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in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O?
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?in the presence of excess of anhydrous srcl2 the amount of water take...
The equilibrium vapor pressure of water in a closed vessel that contains SrCl2.2H2O can be determined using the given equilibrium constant KP=10^12 atm^-4 and the balanced equation for the reaction: SrCl2.2H2O(s) ⇌ SrCl2.6H2O(s) + 4H2O(g)
1. Writing the expression for KP:
KP = (P(H2O))^4 / (P(SrCl2.2H2O) * P(SrCl2.6H2O))
2. Determining the molar concentration of water:
In a closed system, the number of moles of water is constant. Let's assume the initial moles of SrCl2.2H2O is n, and the moles of water vaporized is x. Therefore, the final moles of SrCl2.2H2O is n-x, and the final moles of H2O is 2x.
3. Expressing the equilibrium concentrations in terms of x:
[P(H2O)] = 2x / V
[P(SrCl2.2H2O)] = (n - x) / V
[P(SrCl2.6H2O)] = 1 / V, since it is a solid and its concentration remains constant
4. Substituting the equilibrium concentrations into the KP expression:
KP = (2x / V)^4 / ((n - x) / V) * (1 / V)
KP = (16x^4) / (n - x)
5. Solving for x:
Given that KP = 10^12 atm^-4, we can substitute this value into the equation:
10^12 = (16x^4) / (n - x)
Rearranging the equation, we have:
16x^4 = 10^12(n - x)
x^4 = (10^12/16)(n - x)
x^4 = 6.25 x 10^10(n - x)
Assuming x is small compared to n, we can approximate (n - x) as n:
x^4 = 6.25 x 10^10n
Solving for x, we find:
x = (6.25 x 10^10n)^(1/4)
6. Calculating the equilibrium vapor pressure of water:
The equilibrium vapor pressure of water is given by the ideal gas law:
P(H2O) = nRT / V
Substituting the value of x into the equation, we have:
P(H2O) = ((6.25 x 10^10n)^(1/4)) * RT / V
This equation gives the equilibrium vapor pressure of water in terms of the initial moles of SrCl2.2H2O (n), the gas constant (R), the temperature (T), and the volume (V) of the closed vessel.
By substituting the appropriate values for R, T, and V, you can calculate the equilibrium vapor pressure of water in torr.
1. Writing the expression for KP:
KP = (P(H2O))^4 / (P(SrCl2.2H2O) * P(SrCl2.6H2O))
2. Determining the molar concentration of water:
In a closed system, the number of moles of water is constant. Let's assume the initial moles of SrCl2.2H2O is n, and the moles of water vaporized is x. Therefore, the final moles of SrCl2.2H2O is n-x, and the final moles of H2O is 2x.
3. Expressing the equilibrium concentrations in terms of x:
[P(H2O)] = 2x / V
[P(SrCl2.2H2O)] = (n - x) / V
[P(SrCl2.6H2O)] = 1 / V, since it is a solid and its concentration remains constant
4. Substituting the equilibrium concentrations into the KP expression:
KP = (2x / V)^4 / ((n - x) / V) * (1 / V)
KP = (16x^4) / (n - x)
5. Solving for x:
Given that KP = 10^12 atm^-4, we can substitute this value into the equation:
10^12 = (16x^4) / (n - x)
Rearranging the equation, we have:
16x^4 = 10^12(n - x)
x^4 = (10^12/16)(n - x)
x^4 = 6.25 x 10^10(n - x)
Assuming x is small compared to n, we can approximate (n - x) as n:
x^4 = 6.25 x 10^10n
Solving for x, we find:
x = (6.25 x 10^10n)^(1/4)
6. Calculating the equilibrium vapor pressure of water:
The equilibrium vapor pressure of water is given by the ideal gas law:
P(H2O) = nRT / V
Substituting the value of x into the equation, we have:
P(H2O) = ((6.25 x 10^10n)^(1/4)) * RT / V
This equation gives the equilibrium vapor pressure of water in terms of the initial moles of SrCl2.2H2O (n), the gas constant (R), the temperature (T), and the volume (V) of the closed vessel.
By substituting the appropriate values for R, T, and V, you can calculate the equilibrium vapor pressure of water in torr.
Community Answer
?in the presence of excess of anhydrous srcl2 the amount of water take...
KP =( 1/p4H2O)
PH20 =( 1/Kp)-1/4
PH20 =( 1/Kp)-1/4
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?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O?
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?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O?.
?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O?.
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?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O?, a detailed solution for ?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O? has been provided alongside types of ?in the presence of excess of anhydrous srcl2 the amount of water taken up is goverened by KP=10^12 atm^-4 for the following reaction at 273k SrCl2. 2H2O(s) 4H2O(g) =SrCl2.6H2O(s) What is equilibrium vapour pressure (in torr) of water in closed vessel that contains SrCl2. 2H2O? theory, EduRev gives you an
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