We appreciate your visit to What is the enthalpy of vaporization of a compound that has a vapor pressure of 157 mmHg at 253 K and 0 358 mmHg at. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
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.
Thanks for taking the time to read What is the enthalpy of vaporization of a compound that has a vapor pressure of 157 mmHg at 253 K and 0 358 mmHg at. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
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.
To find more on vaporization enthalpy, refer here:
https://brainly.com/question/29064263
#SPJ1
