Question

Using the given data and Hess's Law, what is the enthalpy of vaporization of water if the enthalpy of combustion for CH₄​(g) to form H₂​O is -891 kJ and the enthalpy of CH₄​ to form H₂​O is -803 kJ?

Answer 1

Using Hess's Law, the enthalpy of vaporization of water is calculated by finding the difference between the enthalpy of combustion of CH₄ to form H₂O gas and liquid, resulting in an enthalpy of vaporization of 88 kJ/mol.

Step-by-step explanation:

To calculate the enthalpy of vaporization of water using Hess's Law and the provided data, we must first understand the relationships between the enthalpies of different states of water.

According to the data, combustion of hydrogen gas (H₂) to form liquid water (H₂O¹) releases 286 kJ/mol.

However, if water gas (H₂O¹) forms, only 242 kJ are released.

Thus, the difference in enthalpy change when water goes from gas to liquid (enthalpy of vaporization) is

286 kJ - 242 kJ = 44 kJ/mol.

Now, we use the given enthalpy changes to find the enthalpy of vaporization of water.

The enthalpy of combustion of CH₄ to form H₂O gas is -891 kJ, and the enthalpy of CH₄ to form H₂O liquid is -803 kJ. The difference between these two values represents the enthalpy of vaporization of water, which is:

-891 kJ + 803 kJ = -88 kJ

Hence, the enthalpy of vaporization of water is 88 kJ/mol.