Question

Observe the following reaction, $ CuS{{O}_{4}}.5{{H}_{2}}O\left( s \right)\rightleftharpoons CuS{{O}_{4}}.3{{H}_{2}}O\left( s \right)+2{{H}_{2}}O\left( g \right) $ ; $ {{K}_{P}}=4\times {{10}^{-4}}at{{m}^{2}} $ . if the vapour pressure of water is $ 38 $ $ torr $ then percentage of relative humidity is: (Assume all data at constant temperature)
(A) $ 4 $
(B) $ 10 $
(C) $ 40 $
(D) None of the above

Answer 1

Hint: equilibrium is defined as the state of a system where the concentration of the reactant and the concentration of the product do not change with time and there will be no change in the properties of the system. A chemical equilibrium is achieved when the rate of forward reaction is equal to rate of backward reaction.

Complete step by step solution:
 It is given that equilibrium constant of a pressure, $ {{K}_{P}}=4\times {{10}^{-4}}at{{m}^{2}} $
Vapour pressure of water is $ 38 $ $ torr $
 $ {{K}_{P}}=\dfrac{{{P}_{P}}}{{{P}_{R}}} $
Where, $ {{K}_{P}} $ is the equilibrium constant of pressure
 $ {{P}_{P}} $ is the partial pressure of products
 $ {{P}_{R}} $ is the partial pressure of the reactants
Therefore, according to the given reaction that is:
 $ CuS{{O}_{4}}.5{{H}_{2}}O\left( s \right)\rightleftharpoons CuS{{O}_{4}}.3{{H}_{2}}O\left( s \right)+2{{H}_{2}}O\left( g \right) $
 $ {{K}_{P}}=P_{{{H}_{2}}O}^{2} $
The partial pressure of $ CuS{{O}_{4}}.5{{H}_{2}}O $ and $ CuS{{O}_{4}}.3{{H}_{2}}O $ should not be included because they are present in solid state.
On substituting the value in the above formula we get,
 $ 4\times {{10}^{-4}}=P_{{{H}_{2}}O}^{2} $
 $ {{P}_{{{H}_{2}}O}}=2\times {{10}^{-2}}atm $
 $ {{P}_{{{H}_{2}}O}}=0.02atm $
 As it is given that the vapour pressure of water $ P_{{{H}_{2}}O}^{1} $ is $ 38 $ $ torr $
We will firstly convert this vapour pressure into atm
We will apply unitary method for converting torr into atm,
 $ 760torr=1atm $
 $ 1torr=\dfrac{1}{760} $
 $ 38torr=\dfrac{1}{760}\times 38 $
 $ 38torr=0.05atm $
To calculate percentage Relative humidity,
 $ \phi =\dfrac{{{P}_{{{H}_{2}}O}}}{P_{{{H}_{2}}O}^{1}} $
Where, $ \phi $ is relative humidity
 $ {{P}_{{{H}_{2}}O}} $ is partial pressure of water
 $ P_{{{H}_{2}}O}^{1} $ vapour pressure of water
Percentage relative humidity = $ \dfrac{{{P}_{{{H}_{2}}O}}}{P_{{{H}_{2}}O}^{1}}\times 100 $
Now on substituting the value from above we get,
 $ \Rightarrow \dfrac{{{P}_{{{H}_{2}}O}}}{P_{{{H}_{2}}O}^{1}}\times 100=\dfrac{0.02}{0.05}\times 100 $
 $ \Rightarrow \dfrac{{{P}_{{{H}_{2}}O}}}{P_{{{H}_{2}}O}^{1}}\times 100=40% $
Percentage relative humidity $ =40% $

Note :
Relative humidity is defined as the amount of moisture present in an environment. It is the ratio of partial pressure of water and vapour pressure of water. If the percentage of relative humidity is high then it is considered as a very humid atmosphere.
Equilibrium constant gives a relationship between the amount of products and the amount of reactant when reaches the equilibrium.