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 find the number of moles of PbBr2 in 2.45 g.
mass of PbBr2 = 2.45 g
molar mass of PbBr2 = 207.2 g/mol (207.2 = 207.2+2(79.9))
number of moles of PbBr2 = mass/molar mass = 2.45 g/207.2 g/mol = 0.0118 mol

Next, we 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 the substance, not the change in entropy. Therefore, we need to make the answer negative because the entropy of a substance cannot be negative.

Entropy of 2.45 g of PbBr2 = -1.90 J/K ≈ -1.07 J/K (to two significant figures)

Therefore, the answer is A, -1.07 J/K.

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

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