To tackle this equilibrium problem, we need to analyze the reaction and the conditions provided. The equilibrium reaction is represented as follows:
Understanding the Reaction
The equilibrium can be expressed as:
- LiCl·3NH3 (s) ⇌ LiCl·NH3 (s) + 2NH3
Here, the solid forms of lithium chloride and ammonia are involved in a dynamic equilibrium. The equilibrium constant, Kp, is given as 9 atm² at 40ºC.
Initial Conditions
We have a 5-liter vessel containing 0.1 mole of LiCl·NH3. To find out how many moles of NH3 need to be added to shift the equilibrium towards the left (favoring the formation of LiCl·3NH3), we can start by calculating the initial concentration of LiCl·NH3.
Calculating Initial Concentration
The concentration of LiCl·NH3 in the vessel can be calculated using the formula:
Concentration (C) = Moles / Volume
Substituting the values:
C = 0.1 moles / 5 L = 0.02 M
Setting Up the Equilibrium Expression
The equilibrium constant expression for this reaction in terms of partial pressures is:
Kp = (P(NH3)^2) / (P(LiCl·NH3))
Since LiCl·NH3 is a solid, its activity is considered to be 1, simplifying our expression to:
Kp = P(NH3)^2
Finding the Required Pressure of NH3
Given that Kp = 9 atm², we can set up the equation:
9 = P(NH3)^2
Taking the square root of both sides gives us:
P(NH3) = 3 atm
Relating Pressure to Moles
To find out how many moles of NH3 correspond to a pressure of 3 atm in a 5-liter vessel, we can use the ideal gas law:
P = (nRT) / V
Rearranging gives us:
n = (PV) / RT
Assuming R = 0.0821 L·atm/(K·mol) and T = 313 K (40ºC), we can substitute:
n = (3 atm * 5 L) / (0.0821 L·atm/(K·mol) * 313 K)
Calculating this gives:
n ≈ 0.6 moles of NH3
Calculating Additional Moles Needed
Since we need to add enough NH3 to reach this pressure, we must consider how much NH3 is already present in the equilibrium. Initially, we have no NH3, as it is not mentioned in the initial conditions. Therefore, to achieve the desired equilibrium, we need to add:
0.6 moles of NH3
Final Summary
To shift the equilibrium towards the left and favor the formation of LiCl·3NH3, you should add approximately 0.6 moles of NH3 to the 5-liter vessel at 40ºC. This will help achieve the necessary conditions for the backward reaction to proceed to completion.
