A certain substance has an enthalpy of fusion of 5.15 kJ/mol and an entropy of fusion of 27.8 J/mol/K. Calculate the approximate melting point of the substance. Report your answer in degrees Celsius.
A certain substance has an enthalpy of fusion of 5.15 kJ/mol and an entropy of fusion of 27.8 J/mol/K. Calculate the approximate melting point of the substance. Report your answer in degrees Celsius.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Problem Statement:**
A certain substance has an enthalpy of fusion of 5.15 kJ/mol and an entropy of fusion of 27.8 J/mol/K. Calculate the approximate melting point of the substance.
Report your answer in degrees Celsius.
**Instructions:**
To find the melting point in degrees Celsius, use the formula:
\[
T = \frac{\Delta H}{\Delta S}
\]
where:
- \( T \) = melting point in Kelvin
- \( \Delta H \) = enthalpy of fusion (in J/mol)
- \( \Delta S \) = entropy of fusion (in J/mol/K)
Convert the enthalpy from kJ/mol to J/mol (1 kJ = 1000 J).
Finally, convert the temperature from Kelvin to Celsius using the conversion:
\[
T(°C) = T(K) - 273.15
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Transcribed Image Text:**Problem Statement:**
A certain substance has an enthalpy of fusion of 5.15 kJ/mol and an entropy of fusion of 27.8 J/mol/K. Calculate the approximate melting point of the substance.
Report your answer in degrees Celsius.
**Instructions:**
To find the melting point in degrees Celsius, use the formula:
\[
T = \frac{\Delta H}{\Delta S}
\]
where:
- \( T \) = melting point in Kelvin
- \( \Delta H \) = enthalpy of fusion (in J/mol)
- \( \Delta S \) = entropy of fusion (in J/mol/K)
Convert the enthalpy from kJ/mol to J/mol (1 kJ = 1000 J).
Finally, convert the temperature from Kelvin to Celsius using the conversion:
\[
T(°C) = T(K) - 273.15
\]
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