From the following data at 2 5 ∘ C Which of the following statement(s) is/are correct: Statement (a) : Δ r H ∘ for the reaction H 2 O ( g ) ⟶ H ( g ) + OH ( g ) is 502 k J / m o l Statement (b) :

Question

From the following data at 2 5 ∘ C Which of the following statement(s) is/are correct: Statement (a) : Δ r ​ H ∘ for the reaction H 2 ​ O ( g ) ⟶ H ( g ) + OH ( g ) is 502 k J / m o l Statement (b) : Δ r ​ H ∘ for the reaction OH ( g ) ⟶ H ( g ) + O ( g ) is 925.5 k J / m o l Statement (c) : Enthalpy of formation of H ( g ) is − 218 k J / m o l Statement (d) : Enthalpy of formation of OH ( g ) is 42 kJ / mol

TestHub · Accepted Answer

Correct statements are a and b. The correct option is therefore not A. I will provide a detailed explanation of each statement.

Statement (a): $Δ_{r} H^{\circ}$ for the reaction $H_{2} O (g) ⟶ H (g) + OH (g)$ is $502 k J / m o l$

To calculate the enthalpy change for this reaction, we can use Hess's Law and the given data. We need to manipulate the given reactions to arrive at the desired reaction.

1. $H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (g) Δ_{r} H^{\circ} = - 242 kJ / mol$

2. $\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ OH (g) Δ_{r} H^{\circ} = 42 kJ / mol$

3. $H_{2} (g) ⟶ 2 H (g) Δ_{r} H^{\circ} = 436 kJ / mol$

We need to reverse reaction (1) and halve reaction (3):

$- H_{2} O (g) ⟶ - H_{2} (g) - \frac{1}{2} O_{2} (g) Δ_{r} H^{\circ} = 242 kJ / mol$

$\frac{1}{2} H_{2} (g) ⟶ H (g) Δ_{r} H^{\circ} = \frac{436}{2} = 218 kJ / mol$

Adding this to reaction (2):

$H_{2} O (g) ⟶ H (g) + OH (g)$

$Δ_{r} H^{\circ} = 242 + 218 + 42 = 502 kJ / mol$

Hence CORRECT

Statement (b): $Δ_{r} H^{\circ}$ for the reaction $OH (g) ⟶ H (g) + O (g)$ is $502 kJ / mol$

We can obtain this by using the following reactions:

1. $\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ OH (g) Δ_{r} H^{\circ} = 42 kJ / mol$

2. $H_{2} (g) ⟶ 2 H (g) Δ_{r} H^{\circ} = 436 kJ / mol$

3. $O_{2} (g) ⟶ 2 O (g) Δ_{r} H^{\circ} = 495 kJ / mol$

We need to reverse reaction (1) and halve reactions (2) and (3):

$- OH (g) ⟶ - \frac{1}{2} H_{2} (g) - \frac{1}{2} O_{2} (g) Δ_{r} H^{\circ} = - 42 kJ / mol$

$\frac{1}{2} H_{2} (g) ⟶ H (g) Δ_{r} H^{\circ} = \frac{436}{2} = 218 kJ / mol$

$\frac{1}{2} O_{2} (g) ⟶ O (g) Δ_{r} H^{\circ} = \frac{495}{2} = 247.5 kJ / mol$

Adding these together:

$OH (g) ⟶ H (g) + O (g)$

$Δ_{r} H^{\circ} = - 42 + 218 + 247.5 = 423.5 kJ / mol$

The provided value of $925.5 k J / m o l$ is incorrect.

Statement (c): Enthalpy of formation of $H (g)$ is $- 218 kJ / mol$

The enthalpy of formation of $H (g)$ is half the enthalpy change for the reaction $H_{2} (g) ⟶ 2 H (g)$ , which is $436 kJ / mol$ . Therefore, the enthalpy of formation of $H (g)$ is $\frac{436}{2} = 218 kJ / mol$ . The given value of $- 218 kJ / mol$ is incorrect.

Statement (d): Enthalpy of formation of $OH (g)$ is $42 kJ / mol$

The enthalpy of formation of $OH (g)$ is given directly as $42 kJ / mol$ for the reaction $\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ OH (g)$ . This statement is correct.

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