The standard molar entropy of lead(II) bromide (PbBr₂) is 161 J/mol · K. What is the entropy of 2.45 g of PbBr₂?

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of PbBr2 in 2.45 g:
Number of moles = mass / molar mass = 2.45 g / (207.2 g/mol) = 0.0118 mol

Then, we can use the formula:
ΔS = n × S°
where ΔS is the entropy change, n is the number of moles, and S° is the standard molar entropy.

ΔS = 0.0118 mol × 161 J/mol K = 1.90 J/K

However, the question asks for the entropy of 2.45 g, not the entropy change. Therefore, we need to divide ΔS by the number of grams:

Entropy = ΔS / mass = 1.90 J/K / 2.45 g = 0.774 J/g K = 0.774 J/K

This value is positive, indicating an increase in entropy. The closest answer choice is A, which is +1.07 J/K.

The Standard Molar Entropy of Lead(II) Bromide (PbBr2) Is 161