The standard molar entropy of lead(II) bromide (PbBr2) is 161 J/mol • K. What is the entropy of 2.45 g of PbBr2?
a. +1.07 J/K d. –161 J/K
b. –1.07 J/K e. 0 J/K
c. +161 J/K
161 J/mol x (2.45/molar mass PbBr2) = ?
+1.07 J/K
To find the entropy of 2.45 g of PbBr2, we need to use the molar entropy as a conversion factor.
First, we need to calculate the number of moles of PbBr2 in 2.45 g.
The molar mass of PbBr2 is:
Pb (207.2 g/mol) + 2 * Br (79.9 g/mol) = 366 g/mol
To find the number of moles, we can use the following formula:
moles = mass / molar mass
moles = 2.45 g / 366 g/mol
moles ≈ 0.00668 mol
Now we can calculate the entropy using the molar entropy as the conversion factor:
entropy = moles * molar entropy
entropy = 0.00668 mol * 161 J/mol • K ≈ 1.07 J/K
Therefore, the entropy of 2.45 g of PbBr2 is approximately +1.07 J/K.
The correct answer is a. +1.07 J/K.
To find the entropy of 2.45 g of PbBr2, we need to first convert the mass given into moles.
Step 1: Convert mass to moles:
First, find the molar mass of PbBr2 by adding up the atomic masses of Pb (lead) and Br (bromine). The atomic masses are:
Pb: 207.2 g/mol
Br: 79.9 g/mol
So, the molar mass of PbBr2 is:
207.2 g/mol (Pb) + (2 * 79.9 g/mol) (2 Br) = 366.0 g/mol
Now, we can calculate the number of moles in 2.45 g of PbBr2:
moles = mass / molar mass = 2.45 g / 366.0 g/mol ≈ 0.00668 mol (rounded to five decimal places)
Step 2: Calculate the entropy:
The entropy of 0.00668 mol of PbBr2 can be calculated by multiplying the number of moles by the molar entropy:
entropy = moles * molar entropy = 0.00668 mol * 161 J/mol • K ≈ 1.07448 J/K (rounded to five decimal places)
Therefore, the entropy of 2.45 g of PbBr2 is approximately +1.07 J/K (option a).