Calculate the value of ΔS when 1.10mol of Br2(l) is vaporized at 58.8 ∘C
To calculate the change in entropy (∆S) when 1.10 mol of Br2(l) is vaporized at 58.8 °C, we need to use the equation:
∆S = n × ΔSvap
where:
∆S is the change in entropy (in J/mol·K)
n is the number of moles of the substance
ΔSvap is the molar entropy of vaporization (in J/mol·K)
First, we need to determine the molar entropy of vaporization (ΔSvap) for Br2(l). We can look up this value in a reliable reference source, such as a chemistry textbook or a database. Let's assume the value of ΔSvap for Br2 is 93.2 J/mol·K.
Now we can plug in the values into the equation:
∆S = 1.10 mol × 93.2 J/mol·K
Calculating this expression:
∆S = 102.52 J/K
Therefore, the change in entropy (∆S) when 1.10 mol of Br2(l) is vaporized at 58.8 °C is 102.52 J/K.
To calculate ΔS (change in entropy) when 1.10 mol of Br2(l) is vaporized at 58.8 ∘C, we need to use the equation:
ΔS = n × ΔSvap
Where:
ΔS is the change in entropy (in J/mol·K).
n is the number of moles of the substance.
ΔSvap is the molar entropy of vaporization (in J/mol·K).
First, we need to find the molar entropy of vaporization, ΔSvap, for Br2. The molar entropy of vaporization can be found in a data table or reference book. For Br2, ΔSvap is approximately 93.7 J/mol·K.
Now, we can calculate ΔS using the formula:
ΔS = n × ΔSvap
ΔS = 1.10 mol × 93.7 J/mol·K
ΔS = 103.07 J/K
So, the change in entropy, ΔS, when 1.10 mol of Br2(l) is vaporized at 58.8 °C is approximately 103.07 J/K.
58.8 C is the boiling point of Br2.
At the boiling point you have equilibrium between the liquid phase and vapor phase and
dG = dH - TdS.
dG = 0 because it is equlibrium.
dH vap-- look that up in tables.
You know T, solve for dS. If you want to read more about this look up delta S vaporization on Google.