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← JEE Main 2022 · 26 Jun · Shift 1
ChemistryInteger+4 / -1Question 81 of 90

JEE Main 2022 · 26 Jun · Shift 1, Chemistry question 81

For complete combustion of methanol
CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)\mathrm{CH_3OH(l)+\tfrac32O_2(g)\rightarrow CO_2(g)+2H_2O(l)}
the amount of heat produced as measured by bomb calorimeter is 726 kJ mol−1^{-1} at 27°C. The enthalpy of combustion for the reaction is −x-x kJ mol−1^{-1}, where xx is ______. (Nearest integer)

(Given : R=8.3R=8.3 J K−1^{-1} mol−1^{-1})

Answer and worked solution

Official answer: 727

Bomb calorimeter gives ΔU=−726\Delta U=-726 kJ mol−1^{-1}.

Gas moles. Δng=1−32=−12\Delta n_g=1-\frac32=-\frac12.

Enthalpy. ΔH=ΔU+ΔngRT=−726−12×8.3×300×10−3≈−727.2\Delta H=\Delta U+\Delta n_gRT=-726-\frac12\times8.3\times300\times10^{-3}\approx-727.2 kJ mol−1^{-1}.

Answer: 727