# How do you calculate the entropy change in the surroundings when 1.00 mol #N_2O_4(g)# is formed from #NO_2(g)# under standard conditions at 298 K?

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[-192 #J K^-1# ]

[-192

From Levine's Physical Chemistry, my textbook:

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The entropy change in the surroundings can be calculated using the equation:

ΔS_surr = -ΔH_sys / T

Where: ΔS_surr = entropy change in the surroundings ΔH_sys = enthalpy change of the system T = temperature in Kelvin

First, you need to determine the enthalpy change of the system (ΔH_sys) for the reaction:

2NO2(g) -> N2O4(g)

Using standard enthalpies of formation: ΔH_sys = ΣΔHf(products) - ΣΔHf(reactants)

ΔH_sys = [0 kJ/mol - (-19.6 kJ/mol)] - [2(33.2 kJ/mol)]

ΔH_sys = -85.2 kJ/mol

Substitute ΔH_sys and the temperature (298 K) into the equation to find ΔS_surr.

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