Calculate the standard entropy change for the reaction of calcium oxide with water to form calcium hydroxide: CaO(s) + H2O(ℓ) → Ca(OH)2(s) at 298 K. ∆S ◦ m ∆H◦ f J K·mol? kJ mol? CaO(s) 39.75 −635.09 H2O(ℓ) 69.91 −285.83 Ca OH 2(s) 83.39 −986.09

Write and balance the equation. Then

dSo rxn = (n*dSo products) - (dSo reactants)
Substitute and solve for dSo rxn.

To calculate the standard entropy change (∆S◦) for the reaction, you can use the standard entropy values (∆S◦m) of the reactants and products.

The standard entropy change (∆S◦) can be calculated using the following equation:

∆S◦ = ∑(∆S◦m products) - ∑(∆S◦m reactants)

Given the standard entropy values (∆S◦m):

CaO(s): 39.75 J/(mol·K)
H2O(ℓ): 69.91 J/(mol·K)
Ca(OH)2(s): 83.39 J/(mol·K)

Using the equation, we can substitute the values:

∆S◦ = (83.39 J/(mol·K)) - ((39.75 J/(mol·K)) + (69.91 J/(mol·K)))

∆S◦ = 83.39 J/(mol·K) - 109.66 J/(mol·K)

∆S◦ = -26.27 J/(mol·K)

The standard entropy change (∆S◦) for the reaction of calcium oxide with water to form calcium hydroxide at 298 K is -26.27 J/(mol·K).

To convert the units to kJ/mol·K, divide by 1000:

-26.27 J/(mol·K) = -0.02627 kJ/(mol·K)

Therefore, the standard entropy change (∆S◦) for the reaction is -0.02627 kJ/(mol·K).

To calculate the standard entropy change (∆S°) for the reaction, we need to use the formula:

∆S° = ∑S°(products) - ∑S°(reactants)

First, we need to find the standard entropies (S°) for each compound involved in the reaction. The values you provided are not the enthalpy of formation values (ΔH°f), but they can be converted into standard entropy values (S°) using a different equation.

The equation to convert between enthalpy of formation and standard entropy is:
ΔH°f = ΔG°f - TΔS°
Where ΔH°f is the enthalpy of formation, ΔG°f is the standard Gibbs free energy of formation, T is the temperature in Kelvin, and ΔS° is the standard entropy change.

Let's calculate the standard entropies (S°) for each compound:

For CaO(s):
ΔH°f = 635.09 kJ/mol
T = 298 K
ΔG°f = ΔH°f + TΔS°
ΔS° = (ΔG°f - ΔH°f) / T
ΔS° = (0 - 635.09) / 298
ΔS° ≈ -2.13 J/(K·mol)

For H2O(l):
ΔH°f = 285.83 kJ/mol
T = 298 K
ΔG°f = ΔH°f + TΔS°
ΔS° = (ΔG°f - ΔH°f) / T
ΔS° = (0 - 285.83) / 298
ΔS° ≈ -0.96 J/(K·mol)

For Ca(OH)2(s):
ΔH°f = 986.09 kJ/mol
T = 298 K
ΔG°f = ΔH°f + TΔS°
ΔS° = (ΔG°f - ΔH°f) / T
ΔS° = (0 - 986.09) / 298
ΔS° ≈ -3.31 J/(K·mol)

Now, we can substitute the respective values into the formula for calculating the standard entropy change (∆S°) for the reaction:

∆S° = ∑S°(products) - ∑S°(reactants)
∆S° = [S°(Ca(OH)2(s))] - [S°(CaO(s)) + S°(H2O(l))]

∆S° = [-3.31] - [-2.13 + (-0.96)]
∆S° ≈ -3.31 + 3.09
∆S° ≈ -0.22 J/(K·mol)

Therefore, the standard entropy change (∆S°) for the reaction of CaO(s) + H2O(ℓ) → Ca(OH)2(s) at 298 K is approximately -0.22 J/(K·mol).