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Thermodynamics question

2025 · 28 Jan · Shift 1 · Q21
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Thermodynamics question

2025 · 28 Jan · Shift 1 · Q21

JEE MainChemistryThermodynamicsNumerical+4 / −1
The formation enthalpies, ΔHf⊖\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}ΔHf⊖​ for H(g)\mathrm{H}_{(\mathrm{g})}H(g)​ and O(g)\mathrm{O}_{(\mathrm{g})}O(g)​ are 220.0 and 250.0 kJ mol−1250.0 \mathrm{~kJ} \mathrm{~mol}^{-1}250.0 kJ mol−1, respectively, at 298.15 K , and ΔHf⊖\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}ΔHf⊖​ for H2O(g)\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}H2​O(g)​ is −242.0 kJ mol−1-242.0 \mathrm{~kJ} \mathrm{~mol}^{-1}−242.0 kJ mol−1 at the same temperature. The average bond enthalpy of the O−H\mathrm{O}-\mathrm{H}O−H bond in water at 298.15 K is ‾\underline{\hspace{2cm}}​kJ mol−1\mathrm{kJ} \mathrm{~mol}^{-1}kJ mol−1 (nearest integer).
Numerical answer
View written solutionFree

Correct answer: 466

  1. Write the required bond dissociation process

For water vapor, the atomization corresponding to breaking both O−H\mathrm{O-H}O−H bonds is:

H2O(g)→2H(g)+O(g)\mathrm{H_2O_{(g)} \rightarrow 2H_{(g)} + O_{(g)}}H2​O(g)​→2H(g)​+O(g)​

If the average O−H\mathrm{O-H}O−H bond enthalpy is DDD, then

2D=ΔH(H2O(g)→2H(g)+O(g))2D = \Delta H\big(\mathrm{H_2O_{(g)} \rightarrow 2H_{(g)} + O_{(g)}}\big)2D=ΔH(H2​O(g)​→2H(g)​+O(g)​)
  1. Use enthalpies of formation

Given:

ΔHf∘(H(g))=220.0 kJ mol−1\Delta H_f^\circ\big(\mathrm{H_{(g)}}\big)=220.0\ \mathrm{kJ\ mol^{-1}}ΔHf∘​(H(g)​)=220.0 kJ mol−1 ΔHf∘(O(g))=250.0 kJ mol−1\Delta H_f^\circ\big(\mathrm{O_{(g)}}\big)=250.0\ \mathrm{kJ\ mol^{-1}}ΔHf∘​(O(g)​)=250.0 kJ mol−1 ΔHf∘(H2O(g))=−242.0 kJ mol−1\Delta H_f^\circ\big(\mathrm{H_2O_{(g)}}\big)=-242.0\ \mathrm{kJ\ mol^{-1}}ΔHf∘​(H2​O(g)​)=−242.0 kJ mol−1

Now,

ΔHreaction∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_\text{reaction}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})ΔHreaction∘​=∑ΔHf∘​(products)−∑ΔHf∘​(reactants)

So for

H2O(g)→2H(g)+O(g)\mathrm{H_2O_{(g)} \rightarrow 2H_{(g)} + O_{(g)}}H2​O(g)​→2H(g)​+O(g)​

we get

ΔH∘=[2×220.0+250.0]−[−242.0]\Delta H^\circ = \left[2\times 220.0 + 250.0\right] - \left[-242.0\right]ΔH∘=[2×220.0+250.0]−[−242.0] =440.0+250.0+242.0= 440.0 + 250.0 + 242.0=440.0+250.0+242.0 =932.0 kJ mol−1= 932.0\ \mathrm{kJ\ mol^{-1}}=932.0 kJ mol−1
  1. Relate this to average bond enthalpy

Since water has two O−H\mathrm{O-H}O−H bonds,

2D=932.02D = 932.02D=932.0 D=932.02=466.0 kJ mol−1D = \frac{932.0}{2} = 466.0\ \mathrm{kJ\ mol^{-1}}D=2932.0​=466.0 kJ mol−1
  1. Nearest integer
466\boxed{466}466​

So, the average bond enthalpy of the O−H\mathrm{O-H}O−H bond in water is 466 kJ mol−1466\ \mathrm{kJ\ mol^{-1}}466 kJ mol−1.

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