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JEE Main 2024 · Chemistry · Chemical Thermodynamics · medium

640 g of naphthalene, C10H8(s) (M=128), is burnt completely in a bomb calorimeter whose heat capacity (including contents) is 9.00 kJ K^-1, and the…

Single correct option

0.640 g0.640\ \mathrm{g} of naphthalene, C10H8(s)\mathrm{C_{10}H_8(s)} (M=128M=128), is burnt completely in a bomb calorimeter whose heat capacity (including contents) is 9.00 kJ K−19.00\ \mathrm{kJ\,K^{-1}}, and the temperature rises by 2.86 K2.86\ \mathrm{K}. Taking R=8.314 J K−1 mol−1R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}} and T=298 KT=298\ \mathrm{K}, the standard enthalpy of combustion of naphthalene (water formed as liquid) is nearly
  1. A. −5153 kJ mol−1-5153\ \mathrm{kJ\,mol^{-1}}
  2. B. −5143 kJ mol−1-5143\ \mathrm{kJ\,mol^{-1}}
  3. C. −5148 kJ mol−1-5148\ \mathrm{kJ\,mol^{-1}}
  4. D. −5158 kJ mol−1-5158\ \mathrm{kJ\,mol^{-1}}
Show answer and solution

Answer: (A) −5153 kJ mol−1-5153\ \mathrm{kJ\,mol^{-1}}

Heat released =CΔT=9.00×2.86=25.74 kJ=C\Delta T=9.00\times2.86=25.74\ \mathrm{kJ} for 0.640/128=0.005000.640/128=0.00500 mol, so ΔU=−25.74/0.00500=−5148 kJ mol−1\Delta U=-25.74/0.00500=-5148\ \mathrm{kJ\,mol^{-1}}. For C10H8(s)+12O2(g)→10CO2(g)+4H2O(l)\mathrm{C_{10}H_8(s)+12O_2(g)\to10CO_2(g)+4H_2O(l)}, Δng=10−12=−2\Delta n_g=10-12=-2. ΔH=ΔU+ΔngRT=−5148+(−2)(8.314×10−3)(298)=−5148−4.96=−5153 kJ mol−1\Delta H=\Delta U+\Delta n_gRT=-5148+(-2)(8.314\times10^{-3})(298)=-5148-4.96=-5153\ \mathrm{kJ\,mol^{-1}}. Option −5148-5148 is ΔU\Delta U and −5143-5143 has the wrong sign of ΔngRT\Delta n_g RT.
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