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Thermodynamics question

2023 · 1 Feb · Shift 1 · Q16
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Thermodynamics question

2023 · 1 Feb · Shift 1 · Q16

JEE MainChemistryThermodynamicsNumerical+4 / −1

At 25∘C25^{\circ} \mathrm{C}25∘C, the enthalpy of the following processes are given :

H2(g)+O2(g)\mathrm{H_2(g)+O_2(g)}H2​(g)+O2​(g) →\to→ 2OH(g)2\mathrm{OH(g)}2OH(g) ΔH∘=78 kJ mol−1\mathrm{\Delta H^\circ=78~kJ~mol^{-1}}ΔH∘=78 kJ mol−1
H2(g)+12O2(g)\mathrm{H_2(g)+\frac{1}{2}O_2(g)}H2​(g)+21​O2​(g) →\to→ H2O(g)\mathrm{H_2O(g)}H2​O(g) ΔH∘=−242 kJ mol−1\mathrm{\Delta H^\circ=-242~kJ~mol^{-1}}ΔH∘=−242 kJ mol−1
H2(g)\mathrm{H_2(g)}H2​(g) →\to→ 2H(g)\mathrm{2H(g)}2H(g) ΔH∘=436 kJ mol−1\mathrm{\Delta H^\circ=436~kJ~mol^{-1}}ΔH∘=436 kJ mol−1
12O2(g)\frac{1}{2}\mathrm{O_2(g)}21​O2​(g) →\to→ O(g)\mathrm{O(g)}O(g) ΔH∘=249 kJ mol−1\mathrm{\Delta H^\circ=249~kJ~mol^{-1}}ΔH∘=249 kJ mol−1

What would be the value of X for the following reaction ? ‾\underline{\hspace{2cm}}​ (Nearest integer)

H2O(g)→H(g)+OH(g) ΔH∘=X kJ mol−1\mathrm{H_2O(g)\to H(g)+OH(g)~\Delta H^\circ=X~kJ~mol^{-1}}H2​O(g)→H(g)+OH(g) ΔH∘=X kJ mol−1

Numerical answer
View written solutionFree

Correct answer: 499

  1. We need ΔH∘\Delta H^\circΔH∘ for H2O(g)→H(g)+OH(g)\mathrm{H_2O(g) \to H(g) + OH(g)}H2​O(g)→H(g)+OH(g)

  2. Use Hess's law via enthalpies of formation.

Step 1: Find ΔHf∘[OH(g)]\Delta H_f^\circ[\mathrm{OH(g)}]ΔHf∘​[OH(g)]

Given: H2(g)+O2(g)→2OH(g)ΔH∘=78 kJ mol−1\mathrm{H_2(g)+O_2(g) \to 2OH(g)} \qquad \Delta H^\circ = 78\,\text{kJ mol}^{-1}H2​(g)+O2​(g)→2OH(g)ΔH∘=78kJ mol−1

So for formation of one mole of OH(g)\mathrm{OH(g)}OH(g) from elements in standard states: 12H2(g)+12O2(g)→OH(g)\frac{1}{2}\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{OH(g)}21​H2​(g)+21​O2​(g)→OH(g) ΔHf∘[OH(g)]=782=39 kJ mol−1\Delta H_f^\circ[\mathrm{OH(g)}] = \frac{78}{2} = 39\,\text{kJ mol}^{-1}ΔHf∘​[OH(g)]=278​=39kJ mol−1

Step 2: Find ΔHf∘[H(g)]\Delta H_f^\circ[\mathrm{H(g)}]ΔHf∘​[H(g)]

Given: H2(g)→2H(g)ΔH∘=436 kJ mol−1\mathrm{H_2(g) \to 2H(g)} \qquad \Delta H^\circ = 436\,\text{kJ mol}^{-1}H2​(g)→2H(g)ΔH∘=436kJ mol−1

Hence, 12H2(g)→H(g)\frac{1}{2}\mathrm{H_2(g)} \to \mathrm{H(g)}21​H2​(g)→H(g) ΔHf∘[H(g)]=4362=218 kJ mol−1\Delta H_f^\circ[\mathrm{H(g)}] = \frac{436}{2} = 218\,\text{kJ mol}^{-1}ΔHf∘​[H(g)]=2436​=218kJ mol−1

Step 3: Use formation enthalpy of water vapor

Given: H2(g)+12O2(g)→H2O(g)ΔH∘=−242 kJ mol−1\mathrm{H_2(g)+\frac{1}{2}O_2(g) \to H_2O(g)} \qquad \Delta H^\circ = -242\,\text{kJ mol}^{-1}H2​(g)+21​O2​(g)→H2​O(g)ΔH∘=−242kJ mol−1

Thus, ΔHf∘[H2O(g)]=−242 kJ mol−1\Delta H_f^\circ[\mathrm{H_2O(g)}] = -242\,\text{kJ mol}^{-1}ΔHf∘​[H2​O(g)]=−242kJ mol−1

Step 4: Compute required reaction enthalpy

For H2O(g)→H(g)+OH(g)\mathrm{H_2O(g) \to H(g)+OH(g)}H2​O(g)→H(g)+OH(g)

By Hess's law, ΔH∘=[ΔHf∘(H)+ΔHf∘(OH)]−ΔHf∘(H2O)\Delta H^\circ = \left[\Delta H_f^\circ(\mathrm{H}) + \Delta H_f^\circ(\mathrm{OH})\right] - \Delta H_f^\circ(\mathrm{H_2O})ΔH∘=[ΔHf∘​(H)+ΔHf∘​(OH)]−ΔHf∘​(H2​O)

Substitute values: ΔH∘=(218+39)−(−242)\Delta H^\circ = (218 + 39) - (-242)ΔH∘=(218+39)−(−242) ΔH∘=257+242=499 kJ mol−1\Delta H^\circ = 257 + 242 = 499\,\text{kJ mol}^{-1}ΔH∘=257+242=499kJ mol−1

Final Answer

X=499X = 499X=499

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