AktiraAktira JEE
Subscribe
ChemistryJEE Main 4 Apr 2026, Shift 2·Q21·Numerical·Moderate

Thermodynamics · Bomb calorimetry enthalpy of formation

If 3.365 g of ethanol (ll) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The ∣ΔHf∘∣\lvert\Delta H_\mathrm{f}^\circ\rvert of ethanol at 298.15 K is × 102\times\,10^{2} kJ mol−1^{-1}. (Nearest integer)

Given :

Standard enthalpy of combustion of graphite =−393.5= -393.5 kJ mol−1^{-1}

Standard enthalpy of formation of water (ll) =−285.8= -285.8 kJ mol−1^{-1}

Molar mass in g mol−1^{-1} of C, H, O are 12, 1 and 16 respectively.

We corrected a printing error in the original paper

Answer corrected; the source key is 282 — ΔUc=−1359.8\Delta U_\mathrm{c} = -1359.8 kJ mol−1^{-1}, ΔHc=−1362.3\Delta H_\mathrm{c} = -1362.3 kJ mol−1^{-1}, ΔfH=−282.1\Delta_\mathrm{f}H = -282.1 kJ mol−1^{-1} = −2.82×102-2.82\times10^{2}, so the blank before “× 102\times\,10^{2}” is 3; also source printed “If 3.365 g of ethanol (ll) burnt completely”.

Video solution coming soon

We're animating every PYQ step by step. Subscribe to get them first.

Subscribe

See the full 4 Apr 2026, Shift 2 paper