*Integrate if possible please (P-Chem Question) *Question #6 *Show work please 5
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*Integrate if possible please (P-Chem Question)
*Question #6
*Show work please
5. What is the Standard entailpy Change or a reaCTUTIOr WTIC LIe equDIUT CUTIslal IL ID doubled when the temperature is increased from 298 K to 308 K? 6. Naphthalene, CioHg, melts at 80.2°C. If the vapor pressure of liquid naphthalene is 10 torr at 85.80C and 40 torr at 1193Ccalculate the enthalpy of vaporization and the normal boiling point of naphthalene using the Clausius-Clapeyron equation. 7. Use the phase diagram for carbon dioxide in your text to state what would be observed when aExplanation / Answer
The Clausius- Clapeyron equation is given by-
ln (P1 / P2) = - (H / R) (1/T1 - 1/T2)
Given here are,
P1 = 10 torr , P2 = 40 torr , T1 = 358.95 K and T2 = 392.45 K
Putting the values in formula as:-
ln (10 torr / 40 torr) = - (H / 8.314 J K-1 mol-1) (1/(358.95 K) - 1/(392.45 K))
ln (.25) = - (H / 8.314 J K-1 mol-1) (0.0027859 K-1 - 0.0025481 K-1)
ln (.25) = - (H / 8.314 J K-1 mol-1) (0.0002378 K-1)
-1.3862 = - (H * 0.0002378 K-1/ 8.314 J K-1 mol-1)
-1.3862 = - (H * 2.86 * 10-5 J-1 mol-1)
1.3862 / 2.86 * 10-5 J-1 mol) = H
or
H = 1.3862 / (2.86 * 10-5 J-1 mol)
H = 4.85 * 104 J mol-1
Now to get the normal boiling point , we consider
T1 = normal boiling point
P1 = 760 torr because its standard pressure for getting normal boiling point of any substance
Now using Clausius- Clapeyron equation,
ln (760 torr / 40 torr) = - (4.85 *104 J mol-1 / 8.314 J K-1 mol-1) (1/(T1 K) - 1/(392.45 K))
2.94 = - (5.83 * 103 K-1 ) (1/(T1 K) - 1/(392.45 K))
5.04 * 10-4 K = - 1/(T1 K) + 1/(392.45 K))
5.04 * 10-4 K - .002548 K = - 1/(T1 K)
- 002044 = -1 / T1
T1 = 489.23 K = 216.1 oC
so, enthalpy of vapourization = 4.85 * 104 J mol-1 = 48.5 kJ mol-1
and normal boiling point = 216.1 oC
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