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Thermodynamics question

2020 · 4 Sep · Shift 2 · Q3
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Thermodynamics question

2020 · 4 Sep · Shift 2 · Q3

JEE MainChemistryThermodynamicsMCQ+4 / −1
The process that is NOT endothermic in nature is :
  1. A
    ArArAr(g) + e−e^-e− →\to→ Ar−Ar^-Ar−(g)
  2. B
    HHH(g) + e−e^-e− →\to→ H−H^-H−(g)
  3. C
    NaNaNa(g) →\to→ Na+Na^+Na+(g) + e−e^-e−
  4. D
    O−O^-O−(g) + e−e^-e− →\to→ O2−O^{2-}O2−(g)
View written solutionFree

Correct answer: B

  1. Identify the thermodynamic nature of each process

We need the process that is NOT endothermic, i.e. the one that is exothermic or not requiring energy input.

Useful facts:

  • Electron affinity (EA): energy released when an electron is added to a gaseous atom.
  • First electron affinity is often exothermic for many nonmetals and some atoms like H.
  • Second electron affinity is always endothermic because an electron is added to an already negatively charged ion.
  • Ionization is always endothermic.
  1. Check each option

Option A

Ar(g)+e−→Ar−(g)Ar(g)+e^- \to Ar^-(g)Ar(g)+e−→Ar−(g) Argon is a noble gas with completely filled shell. Adding an electron is unfavorable and requires energy.

So this process is endothermic.

Option B

H(g)+e−→H−(g)H(g)+e^- \to H^-(g)H(g)+e−→H−(g) This is the first electron affinity of hydrogen. Hydrogen gains an electron and releases energy.

So this process is exothermic, hence not endothermic.

Option C

Na(g)→Na+(g)+e−Na(g) \to Na^+(g)+e^-Na(g)→Na+(g)+e− This is the first ionization enthalpy of sodium. Removing an electron always requires energy.

So this process is endothermic.

Option D

O−(g)+e−→O2−(g)O^-(g)+e^- \to O^{2-}(g)O−(g)+e−→O2−(g) This is the second electron affinity of oxygen. An electron is being added to an already negative ion, so electron-electron repulsion must be overcome.

So this process is endothermic.

  1. Conclusion

Only option B is not endothermic.

B\boxed{B}B​

  1. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They match.

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