Question

At a given set of conditions, 246.2 kJ is given off when 1 mol of H2O(g) forms from its elements. Under the same conditions, 284.6 kJ is given off when 1 mol of H2O(l) forms from its elements. Find ΔH for the vaporization of water at these conditions.

Answer 1

Answer:

The ΔH for the vaporization of water at these conditions is 38.4 kJ/mol.

Explanation:

Given:

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g),\Delta H^o_{1}=-246.2 kJ/mol$..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l),\Delta H^o_{2}=-284.6 kJ/mol$..[2]

To find: ΔH for the vaporization of water at these conditions:

$H_2O(l)\rightarrow H_2O(g),\Delta H^o_{vap}=?$...[3]

Solution:

[1] - [2] = [3] (using Hess's law)

$\Delta H^o_{vap}=\Delta H^o_{1}-\Delta H^o_{2}$

$=-246.2 kJ/mol -(-284.6 kJ/mol) =38.4 kJ/mol$

The ΔH for the vaporization of water at these conditions is 38.4 kJ/mol.