Q.The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol−1 at 298 K. Calculate the enthalpy change for the reaction at 298 K. NH2CN(g)+23O2(g)→N2(g)+CO2(g)+H2O(l)
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Bomb Calorimetry and Enthalpy: From Intuition to Precision
Imagine you want to know exactly how much heat a handful of cashews releases when your body burns it. You could eat them and measure your temperature rise — but that’s messy, slow, and full of biological noise. A bomb calorimeter is the chemist’s clean, controlled way to do the same thing: burn a sample completely in pure oxygen inside a sealed steel container (the “bomb”) submerged in water, and measure the temperature change of that water.
The key insight: everything stays at constant volume. The bomb is rigid — it doesn’t expand or contract. That single fact changes which thermodynamic quantity you measure directly.
What the bomb actually measures
When the sample burns, it releases heat. That heat warms the bomb, the water, and everything around it. From the temperature rise and the known heat capacity of the entire calorimeter, you calculate the heat released at constant volume, denoted qV.
For any process at constant volume with no non-expansion work (like electrical work), the first law of thermodynamics says:
qV=ΔU
where ΔU is the change in internal energy of the system (the burning sample + oxygen + products). So a bomb calorimeter directly gives you ΔU for the combustion reaction.
Constant volume means no PΔV work is done — the system can’t push against the atmosphere. All the energy change appears as heat.
But we usually want enthalpy, not internal energy
In real life — open beakers, industrial furnaces, your body — reactions happen at constant pressure (usually 1 atm). The heat released at constant pressure is called enthalpy change, ΔH. For a combustion reaction:
ΔH=ΔU+Δ(PV)
For solids and liquids, Δ(PV) is tiny. But for reactions involving gases — and combustion almost always does — the volume change matters. If the number of moles of gas changes during the reaction, the system does work on (or receives work from) the surroundings.
For a reaction at constant temperature and pressure:
ΔH=ΔU+ΔngRT
where Δng = (moles of gaseous products) − (moles of gaseous reactants), R = 8.314 J mol⁻¹ K⁻¹, and T is the temperature in Kelvin.
The precise statement
Bomb calorimetry enthalpy is the enthalpy change of a reaction calculated from the internal energy change measured in a bomb calorimeter, corrected for the PΔV work associated with any change in the number of moles of gas.
In practice:
- Measure ΔU from the bomb calorimeter experiment.
- Determine Δng from the balanced chemical equation.
- Compute ΔH=ΔU+ΔngRT.
A common mistake: assuming ΔH=ΔU for all combustion reactions. This is only true when Δng=0 — for example, burning carbon in oxygen:
C(s)+O2(g)→CO2(g) has Δng=0, so ΔH=ΔU.
But burning methane:
CH4(g)+2O2(g)→CO2(g)+2H2O(l) has Δng=1−3=−2, so ΔH=ΔU−2RT.
Why this matters for exams
You will often be given a bomb calorimeter experiment result (temperature rise, heat capacity) and asked for ΔH of combustion. The steps: …
Concept: Bomb Calorimetry Enthalpy
Bomb calorimeters measure ΔU (internal energy change) at constant volume. To find ΔH for a reaction, we use the relation:
ΔH=ΔU+ΔngRT
where Δng is the change in moles of gas (products minus reactants).
Step 1: Identify Δng from the balanced equation.
Gaseous reactants: 23 mol O2
Gaseous products: 1 mol N2+1 mol CO2=2 mol
Δng=2−23=21 mol
Step 2: Calculate ΔngRT at 298 K. …
The bomb calorimeter gives ΔU=−742.7 kJ mol−1; convert with ΔH=ΔU+ΔngRT. For this reaction Δng=+21, giving ΔH=−741.5 kJ mol−1.
Why ΔU=ΔH
A bomb calorimeter is rigid, so it measures heat at constant volume, which equals ΔU. Enthalpy (constant pressure) is related by
ΔH=ΔU+ΔngRT,
where Δng is the change in moles of gas (solids and liquids are neglected).
The reaction
NH2CN(s)+23O2(g)→N2(g)+CO2(g)+H2O(l),ΔU=−742.7 kJ mol−1.
Cyanamide is the solid burnt in the bomb, and water is liquid — neither counts toward Δng.
Change in moles of gas
- Gaseous products: N2+CO2=1+1=2 mol
- Gaseous reactants: 23O2=1.5 mol
Δng=2−23=+21 mol.
Convert to ΔH
With R=8.314 J mol−1K−1 and T=298 K: …
- KEAM 2026Set pha-2026-0418F4 marksMCQQ.1 g of graphite is burnt completely in excess oxygen at 298K and 1 atmospheric pressure in a bomb calorimeter. During the reaction, the temperature raises from 298K to 299K. If the heat capacity of the bomb calorimeter is 20.7 kJ mol−1, what is the enthalpy of combustion of C(gr)? (Atomic mass of carbon is 12 g mol−1) (A) -248 kJ mol−1 (B) +236 kJ mol−1 (C) -236 kJ mol−1 (D) +246 kJ mol−1 (E) -268 kJ mol−1
›Reveal solutionSolution
The enthalpy of combustion of graphite is about −248 kJ mol⁻¹.
Reasoning
Heat evolved for 1 g of graphite:
q=CΔT=20.7 kJ K−1×(299−298) K=20.7 kJ.
This is for 121 mol of carbon, so per mole:
qmolar=20.7×12=248.4 kJ. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.12g of pure graphite is burnt completely in a bomb calorimeter in excess of oxygen at 298 K at 1 atm. pressure. During combustion, the temperature rises from 298 K to 308 K. The heat capacity of the bomb calorimeter is 20.7 kJ K−1. What is the enthalpy change for combustion of 1 mole of graphite (in kJ mol−1) at 298 K and 1 atm. pressure? (R=8.3 JK−1 mol−1) (A) -2070 (B) -207 (C) +2070 (D) +207 (E) +2.07
›Reveal solutionSolution
Heat released at constant volume =20.7×10=207 kJ for one mole of graphite; since Δngas=0, ΔH=ΔU=−207 kJ mol−1.
12 g of graphite is exactly 1 mole (atomic mass 12). The heat absorbed by the calorimeter is:
q=CcalΔT=20.7 kJ K−1×(308−298) K=207 kJ.
Since combustion is exothermic and occurs at constant volume in a bomb calorimeter, ΔU=−207 kJ mol−1. …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.Enthalpy change is always negative for which one of the following processes? (A) Enthalpy of ionisation (B) Enthalpy of sublimation (C) Enthalpy of vapourisation (D) Enthalpy of bond dissolution (E) Enthalpy of combustion
›Reveal solutionSolution
Combustion always releases heat, so the enthalpy of combustion is always negative; the other four processes are endothermic (positive).
Sign analysis of each process:
- Enthalpy of ionisation — energy is absorbed to remove an electron: positive.
- Enthalpy of sublimation — energy needed to convert solid to gas: positive.
- Enthalpy of vaporisation — energy needed to convert liquid to gas: positive.
- Enthalpy of bond dissociation — energy needed to break a bond: positive. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.Enthalpy of combustion of ethylene gas at constant pressure of 1 atm and at 300 K is −1410 kJ mol−1. The enthalpy change for the reaction at constant volume and at the same temperature is about (R=8.3 J K−1 mol−1) (A) −1405 kJ mol−1 (B) −1415 kJ mol−1 (C) −1407.5 kJ mol−1 (D) −1417.5 kJ mol−1 (E) −1402.5 kJ mol−1
›Reveal solutionSolution
At constant volume the heat is ΔU=ΔH−ΔngRT. With Δng=−2, ΔU≈−1405 kJ mol−1.
Balanced combustion (water liquid): C2H4(g)+3O2(g)→2CO2(g)+2H2O(l).
Gaseous moles: products =2 (CO2 only), reactants =1+3=4, so Δng=2−4=−2. …
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.Enthalpy change is always negative for which one of the following processes? (A) Enthalpy of ionisation (B) Enthalpy of sublimation (C) Enthalpy of vapourisation (D) Enthalpy of fusion (E) Enthalpy of combustion
›Reveal solutionSolution
Enthalpy of combustion is always negative because combustion is exothermic.
Concept and Intuition
Burning a substance in oxygen releases energy, so combustion has a negative enthalpy change. Phase changes such as fusion, vaporisation and sublimation, and ionisation, all require input of energy and are endothermic.
Step-by-Step Solution
- Ionisation, sublimation, vaporisation, fusion: absorb energy ⇒ΔH>0. …
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