Final answer:
∆S∘ surroundings = 0.626 J/K. To calculate the entropy change for the surroundings when 2.26 moles of Ca(OH)2(aq) react at standard conditions, we can use the equation ΔH = q + ΔHsys to relate the heat absorbed or released by the reaction to the heat absorbed or released by the surroundings. Then, we can calculate the entropy change for the surroundings using the equation ΔSsurroundings = -q / T, where T is the temperature in Kelvin.
Step-by-step explanation:
To calculate the entropy change for the surroundings, we need to first calculate the heat absorbed or released by the reaction. In this case, the reaction is Ca(OH)₂(aq) + 2HCl(aq) → CaCl₂(s) + 2H₂O(l). We are given that 2.26 moles of Ca(OH)₂(aq) react at standard conditions.
Using the molar heat of solution (ΔHsoln) for calcium chloride of -82.8 kJ/mol, we can calculate the heat absorbed or released by the reaction:
ΔH = ΔHsoln × moles of Ca(OH)₂(aq)
ΔH = -82.8 kJ/mol × 2.26 mol
= -187.008 kJ
Next, we can use the equation ΔH = q + ΔHsys to relate the heat absorbed or released by the reaction to the heat absorbed or released by the surroundings:
ΔH = q + ΔHsys
Since the system is the surroundings in this case, ΔHsys is equal to -ΔH.
Thus, we can rewrite the equation as:
-ΔH = q + ΔHsys
Substituting the value of ΔH and rearranging the equation, we can solve for q, which represents the heat absorbed or released by the surroundings:
q = -ΔH - ΔHsys
To calculate the entropy change for the surroundings, we need to use the equation:
ΔSsurroundings = -q / T, where T is the temperature in Kelvin.
Given that the reaction occurs at 298K, we can calculate the entropy change for the surroundings:
ΔSsurroundings = -q / T
= -(-187.008 kJ) / 298K
= 0.626 J/K