Question

Enthalpy of fusion and enthalpy of vaporization for water respectively are 6.01 kJ mol\(^{-1}\) and 45.07 kJ mol\(^{-1}\) at 0°C. What is enthalpy of sublimation at 0°C?

• A. 27.50 kJ mol\(^{-1}\)
• B. 48.07 kJ mol\(^{-1}\)
• C. 51.08 kJ mol\(^{-1}\)
• D. 39.06 kJ mol\(^{-1}\)

Hint:

To calculate the enthalpy of sublimation, simply add the enthalpies of fusion and vaporization.

Answer 1

The Correct Option is C

Step 1: Enthalpy of Sublimation.
The enthalpy of sublimation is the total heat required to convert one mole of a solid directly into gas. It can be calculated as the sum of the enthalpy of fusion and enthalpy of vaporization: \[ \Delta H_{\text{sublimation}} = \Delta H_{\text{fusion}} + \Delta H_{\text{vaporization}} \] Substituting the given values: \[ \Delta H_{\text{sublimation}} = 6.01 \, \text{kJ/mol} + 45.07 \, \text{kJ/mol} = 51.08 \, \text{kJ/mol} \]
Step 2: Analyzing the options.
(A) 27.50 kJ/mol: Incorrect. This is lower than the actual value of enthalpy of sublimation.
(B) 48.07 kJ/mol: Incorrect. This value is not the correct sum of fusion and vaporization enthalpies.
(C) 51.08 kJ/mol: Correct. This is the sum of the enthalpy of fusion and enthalpy of vaporization, giving the enthalpy of sublimation.
(D) 39.06 kJ/mol: Incorrect. This is not the correct value for the enthalpy of sublimation.

Step 3: Conclusion.
The correct enthalpy of sublimation is 51.08 kJ/mol, which corresponds to option (C).