Calculate the standard entropy change for the dimerization of NO2: 2NO2(g)→N2O4(g)at 298 K, if

S◦ ∆H◦f J/K·mol
�NO2(g) 240.06 33.18
N2O4(g) 304.29 9.16
Answer in units of J/K·mol

-175.8 j/k.mol

To calculate the standard entropy change (ΔS°) for the dimerization of NO2, we can use the equation:

ΔS° = ΣS°(products) - ΣS°(reactants)

Given the standard entropies for NO2 and N2O4:

S°(NO2) = 240.06 J/K·mol
S°(N2O4) = 304.29 J/K·mol

We can substitute these values into the equation:

ΔS° = S°(N2O4) - 2 * S°(NO2)
ΔS° = 304.29 J/K·mol - 2 * 240.06 J/K·mol

Calculating the value:

ΔS° = 304.29 J/K·mol - 480.12 J/K·mol
ΔS° = -175.83 J/K·mol

Therefore, the standard entropy change for the dimerization of NO2 at 298 K is -175.83 J/K·mol.

To calculate the standard entropy change (∆S°) for the dimerization of NO2, you need to use the formula:

∆S° = ΣS°(products) - ΣS°(reactants)

First, you need to find the values of the standard molar entropy (S°) for each substance involved in the reaction. These values are given in J/K·mol.

For the reactants:
S°(NO2) = 240.06 J/K·mol

For the products:
S°(N2O4) = 304.29 J/K·mol

Next, substitute these values into the formula:

∆S° = S°(N2O4) - 2 * S°(NO2)

∆S° = 304.29 J/K·mol - 2 * 240.06 J/K·mol

∆S° = 304.29 J/K·mol - 480.12 J/K·mol

∆S° = -175.83 J/K·mol

Therefore, the standard entropy change for the dimerization of NO2 is -175.83 J/K·mol.

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