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
b. –1.07 J/K
c. +161 J/K
d. –161 J/K
e. 0 J/K
161 J/mol x # mols = ?
mols PbBr2 = grams/molar mass
To find the entropy of 2.45 g of PbBr2, we first need to calculate the number of moles of PbBr2 using its molar mass.
The molar mass of PbBr2 is calculated by adding the atomic masses of Pb (lead) and 2Br (bromine):
Atomic mass of Pb = 207.2 g/mol
Atomic mass of Br = 79.9 g/mol
Molar mass of PbBr2 = Atomic mass of Pb + 2 * Atomic mass of Br
Molar mass of PbBr2 = 207.2 + 2 * 79.9 = 366 g/mol
Next, we calculate the number of moles of PbBr2 using its mass and molar mass:
Number of moles = mass / molar mass
Number of moles = 2.45 g / 366 g/mol
Now, let's calculate:
Number of moles = 0.00669 mol
The formula to calculate entropy is S = n * S°, where S is the entropy, n is the number of moles, and S° is the standard molar entropy.
Substituting the values:
S = 0.00669 mol * 161 J/mol * K
Now, let's calculate:
S = 1.07 J/K
Therefore, the entropy of 2.45 g of PbBr2 is approximately +1.07 J/K.
Hence, the correct option is (a) +1.07 J/K.