From the following data at 25° CReactionΔrH∘kJ/mol12H2(g)+12O2(g)⟶OH(g)42H2(g)+12O2(g)⟶H2O(g)-242H2(g)⟶2H(g)436O2(g)⟶2O(g)495Which of the following statement(s) is/are correct :Statement A: ΔrH∘ for the reaction.H2O(g)⟶2H(g)+O(g) is 925.5 kJ/molStatement B: ΔrH∘ for the reactionOH(g)⟶H(g)+O(g) is 423.5 kJ/molStatement C: Enthalpy of formation of H(g) is -218 kJ/molStatement D: Enthalpy of formation of OH(g) is 42 kJ/mol

Statement A: A:ΔrH∘=242+436+4952= 925.5 kJ/molStatement B: ΔrH∘=−42+12×436+12×495=423.5kJ/molStatement C : ΔfH∘(H)=4362=+218kJ/mol.

From the following data at 25° CReactionΔrH∘kJ/mol12H2(g)+12O2(g)⟶OH(g)42H2(g)+12O2(g)⟶H2O(g)-242H2(g)⟶2H(g)436O2(g)⟶2O(g)495Which of the following statement(s) is/are correct :Statement A: ΔrH∘ for the reaction.H2O(g)⟶2H(g)+O(g) is 925.5 kJ/molStatement B: ΔrH∘ for the reactionOH(g)⟶H(g)+O(g) is 423.5 kJ/molStatement C: Enthalpy of formation of H(g) is -218 kJ/molStatement D: Enthalpy of formation of OH(g) is 42 kJ/mol

Statement A: A:ΔrH∘=242+436+4952= 925.5 kJ/molStatement B: ΔrH∘=−42+12×436+12×495=423.5kJ/molStatement C : ΔfH∘(H)=4362=+218kJ/mol.

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From the following data at 25° C
Reaction $Δ_{r} H^{\circ} kJ / mol$
$\begin{aligned} \frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ OH (g) \end{aligned}$ 42
$\begin{array}{r} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (g) \end{array}$ -242
$\begin{array}{r} H_{2} (g) ⟶ 2 H (g) \end{array}$ 436
$O_{2} (g) ⟶ 2 O (g)$ 495
Which of the following statement(s) is/are correct :
Statement A: $Δ_{r} H^{\circ}$ for the reaction.
$H_{2} O (g) ⟶ 2 H (g) + O (g)$ is 925.5 kJ/mol
Statement B: $Δ_{r} H^{\circ}$ for the reaction
$OH (g) ⟶ H (g) + O (g)$ is 423.5 kJ/mol
Statement C: Enthalpy of formation of H(g) is -218 kJ/mol
Statement D: Enthalpy of formation of OH(g) is 42 kJ/mol

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Statement A, B only

Statement A, B, D

Statement B, C

Statement C

answer is B.

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Detailed Solution

Statement A: $A : Δ_{r} H^{\circ} = 242 + 436 + \frac{495}{2}$

= 925.5 kJ/mol

Statement B: $\begin{aligned} Δ_{r} H^{\circ} = - 42 + \frac{1}{2} \times 436 + \frac{1}{2} \times 495 \end{aligned}$

$= 423.5 kJ / mol$

Statement C : $Δ_{f} H^{\circ} (H) = \frac{436}{2} = + 218 kJ / mol$ .

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