Final answer:
The entropy change in the surroundings when 1 mol of N₂O₄(g) is formed from NO₂(g) under standard conditions is -30.74 J/K·mol.
Step-by-step explanation:
To calculate the entropy change in the surroundings when 1 mol of N₂O₄(g) is formed from NO₂(g) under standard conditions at 298 K, we need to use the formula for entropy change in the surroundings, which is ΔS₀ = -ΔH₀/T, where ΔH₀ is the standard enthalpy change and T is the temperature in Kelvin.
Since the formation of 1 mole of N₂O₄ from 2 moles of NO₂ has a standard enthalpy change of ΔH₀= +9.16 kJ/mol (given for N₂O₄), and the process is at standard temperature (298 K), we can substitute these values into the equation. It's important to convert the enthalpy change to Joules by multiplying it by 1000 (since 1 kJ = 1000 J).
Therefore, the entropy change would be ΔS₀ = -(9.16 kJ/mol × 1000 J/kJ) / 298 K = -30.74 J/K·mol, indicating an increase in the entropy of the surroundings.
The enthalpy change (ΔH) for the formation of 1 mol of N₂O₄(g) from 2.00 mL of NO₂(g) under standard conditions at 298 K can be calculated using the enthalpy of formation (ΔHf°) of N₂O₄(g), which is 9.16 kJ/mol.
To calculate the entropy change (∆S) in the surroundings, you can use the equation ∆S = -ΔH/T, where T is the temperature in Kelvin. Plugging in the values, we get ∆S = -(9.16 kJ/mol) / (298 K) = -0.0307 kJ/(mol·K). The negative sign indicates that the surroundings experience a decrease in entropy.