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

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

2 . 4g coal is burnt in a bomb calorimeter in excess of oxygen at...

2.4g2 . 4g coal is burnt in a bomb calorimeter in excess of oxygen at 298K298K and 1atm1atm pressure.

The temperature of the calorimeter rises from 298K298K to 300K300K. The enthalpy change during the combustion of coal is −xkJmol−1- xkJ{mol}^{- 1}. The value of xx is_____(Given : Heat capacity of bomb calorimeter 20.0kJK−120 . 0{kJK}^{- 1}. Assume coal to be pure carbon)

Correct answer

0

Step-by-step solution

C(s)+O2(g)→CO2(g);ΔH=−xkJ/moleC\left(s\right) + {O}_{2}\left(g\right) \to {CO}_{2}\left(g\right); \Delta H = - xkJ / mole

Q=CΔT=20kJ×2Q = C \Delta T = 20kJ \times 2

40kJ40kJ heat is released for 2.4g2 . 4g of CarbonCarbon

For 11 mole 'CarbonCarbon' Q=402.4×12Q = \dfrac{40}{2 . 4} \times 12

=40024×12=200kJ/mole= \dfrac{400}{24} \times 12 = 200kJ / mole

Q=ΔE=ΔH=200kJ(∵Δng=0)Q = \Delta E = \Delta H = 200kJ\left(∵ {\Delta n}_{g} = 0\right)

x=200x = 200

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