21.900 9.99 kJ QUESTION 9 Given that the vapor pressure of methanol is T5 otott

Question

21.900 9.99 kJ QUESTION 9 Given that the vapor pressure of methanol is T5 otott 152 C, calculate the mele esthalwy of vaporization of methanol. CNote normal boiling point means 1 atam ce 760mmHg i the same for all beiling lvads) which 0.383 kJ/mol 38.0 kJ/mol 3.00 kJMmol 27.5 kJ/mol 125 points Answr QUESTION 10 mol What is vapor pressure of ethanol, in mmig at 309c? is 39 The vapor pressure of ethanol is 400 mmHg at 63.5 c. Its molar heat of vaporization 0.0099 mmHg 200 mmHg 100 mmHg 4.61 mmHg click Sate and submit to save and subrnir. cliel saveAllAnswers to save allenswers.

Explanation / Answer

Question 9.

Normal BP = 64.6 °C, P° = 1 atm = 760 torr

P° = 75 torr, T = 15.2 °C

now, calculate the enthalpy of vaporization

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

ln(75/760) = Hvap/(8.314) * (1/(64.6+273) - 1/(15.2+273))

Hvap = ln(75/760) * 8.314 /  (1/(64.6+273) - 1/(15.2+273))

Hvap = 37921.56 J/mol = 37.92 kJ/mol

best answer will be B), 38 kJ/mol

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