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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Analyzing +4 | -1
At 25∘C25^{\circ} \mathrm{C}, the enthalpy of the following processes are given

H2(g)+O2(g)→2OH(g)ΔH∘=78 kJ mol−1\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =78 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

H2(g)+12O2(g)→H2O(g)ΔH∘=−242 kJ mol−1\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =-242 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

H2(g)→2H(g)ΔH∘=436 kJ mol−1\mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=436 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

12O2(g)→O(g)ΔH∘=249 kJ mol−1\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=249 \,\mathrm{kJ} \,\mathrm{mol}^{-1}
What would be the value of XX for the following reaction? _____ (Nearest integer)
H2O(g)→H(g)+OH(g)ΔH∘=X kJ mol−1\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=\mathrm{X} \,\mathrm{kJ} \,\mathrm{mol}^{-1}
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Correct option: A
499
The correct answer is 499. By manipulating the given equations and applying Hess's Law, we can calculate the enthalpy change for the desired reaction.

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