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Thermodynamics question

2023 · 1 Feb · Shift 1 · Q16
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  5. /2023 · 1 Feb · Shift 1 · Q16

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
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