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Answer :
The enthalpy of vaporization can be determined using the Clausius -Clapeyron equation. For the given conditions, the enthalpy of vaporization is 39 kJ/mol.
What is Clausius -Clapeyron equation?
Clausius -Clapeyron equation relates the enthalpy of vaporization with the change in pressure and temperature by the expression given below:
[tex]\rm ln \frac{P_{2}}{P_{1}} = \frac{\Delta H }{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}} ][/tex]
Given that, P1 = 157 mmHg
P2 = 0.358 mmHg.
T1 = 253 K and T2 = 191 K.
Apply all these values in Clausius -clapeyron equation
[tex]\rm ln \frac{157 mmHg}{0.358 mmHg} =\frac{\Delta H}{8.31 J/(K mol)} [\frac{1}{253 K} - \frac{1}{191 K} ][/tex]
ΔH = 39 KJ/mol.
Therefore, the enthalpy of vaporization of the compound is 39 KJ/mol.
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Final answer:
To find the enthalpy of vaporization, use the Clausius-Clapeyron equation by calculating the temperature difference and solving for the enthalpy. The enthalpy of vaporization for the compound in question is approximately 39.4 kJ/mol.
Explanation:
The enthalpy of vaporization can be calculated using the Clausius-Clapeyron equation. First, determine the temperature difference (ΔT) between the two vapor pressures. Then, use the equation ln(P2/P1) = -ΔHvap/R × (1/T2 - 1/T1) to find the enthalpy of vaporization.
Substitute the given values, rearrange the equation, and solve for ΔHvap to obtain the enthalpy of vaporization of the compound. In this case, the enthalpy of vaporization is approximately 39.4 kJ/mol.
