Answer
Verified
91.5k+ views
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.
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
A soldier with a machine gun falling from an airplane class 11 physics JEE_MAIN
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main