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JEE Main 202226 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
26 June
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2022 · 26 June · Morning Shift · Chemistry · Thermodynamics (C)

For complete combustion of methanol CH _ 3 OH(l) + (3)/(2) O _ 2...

For complete combustion of methanol

CH3OH(l)+32O2(g)→CO2(g)+2H2O(l){CH}_{3}OH\left(l\right) + \dfrac{3}{2}{O}_{2}\left(g\right) \to {CO}_{2}\left(g\right) + 2{H}_{2}O\left(l\right) the amount of heat produced as measured by bomb calorimeter is 726kJmol−1726kJ{mol}^{- 1} at 27°C27 ° C. The enthalpy of combustion for the reaction is −xkJmol−1- xkJ{mol}^{- 1}, where xx is
integer)
(Given : R=8.3JK−1mol−1R = 8 . 3{JK}^{- 1}{mol}^{- 1})

Correct answer

727

Step-by-step solution

CH3OH(l)+32O2(g)→CO2(g)+2H2O(l).{CH}_{3}OH\left(l\right) + \dfrac{3}{2}{O}_{2}\left(g\right) \to {CO}_{2}\left(g\right) + 2{H}_{2}O\left(l\right) .

For chemical reaction : ΔH=ΔU+Δn(g)RT\Delta H = \Delta U + {\Delta n}_{\left(g\right)}RT

\DeltaH=−726+(−12)8.314×10−3×300\DeltaH = - 726 + \left(- \dfrac{1}{2}\right)8 . 314 \times {10}^{- 3} \times 300

=−727.245KJ/mole= - 727 . 245KJ / mole

∴x=727∴ x = 727

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