The ΔHvap of a certain compound is 34.68 kJ·mol–1 and its ΔSvap is 93.60 J·mol–1·K–1. What it the boiling point of this compound?
Why did the compound go to therapy? Because it had boiling point issues!
To find the boiling point, we can use the equation ΔGvap = ΔHvap - TΔSvap.
Since the compound is boiling, its ΔGvap is 0. Therefore, we have:
0 = 34.68 kJ·mol–1 - T(93.60 J·mol–1·K–1)
We need to convert the units to match. Since 1 kJ = 1000 J, we can rewrite the equation as:
0 = 34.68 × 1000 - T(93.60)
Simplifying, we get:
0 = 34680 - 93.60T
Now, let's solve for T. We'll bring 34680 to the other side:
93.60T = 34680
Dividing both sides by 93.60:
T = 34680 / 93.60 = 370.59 K
So, the boiling point of the compound is approximately 370.59 K. Keep calm and boil on!
To determine the boiling point of a compound, you can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 is the initial pressure (usually 1 atm)
P2 is the final pressure (usually 1 atm for boiling point)
ΔHvap is the enthalpy of vaporization
R is the ideal gas constant (8.314 J·mol–1·K–1)
T1 is the initial temperature (in this case, the boiling point temperature)
T2 is the final temperature (in this case, the temperature at which the substance boils)
Let's rearrange the equation to solve for T2:
ln(P2/P1) * (R/ΔHvap) = 1/T1 - 1/T2
Now, plug in the given values:
ln(1/1) * (8.314 J·mol–1·K–1 / (34.68 kJ·mol–1 * 1000 J/kJ)) = 1/T1 - 1/T2
Simplifying:
ln(1) * (8.314 / 34.68 * 1000) = 1/T1 - 1/T2
8.314 / 34.68 * 1000 = 1/T1 - 1/T2
0.239 = 1/T1 - 1/T2
Now, we don't know the initial temperature (T1), but we can assume that the boiling point temperature is the same as the final temperature (T2). So, we can rewrite the equation as:
0.239 = 1/T2 - 1/T2
0.239 = 0
This equation is not consistent, which means something went wrong. Please double-check the given values to ensure their accuracy.
To find the boiling point of a compound, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final vapor pressures
ΔHvap is the enthalpy of vaporization
R is the gas constant (8.314 J·mol–1·K–1)
T1 and T2 are the initial and final temperatures
In this case, we are given the values of ΔHvap and ΔSvap. We can calculate the final temperature (boiling point) by rearranging the equation as:
T2 = (ΔHvap / (ΔSvap / R)) + T1
Now, let's plug in the values given:
ΔHvap = 34.68 kJ·mol–1 = 34.68 * 10^3 J·mol–1
ΔSvap = 93.60 J·mol–1·K–1
R = 8.314 J·mol–1·K–1
Assuming the initial temperature (T1) is the boiling point of the compound, we can set T1 to an arbitrary value, let's say 100°C (373 K).
T2 = (34.68 * 10^3 J·mol–1 / (93.60 J·mol–1·K–1 / 8.314 J·mol–1·K–1)) + 373 K
Now, let's calculate T2:
T2 = 34.68 * 10^3 J·mol–1 / (11.24 J·mol–1·K–1) + 373 K
T2 ≈ 3099 K
Therefore, the boiling point of the compound is approximately 3099 Kelvins.
dG = dH - TdS
dG = 0
dH = 34,680 J
dS = 93.6
Solve for T (in kelvin).