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

2021 · 25 Feb · Shift 1 · Q17
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. /2021 · 25 Feb · Shift 1 · Q17

Thermodynamics question

2021 · 25 Feb · Shift 1 · Q17

JEE MainChemistryThermodynamicsNumerical+4 / −1
The ionization enthalpy of Na+Na^+Na+ formation from NaNaNa(g) is 495.8 kJ mol −-− 1, while the electron gain enthalpy of BrBrBr is −-− 325.0 kJ mol −-− 1. Given the lattice enthalpy of NaBrNaBrNaBr is −-− 728.4 kJ mol −-− 1. The energy for the formation of NaBrNaBrNaBr ionic solid is (−-−) ‾×\underline{\hspace{2cm}}\times​× 10 −-− 1 kJ mol −-− 1.
Numerical answer
View written solutionFree

Correct answer: 5576

  1. We need the enthalpy change for forming solid NaBrNaBrNaBr from gaseous atoms:

Na(g)+Br(g)→NaBr(s)Na(g) + Br(g) \rightarrow NaBr(s)Na(g)+Br(g)→NaBr(s)

Using the Born–Haber type steps:

  • Ionization of sodium: Na(g)→Na+(g)+e−ΔH1=+495.8 kJ mol−1Na(g) \rightarrow Na^+(g) + e^- \qquad \Delta H_1 = +495.8\ \text{kJ mol}^{-1}Na(g)→Na+(g)+e−ΔH1​=+495.8 kJ mol−1

  • Electron gain by bromine: Br(g)+e−→Br−(g)ΔH2=−325.0 kJ mol−1Br(g) + e^- \rightarrow Br^-(g) \qquad \Delta H_2 = -325.0\ \text{kJ mol}^{-1}Br(g)+e−→Br−(g)ΔH2​=−325.0 kJ mol−1

  • Lattice formation: Na+(g)+Br−(g)→NaBr(s)ΔH3=−728.4 kJ mol−1Na^+(g) + Br^-(g) \rightarrow NaBr(s) \qquad \Delta H_3 = -728.4\ \text{kJ mol}^{-1}Na+(g)+Br−(g)→NaBr(s)ΔH3​=−728.4 kJ mol−1

  1. Add these enthalpy changes:

ΔH=ΔH1+ΔH2+ΔH3\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3ΔH=ΔH1​+ΔH2​+ΔH3​

ΔH=495.8−325.0−728.4\Delta H = 495.8 - 325.0 - 728.4ΔH=495.8−325.0−728.4

ΔH=170.8−728.4=−557.6 kJ mol−1\Delta H = 170.8 - 728.4 = -557.6\ \text{kJ mol}^{-1}ΔH=170.8−728.4=−557.6 kJ mol−1

  1. The question writes the answer as:

(−) ‾×10−1 kJ mol−1(-)\,\underline{\hspace{1cm}} \times 10^{-1}\ \text{kJ mol}^{-1}(−)​×10−1 kJ mol−1

So we express:

−557.6=−(5576)×10−1-557.6 = -(5576) \times 10^{-1}−557.6=−(5576)×10−1

Hence the required integer is:

5576\boxed{5576}5576​

PreviousNext

More from Thermodynamics

  • Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is ​ kJ mol − 1.…2021 · Numerical
  • At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ mol − 1 and the enthalpy of vaporisation of the liquid (X) is 98.2 kJ mol − 1. The enthalpy of sublimation of the substance (X) in kJ mol − 1 is ​.…2021 · Numerical
  • A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is ​ J. (Nearest integer)2021 · Numerical
  • The Born-Haber cycle for KCl is evaluated with the following data : Δf​HΘ for KCl =− 436.7 kJ mol − 1 ; Δsub​HΘ for K = 89.2 kJ mol − 1 ; Δionization​HΘ for K = 419.0…2021 · Numerical
  • For water Δ vap H = 41 kJ mol − 1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ​…2021 · Numerical
  • For a chemical reaction A + B ⇌ C + D (Δr​HΘ = 80 kJ mol − 1) the entropy change Δr​SΘ depends on the temperature T (in K) as Δr​SΘ = 2T (J K − 1mol − 1). Minimum temperature…2021 · Numerical
  • An exothermic reaction X → Y has an activation energy 30 kJ mol − 1. If energy change Δ E during the reaction is − 20 kJ, then the activation energy for the reverse reaction in kJ is ​. (Integer…2021 · Numerical
  • The average S-F bond energy in kJ mol − 1 of SF6​ is ​. (Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of SF6​(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol − 1…2021 · Numerical