Nh4hs dissociates in a vessel at temp T K in the presence of 0.5 atm nh3. At equilibrium the pressure due to h2s is 1.5 atm.In a vessel V2 ( initially evacuated) the solid dissociates at T K.The total pressure at equilibrium in vessel V2 is P.and the weight ratio of nh3 and h2s is W. The value of P and W are? Options are 1.(2√3 ATM,2/3) 2.( √3 ATM,2/3) 3.(2√3 ATM,3/2)
Why did the molecule go to therapy?
Because it had dissociation issues!
Now, let's find the value of P and W in vessel V2.
For the dissociation of NH4HS, the balanced equation is:
NH4HS ⇌ NH3 + H2S
Given that the pressure due to NH3 at equilibrium is 0.5 atm and the pressure due to H2S is 1.5 atm, the total pressure at equilibrium in vessel V2 will be the sum of these pressures, which is 0.5 atm + 1.5 atm = 2 atm.
So, the value of P is 2 atm.
Now, let's find the weight ratio of NH3 and H2S. Since the balanced equation shows a 1:1 molar ratio between NH3 and H2S, it means that the weight ratio will also be 1:1.
Therefore, the value of W is 1.
Hence, the correct option is 1. (2√3 ATM, 2/3).
To find the value of P and W, we'll first write the balanced chemical equation for the dissociation of NH4HS in the presence of NH3:
NH4HS ⇌ NH3 + H2S
From the given information, we know that the pressure due to NH3 at equilibrium is 0.5 atm and the pressure due to H2S at equilibrium is 1.5 atm. Now, let's solve for P and W step-by-step.
Step 1: Calculate the initial moles of NH4HS in vessel V2.
Since vessel V2 is initially evacuated, there are no NH4HS moles present. Thus, the initial moles of NH4HS in vessel V2 is 0.
Step 2: Write the balanced chemical equation for the dissociation of NH4HS in vessel V2.
NH4HS ⇌ NH3 + H2S
Step 3: Express the equilibrium pressure of NH3 and H2S in vessel V2 in terms of P.
Since there are no initial moles of NH4HS in vessel V2, at equilibrium, all the NH3 and H2S is formed from the dissociation of NH4HS. Thus, the equilibrium pressure of NH3 and H2S in vessel V2 is directly proportional to the moles of NH3 and H2S formed. Therefore, the pressure due to NH3 in vessel V2 is 0.5P, and the pressure due to H2S in vessel V2 is 1.5P.
Step 4: Calculate the moles of NH3 and H2S formed in vessel V2.
Using the pressure-volume-temperature (PVT) relationship, we know that P1V1/T1 = P2V2/T2, where P1, V1, and T1 are the initial conditions, and P2, V2, and T2 are the conditions at equilibrium. Since vessel V2 is initially evacuated, the initial pressure is 0. Therefore, we can write the equation as:
0 * V2 / T = (0.5P + 1.5P) * V2 / T
Simplifying the equation, we get:
0 = 2P * V2 / T
Since V2 and T are both non-zero values, this equation can only be true if P = 0.
Therefore, the value of P is 0. Option 4 is not listed as an option, so none of the given options match the correct value for P.
As for the weight ratio W, it is not possible to determine without further information.
To find the value of P and W, we need to understand the dissociation equation of NH4HS and the equilibrium conditions.
The dissociation equation of NH4HS is as follows:
NH4HS ⇌ NH3 + H2S
At equilibrium, the pressure due to NH3 is given as 0.5 atm, and the pressure due to H2S is given as 1.5 atm. We also know that the vessel V2 is initially evacuated, which means there is no pressure from any species in the beginning.
Now, let's consider the equilibrium conditions in vessel V2. Since the solid NH4HS is dissociating, it will contribute to the total pressure at equilibrium. Let's assume the partial pressure of NH4HS in vessel V2 is P(NH4HS). Since the vessel was initially evacuated, the total pressure at equilibrium in vessel V2 will be P(NH4HS).
To find the value of P(NH4HS), we can use the principle of partial pressures. The sum of partial pressures of all species at equilibrium should be equal to the total pressure. Therefore,
P(NH4HS) + P(NH3) + P(H2S) = P
However, we know that P(NH3) is given as 0.5 atm, and P(H2S) is given as 1.5 atm. So, substituting these values into the equation, we get:
P(NH4HS) + 0.5 atm + 1.5 atm = P
Simplifying the equation, we have:
P(NH4HS) + 2 atm = P
Therefore, we can conclude that the value of P is given by 2 atm.
To find the weight ratio of NH3 and H2S, we need to consider the stoichiometry of the dissociation reaction. From the equation NH4HS ⇌ NH3 + H2S, we can see that the molar ratio of NH3 to H2S is 1:1.
Since the weight ratio is directly proportional to the molar ratio, we can conclude that the weight ratio of NH3 to H2S is also 1:1. Therefore, the value of W is 1/1, which is equal to 1.
Hence, the correct option is 1. (2√3 ATM, 2/3).