Answer
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Hint :When a solid dissociates into two gases, we can assume that the pressure of both the gases is the same. So, do that and then try and find the total pressure using equilibrium constant using the pressure of gases.
Complete answer:
We are given that two solids dissociate simultaneously to form gases. We will try and find the total pressure of the mixture of gases.
- In the dissociation reaction of A, we can see that two gases B and C are formed. Both of these gases will have the same pressure because they are produced in the same amounts. So, assume that gases B and C have pressure ${P_1}$ .
- Same way gases C and E are formed from D. Thus, we can say that the pressure of C and E will be the same. We assume that the pressure of C and E is ${P_2}$.
Now, we can say that total pressure of C = ${P_1} + {P_2}$
- Now, for reaction (1), we can write its equilibrium constant in terms of pressure as
\[{K_{{P_1}}} = \dfrac{{[{\text{Product]}}}}{{[{\text{Reactant]}}}}\]
But we know that the reactant is solid and it does not have any pressure. So, we can write that
\[{K_{{P_1}}} = [{P_B}][{P_C}]\]
But we already assumed that ${P_B} = {P_1}$ and ${P_C} = {P_1} + {P_2}$
So, we can write that
\[{K_{{P_1}}} = ({P_1})({P_1} + {P_2})\]
But we are given that ${K_{{P_1}}} = x$ So,
\[x = ({P_1})({P_1} + {P_2}){\text{ }}......{\text{(1)}}\]
For reaction (2), we can also write that
\[{K_{{P_2}}} = y = {P_c} \cdot {P_E}\]
So, we obtain
\[y = ({P_1} + {P_2})({P_2}){\text{ }}...{\text{(2)}}\]
Adding equations (1) and (2), we get
\[x + y = {({P_1} + {P_2})^2}\]
Now, total pressure can be given as
\[{P_T} = {P_C} + {P_B} + {P_E}\]
Putting the available values into above equation, we get
\[{P_T} = ({P_1} + {P_2}) + {P_1} + {P_2} = 2({P_1} + {P_2})\]
But we obtained that \[x + y = {({P_1} + {P_2})^2}\] . So, we can put its value in above equation and we will get
\[{P_T} = 2\sqrt {x + y} \]
Thus, the correct answer of this question is (B).
Note:
Remember that in the expression of equilibrium constant of the reaction, we can either use molar concentration or partial pressures. We cannot use both at the same time. The equilibrium constant is independent of the inert gases present in the mixture.
Complete answer:
We are given that two solids dissociate simultaneously to form gases. We will try and find the total pressure of the mixture of gases.
- In the dissociation reaction of A, we can see that two gases B and C are formed. Both of these gases will have the same pressure because they are produced in the same amounts. So, assume that gases B and C have pressure ${P_1}$ .
- Same way gases C and E are formed from D. Thus, we can say that the pressure of C and E will be the same. We assume that the pressure of C and E is ${P_2}$.
Now, we can say that total pressure of C = ${P_1} + {P_2}$
- Now, for reaction (1), we can write its equilibrium constant in terms of pressure as
\[{K_{{P_1}}} = \dfrac{{[{\text{Product]}}}}{{[{\text{Reactant]}}}}\]
But we know that the reactant is solid and it does not have any pressure. So, we can write that
\[{K_{{P_1}}} = [{P_B}][{P_C}]\]
But we already assumed that ${P_B} = {P_1}$ and ${P_C} = {P_1} + {P_2}$
So, we can write that
\[{K_{{P_1}}} = ({P_1})({P_1} + {P_2})\]
But we are given that ${K_{{P_1}}} = x$ So,
\[x = ({P_1})({P_1} + {P_2}){\text{ }}......{\text{(1)}}\]
For reaction (2), we can also write that
\[{K_{{P_2}}} = y = {P_c} \cdot {P_E}\]
So, we obtain
\[y = ({P_1} + {P_2})({P_2}){\text{ }}...{\text{(2)}}\]
Adding equations (1) and (2), we get
\[x + y = {({P_1} + {P_2})^2}\]
Now, total pressure can be given as
\[{P_T} = {P_C} + {P_B} + {P_E}\]
Putting the available values into above equation, we get
\[{P_T} = ({P_1} + {P_2}) + {P_1} + {P_2} = 2({P_1} + {P_2})\]
But we obtained that \[x + y = {({P_1} + {P_2})^2}\] . So, we can put its value in above equation and we will get
\[{P_T} = 2\sqrt {x + y} \]
Thus, the correct answer of this question is (B).
Note:
Remember that in the expression of equilibrium constant of the reaction, we can either use molar concentration or partial pressures. We cannot use both at the same time. The equilibrium constant is independent of the inert gases present in the mixture.
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