Question

nitrogen dioxide, no₂(g) (δhᵢ = 33.84 kj/mol), is decomposed according to the following reaction:
2no₂(g) → n₂(g) + 2o₂(g)
what is the enthalpy change when 2.50 mol of nitrogen dioxide decomposes?
use δhᵣₓₙ = σ(δhᵢ,ₚᵣₒdᵤcₜₛ) − σ(δhᵢ,ᵣₑₐcₜₐₙₜₛ).
○ 13.5 kj of energy released
○ 13.5 kj of energy absorbed
○ 84.6 kj of energy released
○ 84.6 kj of energy absorbed

Explanation:

Step1: Find ΔHᵣₓₙ for the reaction

The standard enthalpy of formation ($\Delta H_f$) for elements in their standard states (like $N_2(g)$ and $O_2(g)$) is 0 kJ/mol. For $NO_2(g)$, $\Delta H_f = 33.84$ kJ/mol.
Using the formula $\Delta H_{rxn} = \sum (\Delta H_{f, products}) - \sum (\Delta H_{f, reactants})$:
Products: $N_2(g)$ (0 kJ/mol) and $2O_2(g)$ (0 kJ/mol). Sum: $0 + 0 = 0$.
Reactants: $2NO_2(g)$. Sum: $2 \times 33.84 = 67.68$ kJ/mol.
So, $\Delta H_{rxn} = 0 - 67.68 = -67.68$ kJ (negative means energy is released for 2 mol of $NO_2$ decomposing).

Step2: Calculate ΔH for 2.50 mol of $NO_2$

For 2 mol of $NO_2$, $\Delta H = -67.68$ kJ (released).
For 1 mol, $\Delta H = \frac{-67.68}{2} = -33.84$ kJ (released per mol? Wait, no—wait, the reaction is $2NO_2
ightarrow N_2 + 2O_2$, so when 2 mol $NO_2$ decompose, 67.68 kJ is released (since $\Delta H_{rxn}$ is negative, exothermic). Wait, maybe I mixed up signs. Let's re-express:
$\Delta H_{rxn} = [\Delta H_f(N_2) + 2\Delta H_f(O_2)] - [2\Delta H_f(NO_2)] = (0 + 0) - 2(33.84) = -67.68$ kJ (for 2 mol $NO_2$). So per mol $NO_2$, $\Delta H = \frac{-67.68}{2} = -33.84$ kJ/mol (released).
Now, for 2.50 mol: $\Delta H = 2.50 \, \text{mol} \times (-33.84 \, \text{kJ/mol}) = -84.6$ kJ. The negative sign means energy is released. Wait, but let's check the options. Wait, maybe I made a mistake in sign interpretation. Wait, the formation enthalpy of $NO_2$ is positive (it's an endothermic formation), so decomposing $NO_2$ (reverse of formation) would be exothermic (releasing energy). Wait, the reaction is $2NO_2
ightarrow N_2 + 2O_2$, which is the reverse of $N_2 + 2O_2
ightarrow 2NO_2$ (which has $\Delta H = 2 \times 33.84 = 67.68$ kJ, endothermic). So the reverse reaction (decomposition) has $\Delta H = -67.68$ kJ (exothermic, releasing energy) for 2 mol $NO_2$.
To find for 2.50 mol:
First, find ΔH per mol of $NO_2$: $\frac{-67.68 \, \text{kJ}}{2 \, \text{mol}} = -33.84 \, \text{kJ/mol}$ (released per mol).
Then, for 2.50 mol: $2.50 \, \text{mol} \times (-33.84 \, \text{kJ/mol}) = -84.6$ kJ. The negative sign indicates energy is released. So the enthalpy change is -84.6 kJ (84.6 kJ released).

Answer:

84.6 kJ of energy released (corresponding to the option: 84.6 kJ of energy released)