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JEE Main 2016 · Chemistry · Chemical Thermodynamics · medium

The combustion of liquid ethanol, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l), in a bomb calorimeter at 300 K gives Δ U=-1364.0 kJ mol^-1. Given ΔfH^°(CO2)=-394…

Single correct option

The combustion of liquid ethanol, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)\mathrm{C_2H_5OH(l)+3O_2(g)\to2CO_2(g)+3H_2O(l)}, in a bomb calorimeter at 300 K300\ \mathrm{K} gives ΔU=−1364.0 kJ mol−1\Delta U=-1364.0\ \mathrm{kJ\,mol^{-1}}. Given ΔfH∘(CO2)=−394 kJ mol−1\Delta_fH^\circ(\mathrm{CO_2})=-394\ \mathrm{kJ\,mol^{-1}}, ΔfH∘(H2O,l)=−286 kJ mol−1\Delta_fH^\circ(\mathrm{H_2O,l})=-286\ \mathrm{kJ\,mol^{-1}} and R=8.314 J mol−1 K−1R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}, the standard enthalpy of formation of liquid ethanol (in kJ mol−1\mathrm{kJ\,mol^{-1}}) is nearly
  1. A. −282.0-282.0
  2. B. −279.5-279.5
  3. C. −284.5-284.5
  4. D. −1366.5-1366.5
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Answer: (B) −279.5-279.5

Only gases count in Δng\Delta n_g: 2−3=−12-3=-1. ΔH=ΔU+ΔngRT=−1364.0+(−1)(8.314×300)/1000=−1366.5\Delta H=\Delta U+\Delta n_gRT=-1364.0+(-1)(8.314\times300)/1000=-1366.5 kJ/mol. By Hess's law, ΔHc=2(−394)+3(−286)−ΔfH(C2H5OH)\Delta H_c=2(-394)+3(-286)-\Delta_fH(\mathrm{C_2H_5OH}), since O2\mathrm{O_2} has zero formation enthalpy. So ΔfH=−1646+1366.5=−279.5\Delta_fH=-1646+1366.5=-279.5 kJ/mol. Using ΔU\Delta U directly gives −282.0-282.0; a wrong sign on ΔngRT\Delta n_gRT gives −284.5-284.5; the option −1366.5-1366.5 is merely ΔHc\Delta H_c.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake