Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1 degrees celsius. At what temperature does benzene boil when the external pressure is 480 torr? Can someone show me an equation and how to solve this?
Use the Clausius-Clapeyron equation. Remember the definition of boiling point; i.e., the temperature at which the vapor pressure of the liquid equals atmospheric pressure. So you know p1 and p2 and T1. T2 is the only unknown.
I know im using that equation but would you set it up like this
1/t2= (760mm hg/480torr)/(30720j/8.314)+(1/353.25degrees celsius) when i do it out i get a number that does not make sense.
I didn't work it out but I suspect the problem is that you are trying to solve for 1/t2 from the get go. It is far simpler if you plug the numbers into the equation and solve that way.
ln(P2/p1) = DHvap/R (1/T1 - 1/T2)
ln(480/760) = (30720/8.314)(1/353.25 - 1/T2)
First term = ln(480/760) = -0.4595
30720/8.314 = 3694.97
1/353.25 = 0.002831
1/t2 = 1/t2
-0.4595 = 3694.97(0.002831-1/T2)
-0.4595 = (10.4599 - 3694.97/T2)
Check my work. I'll let you finish.
I have -.4595= -3684.51/T2. Would you then just divide -3684.51/-.4595? and how did you het 10.4599 in the last equation?
No. I think your algebra is cockeyed.
Assuming my work is ok to where I stopped, then multiply through by T2 to get rid of the T2 term in the denominator.
-0.4595 = (10.4599 - 3694.97/T2)
-0.4595T2 = 10.4599T2 - 3694.97
-10.919T2 = -3694.97
T2 = 338.4 or thereabouts which is approximately 65 C. Check my work but that sounds reasonable to me. Lower pressure means lower boiling point. I can see why 8,000 K didn't sound right to you.
Isn't the formula suppose to be ln(p2/p1)=Dvap/R(1/T2 -1/T1) ? That is what my textbook says.
To solve this problem, we can use the Clausius-Clapeyron equation, which relates the boiling point of a substance to its heat of vaporization and the external pressure. The equation is as follows:
ln(P1/P2) = (∆Hvap/R) * (1/T2 - 1/T1)
where:
P1 is the initial pressure (in this case, the vapor pressure at the normal boiling point, which can be found in a reference book),
P2 is the final pressure (480 torr in this case),
∆Hvap is the heat of vaporization of the substance (30.72 kJ/mol for benzene),
R is the ideal gas constant (8.314 J/(mol*K)),
T1 is the initial temperature (80.1 degrees Celsius in this case, which needs to be converted to Kelvin),
T2 is the final temperature (which needs to be solved for).
First, we need to convert the initial temperature from degrees Celsius to Kelvin by adding 273.15:
T1 = 80.1 + 273.15 = 353.25 K
Now, let's plug in the known values into the Clausius-Clapeyron equation and solve for T2:
ln(1/P2) = (30.72 * 10^3 J/mol) / (8.314 J/(mol*K)) * (1/T2 - 1/353.25)
To isolate T2, we can rearrange the equation as follows:
1/T2 = (ln(1/P2) * 8.314 J/(mol*K)) / (30.72 * 10^3 J/mol) + 1/353.25
Now, let's calculate the value for 1/T2:
1/T2 = (ln(1/480) * 8.314 J/(mol*K)) / (30.72 * 10^3 J/mol) + 1/353.25
Next, we'll solve for T2 by taking the reciprocal of both sides:
T2 = 1 / [(ln(1/480) * 8.314 J/(mol*K)) / (30.72 * 10^3 J/mol) + 1/353.25]
By plugging in the values and evaluating the equation, we can find the boiling point of benzene at an external pressure of 480 torr.