Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Thermodynamics question

2022 · 26 Jul · Shift 2 · Q14
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Thermodynamics
  5. /2022 · 26 Jul · Shift 2 · Q14

Thermodynamics question

2022 · 26 Jul · Shift 2 · Q14

JEE MainChemistryThermodynamicsNumerical+4 / −1
For the reaction H2F2( g)→H2( g)+F2( g)\mathrm{H}_{2} \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})H2​F2​( g)→H2​( g)+F2​( g)ΔU=−59.6 kJ mol−1\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}ΔU=−59.6 kJ mol−1 at 27∘C27^{\circ} \mathrm{C}27∘C. The enthalpy change for the above reaction is (−-−) ‾kJ mol−1\underline{\hspace{2cm}}\mathrm{kJ} \,\mathrm{mol}^{-1}​kJmol−1 [nearest integer] Given: R=8.314 J K−1 mol−1\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}R=8.314 J K−1 mol−1.
Numerical answer
View written solutionFree

Correct answer: 57

  1. Use the relation between enthalpy and internal energy

For gaseous reactions,

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RTΔH=ΔU+Δng​RT

where Δng\Delta n_gΔng​ is the change in moles of gaseous species.

  1. Calculate Δng\Delta n_gΔng​

Reaction:

H2F2(g)→H2(g)+F2(g)\mathrm{H_2F_2(g)} \rightarrow \mathrm{H_2(g)} + \mathrm{F_2(g)}H2​F2​(g)→H2​(g)+F2​(g)

Moles of gaseous products =2= 2=2

Moles of gaseous reactants =1= 1=1

So,

Δng=2−1=1\Delta n_g = 2 - 1 = 1Δng​=2−1=1
  1. Substitute the given values

Given:

  • ΔU=−59.6 kJ mol−1\Delta U = -59.6\,\mathrm{kJ\,mol^{-1}}ΔU=−59.6kJmol−1
  • T=27∘C=300 KT = 27^\circ \mathrm{C} = 300\,\mathrm{K}T=27∘C=300K
  • R=8.314 J K−1 mol−1R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}R=8.314JK−1mol−1

First calculate RTRTRT:

RT=8.314×300=2494.2 J mol−1=2.4942 kJ mol−1RT = 8.314 \times 300 = 2494.2\,\mathrm{J\,mol^{-1}} = 2.4942\,\mathrm{kJ\,mol^{-1}}RT=8.314×300=2494.2Jmol−1=2.4942kJmol−1

Therefore,

ΔH=−59.6+1(2.4942)\Delta H = -59.6 + 1(2.4942)ΔH=−59.6+1(2.4942) ΔH=−57.1058 kJ mol−1\Delta H = -57.1058\,\mathrm{kJ\,mol^{-1}}ΔH=−57.1058kJmol−1
  1. Nearest integer
ΔH≈−57 kJ mol−1\Delta H \approx -57\,\mathrm{kJ\,mol^{-1}}ΔH≈−57kJmol−1

Since the question asks for the value after the minus sign, the required integer is:

575757
PreviousNext

More from Thermodynamics

  • For complete combustion of methanol CH3​OH(I) + 23​ O2​(g) → CO2​(g) + 2H2​O(I) the amount of heat produced as measured by bomb calorimeter is 726 kJ mol − 1 at 27 ∘ C. The enthalpy of combustion for the…2022 · Numerical
  • A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100 ∘ C, then the internal energy for vaporization in kJ mol − 1 is ​…2022 · Numerical
  • The molar heat capacity for an ideal gas at constant pressure is 20.785 J K−1 mol−1. The change in internal energy is 5000 J upon heating it from 300 K to 500 K.…2022 · Numerical
  • A gas (Molar mass = 280  g mol−1) was burnt in excess O2​ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45K.…2022 · Numerical
  • Match List-I with List-II. Choose the correct answer from the options given below : Includes table2022 · MCQ
  • When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is ​ J. [nearest integer] (Given : R = 8.3 J K − 1 mol − 1 and log 2 =…2022 · Numerical
  • Which of the following relation is not correct?2022 · MCQ
  • Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.…2022 · MCQ