Question

Anhydrous calcium chloride is often used as a desiccant. In the presence of an excess of \[{\text{CaC}}{{\text{l}}_{\text{2}}}\] ​, the amount of the water taken up is governed by \[{{\text{K}}_{\text{p}}}{\text{ = 6}}{\text{.4 \times 1}}{{\text{0}}^{\text{8}}}^{\text{5}}\] for the following reaction at room temperature, \[{\mathbf{CaC}}{{\mathbf{l}}_{\mathbf{2}}}\left( {\mathbf{s}} \right) + {\mathbf{6}}{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\left( {\mathbf{g}} \right) \rightleftharpoons {\mathbf{CaC}}{{\mathbf{l}}_{\mathbf{2}}}.{\mathbf{6H2}}{\mathbf{O}}\left( {\mathbf{s}} \right)\] . What is the equilibrium vapour pressure of water in a closed vessel that contains \[{\text{CaC}}{{\text{l}}_{\text{2}}}\] (s)?
A.\[{{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}} = 5 \times {10^{ - 15}}{\text{atm}}\]
B.\[{{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}} = 6 \times {10^{ - 15}}{\text{atm}}\]
C.\[{{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}} = 7 \times {10^{ - 15}}{\text{atm}}\]
D.\[{{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}} = 9 \times {10^{ - 15}}{\text{atm}}\]

Answer 1

Hint: To answer this question, you should recall the concept of pressure equilibrium constant of a gas. Write the relation for the pressure constant to calculate the value of pressure of components. Use these values to answer this question.

Formula used: \[{{\text{K}}_{\text{p}}}{\text{ }}{\text{ = }}\dfrac{{{{\left( {{{\text{p}}_{\text{A}}}} \right)}^{\text{a}}}{{\left( {{{\text{p}}_{\text{B}}}} \right)}^{\text{b}}}}}{{{{\left( {{{\text{p}}_{\text{C}}}} \right)}^{\text{c}}}{{\left( {{{\text{p}}_{\text{D}}}} \right)}^{\text{d}}}}}\] where, ${{\text{K}}_{\text{p}}}$ is equilibrium constant, ${{\text{p}}_{\text{A}}}$ and ${{\text{p}}_{\text{B}}}$ are pressure of the reactants and a and b are the coefficients in the reaction, ${{\text{p}}_{\text{C}}}$ and ${{\text{p}}_{\text{D}}}$ are pressure of the products and c and d are its coefficients in the reaction.

Complete Step by step solution:
We know that the equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. We know that the reaction given is
\[{\mathbf{CaC}}{{\mathbf{l}}_{\mathbf{2}}}\left( {\mathbf{s}} \right) + {\mathbf{6}}{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\left( {\mathbf{g}} \right) \rightleftharpoons {\mathbf{CaC}}{{\mathbf{l}}_{\mathbf{2}}}.{\mathbf{6H2}}{\mathbf{O}}\left( {\mathbf{s}} \right).\;\]
The concentration of solids is taken to be one. The relation can be written as:
\[{{\text{K}}_{\text{p}}} = \dfrac{1}{{{{({{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}})}^6}}}\]
\[{{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}} = \dfrac{1}{{{{\text{K}}_{\text{p}}}^{\dfrac{1}{6}}}}\],
Solving and rearranging we get:
 \[{{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}} = \dfrac{1}{{{{({\mathbf{6}}.{\mathbf{4}} \times {\mathbf{1}}{{\mathbf{0}}^{\mathbf{8}}}^{\mathbf{5}})}^{\dfrac{1}{6}}}}}\].
The final value which we get is \[{{\text{p}}_{{{\text{H}}_{\text{2}}}{\text{O}}}} = 5 \times {10^{ - 15}}\]
Therefore, we can conclude that the correct answer to this question is option A.

Note: Drying agents can be defined as any hygroscopic substance that induces or sustains a state of dryness in its vicinity. strong bases such as soda-lime, sodium hydroxide, potassium hydroxide, and lithium hydroxide can remove carbon dioxide by chemically reacting with it. You should remember the mechanism of action of drying agents. Important drying agents include anhydrous sodium sulfate, magnesium sulfate, or calcium chloride. These drying agents when added to a solution of any compound pick up any extra water from the compound solution and become hydrated. The hydrated salt now clumps and further can be filtered out or left behind during decanting.