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Thermodynamics question

2020 · 5 Sep · Shift 2 · Q9
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  5. /2020 · 5 Sep · Shift 2 · Q9

Thermodynamics question

2020 · 5 Sep · Shift 2 · Q9

JEE MainChemistryThermodynamicsMCQ+4 / −1
Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol–1, and 4 kJ mol–1, respectively. The hydration enthalpy of NaCl is :
  1. A
    –780 kJ mol–1
  2. B
    –784 kJ mol–1
  3. C
    780 kJ mol–1
  4. D
    784 kJ mol–1
View written solutionFree

Correct answer: B

  1. Use the relation for enthalpy of solution

For an ionic solid,

ΔHsolution=ΔHlattice dissociation+ΔHhydration\Delta H_{\text{solution}} = \Delta H_{\text{lattice dissociation}} + \Delta H_{\text{hydration}}ΔHsolution​=ΔHlattice dissociation​+ΔHhydration​

Here:

  • Lattice enthalpy of NaCl =788 kJ mol−1= 788\,\text{kJ mol}^{-1}=788kJ mol−1
  • Enthalpy of solution =4 kJ mol−1= 4\,\text{kJ mol}^{-1}=4kJ mol−1
  • Hydration enthalpy =?= ?=?
  1. Substitute the values

4=788+ΔHhydration4 = 788 + \Delta H_{\text{hydration}}4=788+ΔHhydration​

  1. Solve for hydration enthalpy

ΔHhydration=4−788=−784 kJ mol−1\Delta H_{\text{hydration}} = 4 - 788 = -784\,\text{kJ mol}^{-1}ΔHhydration​=4−788=−784kJ mol−1

  1. Match with the options

The correct option is:

B: −784 kJ mol−1\boxed{\text{B: } -784\,\text{kJ mol}^{-1}}B: −784kJ mol−1​

  1. Comparison with stored answer

Stored correct answer: B

My derived answer: B

So, the answer agrees with the stored correct answer.

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