simplify: 3047−845=5ΔH(C-H) 2202=5ΔH(C-H)
- Solve: …
- AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQQ.Match the following List-I (Bond) — List-II (Bond enthalpy, in kJ mol−1) A. Si–Si — I. 240 B. C–C — II. 297 C. Sn–Sn — III. 348 D. Ge-Ge — IV. 260 The correct answer is (A) A-II, B-III, C-I, D-IV (B) A-II, B-IV, C-III, D-I (C) A-III, B-II, C-I, D-IV (D) A-III, B-I, C-IV, D-II
›Reveal solutionSolution
Group 14 catenation bond enthalpies decrease steadily down the group (C > Si > Ge > Sn); matching the given numbers to this trend gives A-II, B-III, C-I, D-IV — option (A).
Concept and Intuition
Catenation (self-linking) ability in Group 14 falls sharply as we go down the group, because atomic size increases and orbital overlap for the E–E bond becomes progressively weaker and longer. So we expect the bond dissociation enthalpy order C–C > Si–Si > Ge–Ge > Sn–Sn, and we can match the four given numerical values to this monotonically decreasing order.
Step-by-Step Solution
- Recall (or derive from the physical trend) the standard bond-enthalpy values: C–C = 348, Si–Si = 297, Ge–Ge = 260, Sn–Sn = 240 kJ/mol — strictly decreasing down the group.
- Rank the given List-II values in decreasing order: III (348) > II (297) > IV (260) > I (240). …
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.If the mean bond dissociation energies of C-C and C-H are 80 and 90 kJmol−1 respectively, the standard enthalpy of atomization of n-butane(g) (in kJmol−1) is (A) -980 (B) 1040 (C) 1140 (D) -1140
›Reveal solutionSolution
Summing the bond energies of all 3 C-C and 10 C-H bonds in n-butane gives an atomization enthalpy of +1140kJ mol−1.
Concept and Intuition
Atomization enthalpy is the energy needed to break a mole of a compound completely into its constituent gaseous atoms — it is always endothermic (positive), equal to the sum of all the bond energies present in the molecule (using mean bond enthalpies here since actual bond strengths vary slightly by position).
Step-by-Step Solution
- n-Butane structure: CH3−CH2−CH2−CH3, which has 3 C-C bonds and 10 C-H bonds (3 CH3/CH2 groups contribute: terminal CH3 ×2 = 6 C-H, middle CH2 ×2 = 4 C-H, total 10 C-H; plus 3 C-C linking bonds).
- Total bond energy = 3×80+10×90=240+900=1140kJ mol−1. …
- AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQQ.The standard enthalpy of atomization of ethane according to the equation C2H6(g)→2C(g)+6H(g) is 622 kJ mol−1. If standard mean C-H bond dissociation enthalpy is 90 kJ mol−1, the standard mean dissociation enthalpy of C-C bond (in kJ mol−1) is (A) 540 (B) 90 (C) 85 (D) 82
›Reveal solutionSolution
Subtracting the energy used to break 6 C–H bonds (540 kJ/mol) from the total atomization enthalpy (622 kJ/mol) leaves 82 kJ/mol for the single C–C bond.
Concept and Intuition
Atomization enthalpy is the total energy needed to break all bonds in a molecule into gaseous atoms. Ethane has 6 C–H bonds and 1 C–C bond; the sum of the individual bond dissociation enthalpies must equal the total atomization enthalpy.
Step-by-Step Solution
- Total atomization enthalpy of C2H6: 622 kJ/mol (breaks all bonds: 6 C–H + 1 C–C).
- Energy for 6 C–H bonds: 6×90=540 kJ/mol. …
- AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQQ.H-X mean single bond enthalpy is maximum when X is (A) O (B) F (C) Cl (D) I
›Reveal solutionSolution
H–F has the largest mean single-bond enthalpy of the given hydrides, so X = F.
Concept and Intuition
Bond enthalpy measures the energy needed to break one mole of a bond in the gas phase. A shorter bond formed by a small, highly electronegative atom generally gives stronger, more efficient orbital overlap with hydrogen's 1s orbital, hence a higher bond enthalpy. Fluorine is the smallest halogen here, so H–F is short and strong.
Step-by-Step Solution
- List the mean H–X single-bond enthalpies (kJ/mol): H–F ≈ 565, H–O ≈ 463, H–Cl ≈ 431, H–I ≈ 299.
- Down a group the atom gets larger, the bond longer, and the bond enthalpy falls: H–F > H–Cl > H–Br > H–I.
- Compare the four given: 565 (F) > 463 (O) > 431 (Cl) > 299 (I).
- The maximum belongs to X = F.
Common Mistakes …
- AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQQ.Bond enthalpy of Ge-Ge bond is 260 kJ mol−1. The bond enthalpies of Si-Si and Sn-Sn bonds in kJ mol−1 are respectively, (A) 240, 270 (B) 297, 297 (C) 297, 240 (D) 200, 348
›Reveal solutionSolution
Group-14 element-element single bond enthalpies steadily decrease down the group (C>Si>Ge>Sn); the standard values are 348, 297, 260, 240 kJ/mol respectively, giving Si–Si =297 and Sn–Sn =240.
Concept and Intuition
Down group 14, atomic size increases and the valence orbitals become more diffuse, so orbital overlap between bonded atoms in the element-element single bond weakens progressively. This makes catenation (chain formation) strongest for carbon and weakest for the heavier elements, and is reflected quantitatively in decreasing bond dissociation enthalpies: C–C > Si–Si > Ge–Ge > Sn–Sn.
Step-by-Step Solution
- Recall the standard trend (NCERT data, kJ/mol): C–C ≈348, Si–Si ≈297, Ge–Ge ≈260, Sn–Sn ≈240. …
- AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQQ.The atomization enthalpy of CH4 is 1660 kJmol−1. The C-H bond enthalpy of each successive step in CH4→CH3→CH2→CH are +15, +30 and +45 kJmol−1 higher than the mean bond enthalpy of CH bonds, respectively. The bond enthalpy of the last C-H unit is (A) 400 kJmol−1 (B) 325 kJmol−1 (C) 475 kJmol−1 (D) 385 kJmol−1
›Reveal solutionSolution
Subtracting the first three (mean + offset) bond enthalpies from the total atomization enthalpy leaves 325 kJ/mol for the last C–H bond.
Concept and Intuition
The atomization enthalpy of methane (breaking all four C–H bonds completely, CH4→C+4H) is the sum of the four successive, individually distinct bond dissociation enthalpies — these differ from each other because each successive dissociation happens from a different, changing molecular species (CH4→CH3→CH2→CH→C), even though the "mean bond enthalpy" often quoted is just their average. We're given each step's enthalpy relative to that mean, and need the last (4th) step's actual value.
Step-by-Step Solution
- Total atomization enthalpy = sum of 4 individual C–H bond enthalpies = 1660 kJ/mol.
- Mean bond enthalpy =41660=415 kJ/mol.
- Step 1 (CH4→CH3): 415+15=430 kJ/mol.
- Step 2 (CH3→CH2): 415+30=445 kJ/mol. …
- AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQQ.Find the ΔHo for the reaction H2O(g)+Br2(g)→HBr(g)+HOBr(g) using: Bond energy of O−H bond in Water =463 kJ Bond energy of Br−Br bond in Br2=192 kJ Bond energy of O−Br bond in HOBr=234 kJ Bond energy of H−Br bond in HBr=364 kJ (A) 655 kJ.mol−1 (B) 598 kJ.mol−1 (C) 57 kJ.mol−1 (D) 192 kJ.mol−1
›Reveal solutionSolution
A bond-enthalpy (Hess's law style) calculation: ΔH=Σ(bonds broken)−Σ(bonds formed), giving 57 kJ.mol−1.
Concept and Intuition
ΔHo for a gas-phase reaction can be estimated from bond dissociation
energies: energy is absorbed to break bonds in reactants, and released when new
bonds form in products.
ΔH=∑BE(reactants)−∑BE(products)
The key is to identify exactly which bonds are actually broken and which are
newly formed — bonds that are simply "carried over" unchanged from reactant to
product don't enter the calculation.
Step-by-Step Solution
- Write the reaction with structures: H−O−H(g)+Br−Br(g)→H−Br(g)+H−O−Br(g).
- Track the atoms: one H atom leaves water and combines with one Br atom to form HBr. The remaining −OH fragment combines with the other Br atom to form HOBr.
- Bonds broken: one O–H bond of water (the other O–H bond survives intact in HOBr) and the Br–Br bond: 463+192=655 kJ. …
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