The normal boiling point of bromine (Br2) is 58.8°C, and its molar enthalpy of vaporization is ΔHvap = 29.6 kJ/mol.
Calculate the value of ΔS when 1.00 mol of Br2(l) is vaporized at 58.8°C.?
dG = dH - TdS
At equilibrium (which you have at the boiling point), dG = 0, then
0 = dH -TdS
TdS = dH
dS = dH/T = ?
Remember T must be in kelvin; also dH in kJ gives dS in kJ or dH in J gives dS in J.
To calculate the value of ΔS when 1.00 mol of Br2(l) is vaporized at 58.8°C, you can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R)(1/T1 - 1/T2)
Where:
P1 = initial pressure (usually 1 atm)
P2 = final pressure (also 1 atm since it is in the vapor phase)
T1 = initial temperature (the normal boiling point, 58.8°C converted to Kelvin, K)
T2 = final temperature (also 58.8°C converted to K)
R = ideal gas constant (8.314 J/mol·K)
ΔHvap = molar enthalpy of vaporization (29.6 kJ/mol converted to J/mol)
Now, let's plug in the values and solve for ΔS.
ln(1/1) = (29.6 × 10^3 J/mol / 8.314 J/mol·K) (1/(58.8 + 273.15) K - 1/(58.8 + 273.15) K)
ln(1/1) = (29.6 × 10^3 J/mol / 8.314 J/mol·K) (1/331.95 K - 1/331.95 K)
ln(1/1) = 3.558 × (0 - 0)
ln(1/1) = 0
Therefore, the value of ΔS when 1.00 mol of Br2(l) is vaporized at 58.8°C is 0.
To calculate the value of ΔS (change in entropy) when 1.00 mol of Br2(l) is vaporized at 58.8°C, we need to use the following formula:
ΔS = ΔHvap / T
Where:
ΔS = change in entropy
ΔHvap = molar enthalpy of vaporization
T = temperature in Kelvin
Step 1: Convert the boiling point from Celsius to Kelvin
To convert from Celsius to Kelvin, use the formula:
T(K) = T(°C) + 273.15
For bromine, whose boiling point is 58.8°C:
T(K) = 58.8°C + 273.15 = 331.95 K
Step 2: Calculate ΔS using the formula:
ΔS = ΔHvap / T
Plug in the values:
ΔS = 29.6 kJ/mol / 331.95 K
Step 3: Convert the units to J/(mol·K)
To convert from kJ/mol to J/(mol·K), multiply by 1000:
ΔS = (29.6 kJ/mol / 331.95 K) * 1000 = 89.19 J/(mol·K)
Therefore, the value of ΔS when 1.00 mol of Br2(l) is vaporized at 58.8°C is 89.19 J/(mol·K).