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JEE Main 202325 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2023 · 25 January · Evening Shift · Chemistry · Thermodynamics (C)

28 . 0 L of CO _ 2 is produced on complete combustion of 16 . 8L...

28.028 . 0 L of CO2{CO}_{2} is produced on complete combustion of 16.8L16 . 8L gaseous mixture of ethene and methane at 25°C25 ° C and 11 atm. Heat evolved during the combustion process is kJkJ

Given:ΔHC(CH4)=−900kJmol−1: {\Delta H}_{C}\left({CH}_{4}\right) = - 900kJ{mol}^{- 1}

ΔHC(C2H4)=−1400kJmol−1{\Delta H}_{C}\left({C}_{2}{H}_{4}\right) = - 1400kJ{mol}^{- 1}.

Correct answer

847

Step-by-step solution

Complete combustion of ethene and methane mixture produces carbon dioxide.
Let, Volume of C2H4{C}_{2}{H}_{4} is xx litre

C2H4+3O2→2CO2+2H2O{C}_{2}{H}_{4} + 3{O}_{2} \to 2{CO}_{2} + 2{H}_{2}O

Initial volume of carbon dioxide 0
Final volume of carbon dioxide 2x2x

CH4+2O2→CO2+2H2O{CH}_{4} + 2{O}_{2} \to {CO}_{2} + 2{H}_{2}O

Initial volume of carbon dioxide is 0

Final volume of carbon dioxide is (16.8−x)( 16 . 8 - x )

Total volume of CO2=2x+16.8−x{CO}_{2} = 2x + 16 . 8 - x

⇒28=16.8+x⇒ 28 = 16 . 8 + x

x=11.2Lx = 11 . 2L

nCH4=PVRT=1×5.60.082×298=0.229mole{n}_{{CH}_{4}} = \dfrac{PV}{RT} = \dfrac{1 \times 5 . 6}{0 . 082 \times 298} = 0 . 229mole

nC2H4=11.20.082×298=0.458mole{n}_{{C}_{2}{H}_{4}} = \dfrac{11 . 2}{0 . 082 \times 298} = 0 . 458mole

∴∴ Heat evolved using standard enthalpy of combustion =0.229×900+0.458×1400= 0 . 229 \times 900 + 0 . 458 \times 1400

=206.1+641.2= 206 . 1 + 641 . 2

=847.3kJ= 847 . 3kJ

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