Question

Equilibrium constants is given (in atm) for the following reaction ${{0}^{0}}C$:$N{{a}_{2}}HP{{O}_{4}}.12{{H}_{2}}O(s)N{{a}_{2}}HP{{O}_{4}}.7{{H}_{2}}O(s)+5{{H}_{2}}O(g)$ ; ${{K}_{p}}=2.43\times {{10}^{-13}}$ . The vapour pressure of water at ${{0}^{0}}C$ is 4.56 torr. At what relative humidity will $N{{a}_{2}}HP{{O}_{4}}.12{{H}_{2}}O(s)$ be efflorescent when exposed to air at ${{0}^{0}}C$?

Answer 1

Hint:. The answer to this question is based on the concept of relative humidity calculation which is given by ratio of partial pressure of water vapour to that of equilibrium vapour pressure of water that is $\phi =\dfrac{{{p}_{{{H}_{2}}O}}}{{{p}^{*}}_{{{H}_{2}}O}}$.


Complete step by step answer:
We have come across the concept about the vapour pressure and some of the terms which are linked to the vapour pressure in our previous classes.

Now, let us see what actually vapour pressure is
- Vapour pressure is defined as ‘the measure of tendency of any material to change into its vapour phase or the gaseous phase’.
- Vapour pressure is dependent on temperature and it increases with increase in the temperature.
- Humidity on the other hand is nothing but the amount of vapour pressure in the air.
Therefore, relative humidity is given by the ratio of partial pressure of water vapour to that of equilibrium vapour pressure of water and is denoted by the symbol$\phi $ which is given by,
$\phi =\dfrac{{{p}_{{{H}_{2}}O}}}{{{p}^{*}}_{{{H}_{2}}O}}$ ………(1)
Now, according to the data given ${{K}_{p}} = 2.43\times {{10}^{-13}}$ …..(2)
Here, in the reaction $N{{a}_{2}}HP{{O}_{4}}.12{{H}_{2}}O(s) \longrightarrow N{{a}_{2}}HP{{O}_{4}}.7{{H}_{2}}O(s)+5{{H}_{2}}O(g)$, five moles of water is being released in vapour form.
Therefore, ${{K}_{p}}$ for this is ${{K}_{p}} = {{(p{{H}_{2}}O)}^{5}}$

By substituting the value of ${{K}_{p}}$ from equation number (2) we have,
${{(p{{H}_{2}}O)}^{5}} = 2043\times {{10}^{-13}}$
\[\Rightarrow p{{H}_{2}}O = 0.003 atm\]
\[\Rightarrow p{{H}_{2}}O = 2.28 torr\]
Therefore, if the vapour pressure in air is less than the above value then there will be efflorescence.
Thus, relative humidity is given by equation (1) and by substituting the values,
$\phi =\dfrac{2.28}{4.56}\times 100$ [in form of percentage]

$\phi =50%$
Thus, relative humidity is less than 50% and therefore $N{{a}_{2}}HP{{O}_{4}}.12{{H}_{2}}O(s)$ is efflorescent.

Note: Note that the question is asked on the basis of pressure unit in terms of torr and thus while solving make sure that you convert the value into torr where 1 atm = 760 torrs and do not write the direct answer in form of atmospheric pressure unit.