Question

consider the following thermochemical equations.
no(g) + o₃(g) → no₂(g) + o₂(g)
δh = -198.9 kj/mol
o₃(g) → 3/2 o₂(g)
δh = -142.3 kj/mol
o₂(g) → 2 o(g)
δh = +495 kj/mol
determine the enthalpy change, in kj/mol, for the reaction
2 no₂(g) → 2 no(g) + o₂(g)

Explanation:

Let's denote the given reactions as follows:

Reaction 1: $\ce{NO(g) + O_{3}(g) -> NO_{2}(g) + O_{2}(g)}$ $\Delta H_1 = -198.9\ \text{kJ/mol}$

Reaction 2: $\ce{O_{3}(g) -> \frac{3}{2} O_{2}(g)}$ $\Delta H_2 = -142.3\ \text{kJ/mol}$ (Wait, no, the given $\Delta H$ for this reaction? Wait, no, the user's image: Wait, the second reaction: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$? Wait, no, the third reaction: $\ce{O_{2}(g) -> 2 O(g)}$ $\Delta H_3 = +495\ \text{kJ/mol}$? Wait, no, the target reaction is $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$

Let's re-express the given reactions:

Reaction 1: $\ce{NO(g) + O_{3}(g) -> NO_{2}(g) + O_{2}(g)}$ $\Delta H_1 = -198.9\ \text{kJ/mol}$

Reaction 2: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ Wait, no, the second reaction's $\Delta H$? Wait, the user's image: Let's check again.

Wait, the three reactions (from the image):

1. $\ce{NO(g) + O_{3}(g) -> NO_{2}(g) + O_{2}(g)}$ $\Delta H = -198.9\ \text{kJ/mol}$

2. $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ $\Delta H = -142.3\ \text{kJ/mol}$? Wait, no, the sign? Wait, no, the user's image: Wait, the second reaction's $\Delta H$ is -142.3? Wait, no, maybe I misread. Wait, the third reaction: $\ce{O_{2}(g) -> 2 O(g)}$ $\Delta H = +495\ \text{kJ/mol}$

But the target reaction is $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$

Let's manipulate the given reactions to get the target.

First, reverse Reaction 1 and multiply by 2:

Reversed Reaction 1 (times 2): $\ce{2 NO_{2}(g) + 2 O_{2}(g) -> 2 NO(g) + 2 O_{3}(g)}$ $\Delta H_{1r} = 2 \times (198.9\ \text{kJ/mol}) = +397.8\ \text{kJ/mol}$

Now, take Reaction 2: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ $\Delta H_2 = -142.3\ \text{kJ/mol}$ (Wait, no, the sign? Wait, the user's image: Let's check the original problem again. Wait, the second reaction: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ with $\Delta H = -142.3$? Wait, maybe I made a mistake. Wait, no, let's check the target reaction.

Wait, maybe the second reaction is $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ with $\Delta H = -142.3$? No, that can't be. Wait, perhaps the second reaction is $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ with $\Delta H = -142.3$? Wait, no, let's think again.

Wait, the target reaction is $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$

Let's express the target reaction as a combination of the given reactions.

Let's list the given reactions properly:

Reaction A: $\ce{NO(g) + O_{3}(g) -> NO_{2}(g) + O_{2}(g)}$ $\Delta H_A = -198.9\ \text{kJ/mol}$

Reaction B: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ $\Delta H_B = -142.3\ \text{kJ/mol}$ (Wait, no, the user's image: Wait, the second reaction's $\Delta H$ is -142.3? Wait, maybe the second reaction is $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ with $\Delta H = -142.3$? No, that would be exothermic, but ozone decomposition is endothermic? Wait, maybe the sign is wrong. Wait, no, the user's image: Let's check the third reaction: $\ce{O_{2}(g) -> 2 O(g)}$ $\Delta H = +495\ \text{kJ/mol}$ (endothermic, correct, since breaking O=O bond).

But maybe I misread Reaction B. Wait, the user's image: Let's re-express the reactions:

1. $\ce{NO(g) + O_{3}(g) -> NO_{2}(g) + O_{2}(g)}$ $\Delta H = -198.9\ \text{kJ/mol}$

2. $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ $\Delta H = -142.3\ \text{kJ/mol}$? No, that can't be. Wait, maybe the second reaction is $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ with $\Delta H = +142.3$? Wait, the user's image: Let's check the original problem again.

Wait, the user's image:

First reaction: $\ce{NO(g) + O_{3}(g) -> NO_{2}(g) + O_{2}(g)}$ $\Delta H = -198.9\ \text{kJ/mol}$

Second reaction: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ $\Delta H = -142.3\ \text{kJ/mol}$? No, that's exothermic, but ozone decomposition to O2 is endothermic? Wait, maybe the sign is reversed. Wait, no, let's proceed with the given values.

Wait, the target reaction is $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$

Let's reverse Reaction A and multiply by 2:

Reversed A (×2): $\ce{2 NO_{2}(g) + 2 O_{2}(g) -> 2 NO(g) + 2 O_{3}(g)}$ $\Delta H_{A_r} = 2 \times (198.9) = +397.8\ \text{kJ/mol}$

Now, take Reaction B: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ $\Delta H_B = -142.3\ \text{kJ/mol}$ (assuming the given value)

Multiply Reaction B by 2: $\ce{2 O_{3}(g) -> 3 O_{2}(g)}$ $\Delta H_{B×2} = 2 \times (-142.3) = -284.6\ \text{kJ/mol}$

Now, add the reversed A (×2) and B×2:

$\ce{2 NO_{2}(g) + 2 O_{2}(g) + 2 O_{3}(g) -> 2 NO(g) + 2 O_{3}(g) + 3 O_{2}(g)}$

Simplify: $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$ (since 2 O2 + 3 O2? Wait, no, 2 O2 on left, 3 O2 on right: 3 O2 - 2 O2 = O2 on right. And 2 O3 on left and right cancel. So yes, that's the target reaction.

Now, calculate the total $\Delta H$:

$\Delta H_{total} = \Delta H_{A_r} + \Delta H_{B×2} = 397.8 + (-284.6) = 113.2\ \text{kJ/mol}$? Wait, that doesn't seem right. Wait, maybe I messed up Reaction B.

Wait, maybe Reaction B is $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ with $\Delta H = +142.3\ \text{kJ/mol}$ (endothermic). Let's try that.

Then, reversed A (×2): +397.8

Reaction B (×2): $\ce{2 O_{3}(g) -> 3 O_{2}(g)}$ $\Delta H = 2 × 142.3 = +284.6$

Now, add reversed A (×2) and B×2:

$\ce{2 NO_{2}(g) + 2 O_{2}(g) + 2 O_{3}(g) -> 2 NO(g) + 2 O_{3}(g) + 3 O_{2}(g)}$

Simplify: $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$

$\Delta H = 397.8 + 284.6 = 682.4$? No, that's too big.

Wait, maybe the third reaction is involved? Wait, the third reaction is $\ce{O_{2}(g) -> 2 O(g)}$ $\Delta H = +495\ \text{kJ/mol}$, but that's not needed for the target reaction.

Wait, maybe I misread the second reaction. Let's check the original problem again.

Wait, the user's image:

First reaction: $\ce{NO(g) + O_{3}(g) -> NO_{2}(g) + O_{2}(g)}$ $\Delta H = -198.9\ \text{kJ/mol}$

Second reaction: $\ce{O_{3}(g) -> 3/2 O_{2}(g)}$ $\Delta H = -142.3\ \text{kJ/mol}$ (exothermic)

Third reaction: $\ce{O_{2}(g) -> 2 O(g)}$ $\Delta H = +495\ \text{kJ/mol}$

But the target reaction is $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$

Let's try another approach. Let's express the target reaction as 2*(Reversed Reaction 1) + 2*(Reaction B) - something? Wait, no.

Wait, Reaction 1: $\ce{NO + O3 -> NO2 + O2}$ $\Delta H1 = -198.9$

Reaction 2: $\ce{O3 -> 3/2 O2}$ $\Delta H2 = -142.3$ (given)

Target: $\ce{2 NO2 -> 2 NO + O2}$

Let's reverse Reaction 1: $\ce{NO2 + O2 -> NO + O3}$ $\Delta H1_r = +198.9$

Multiply reversed Reaction 1 by 2: $\ce{2 NO2 + 2 O2 -> 2 NO + 2 O3}$ $\Delta H1_r×2 = 2×198.9 = +397.8$

Now, take Reaction 2 and multiply by 2: $\ce{2 O3 -> 3 O2}$ $\Delta H2×2 = 2×(-142.3) = -284.6$

Now, add the two modified reactions:

$\ce{2 NO2 + 2 O2 + 2 O3 -> 2 NO + 2 O3 + 3 O2}$

Simplify: $\ce{2 NO2 -> 2 NO + O2}$ (since 2 O2 + 3 O2? Wait, no: left side O2: 2, right side O2: 3. So 3 O2 - 2 O2 = O2 on right. And 2 O3 on left and right cancel. So yes, the target reaction.

Now, calculate $\Delta H$: 397.8 + (-284.6) = 113.2\ \text{kJ/mol}

Wait, but that seems low. Alternatively, maybe the second reaction's $\Delta H$ is +142.3. Let's check the sign of Reaction 2. Ozone decomposition to O2: O3(g) → 3/2 O2(g). The enthalpy of formation of O3 is about +142.7 kJ/mol, so the reaction O3 → 3/2 O2 would have ΔH = -ΔHf(O3) = -142.7 kJ/mol (exothermic). So the given ΔH for Reaction 2 is -142.3, which is close. So that's correct.

So then the calculation is:

Reversed Reaction 1 (×2): ΔH = 2×(198.9) = +397.8

Reaction 2 (×2): ΔH = 2×(-142.3) = -284.6

Total ΔH = 397.8 - 284.6 = 113.2\ \text{kJ/mol}

Wait, but let's verify with another method.

Alternatively, let's use Hess's law. The target reaction is 2 NO2 → 2 NO + O2.

Let's express this as 2*(NO → NO2 + O2 - O3) reversed? Wait, no.

Wait, Reaction 1: NO + O3 → NO2 + O2 ΔH1 = -198.9

So, NO2 + O2 → NO + O3 ΔH1_r = +198.9

Multiply by 2: 2 NO2 + 2 O2 → 2 NO + 2 O3 ΔH = 2×198.9 = +397.8

Now, Reaction 2: O3 → 3/2 O2 ΔH2 = -142.3

Multiply by 2: 2 O3 → 3 O2 ΔH = 2×(-142.3) = -284.6

Now, add the two equations:

2 NO2 + 2 O2 + 2 O3 → 2 NO + 2 O3 + 3 O2

Cancel 2 O3 from both sides, and subtract 2 O2 from both sides:

2 NO2 → 2 NO + O2

ΔH = 397.8 - 284.6 = 113.2\ \text{kJ/mol}

Yes, that seems correct.

Answer:

The enthalpy change for the reaction $\ce{2 NO_{2}(g) -> 2 NO(g) + O_{2}(g)}$ is $\boxed{+113.2\ \text{kJ/mol}}$ (or check the calculation again for possible sign errors, but based on the given values, this is the result).