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JEE Main 2017 · Chemistry · Chemical Thermodynamics · medium

In a bomb calorimeter of heat capacity 17.0 kJ K^-1, combustion of 1.15 g of ethanol (M=46) raises the temperature by 2.00 K. For…

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In a bomb calorimeter of heat capacity 17.0 kJ K−117.0\ \mathrm{kJ\,K^{-1}}, combustion of 1.15 g1.15\ \mathrm{g} of ethanol (M=46M=46) raises the temperature by 2.00 K2.00\ \mathrm{K}. For C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)\mathrm{C_2H_5OH(l)}+3\mathrm{O_2(g)}\to2\mathrm{CO_2(g)}+3\mathrm{H_2O(l)} at 300 K300\ \mathrm{K}, ΔH\Delta H is nearly (R=8.314 J K−1mol−1R=8.314\ \mathrm{J\,K^{-1}mol^{-1}})
  1. A. −1360 kJ mol−1-1360\ \mathrm{kJ\,mol^{-1}}
  2. B. −34 kJ mol−1-34\ \mathrm{kJ\,mol^{-1}}
  3. C. −1357.5 kJ mol−1-1357.5\ \mathrm{kJ\,mol^{-1}}
  4. D. −1362.5 kJ mol−1-1362.5\ \mathrm{kJ\,mol^{-1}}
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Answer: (D) −1362.5 kJ mol−1-1362.5\ \mathrm{kJ\,mol^{-1}}

Heat released =17.0×2.00=34.0=17.0\times2.00=34.0 kJ for 1.15/46=0.02501.15/46=0.0250 mol, so ΔU=−34.0/0.0250=−1360 kJ mol−1\Delta U=-34.0/0.0250=-1360\ \mathrm{kJ\,mol^{-1}} (constant volume). Δng=2−3=−1\Delta n_g=2-3=-1, so ΔH=ΔU+ΔngRT=−1360+(−1)(8.314×10−3)(300)=−1362.5 kJ mol−1\Delta H=\Delta U+\Delta n_gRT=-1360+(-1)(8.314\times10^{-3})(300)=-1362.5\ \mathrm{kJ\,mol^{-1}}. A sign slip gives −1357.5-1357.5; −1360-1360 ignores the gas correction.
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