One mole of solid NH4Cl (s) isvheated in a closed box to decompose it partially according to the following reaction:
NH4Cl (s) <=> NH3 (g) +HCl (g)
A) calculate Kp of the above system at the temperature of the experiment if the partial pressure of ammonia at equilibrium is 1.6atm.
B) what happens to Kp and the partial pressure of HCl (g) if:
-some NH3 gas is pumped into the box at same T.
-More NH4Cl is added without changing T
-the volume of the box is reduced 50% at same T.
To calculate Kp for the given reaction, you need to use the expression for Kp:
Kp = (P(NH3) * P(HCl)) / P(NH4Cl)
Where P(NH3) represents the partial pressure of ammonia, P(HCl) represents the partial pressure of hydrogen chloride, and P(NH4Cl) represents the partial pressure of ammonium chloride.
A) To calculate Kp, you are given the partial pressure of ammonia at equilibrium, which is 1.6 atm. However, you don't have the partial pressure of hydrogen chloride or ammonium chloride. To solve for Kp, you need to find these missing values or determine their relationship to the known value of P(NH3).
B) Let's consider the effects of different changes on Kp and the partial pressure of HCl (g):
1. If some NH3 gas is pumped into the box at the same temperature (T):
- The increase in NH3 concentration will shift the equilibrium to the left side, favoring the formation of NH4Cl (s).
- As a result, the partial pressure of HCl (g) will decrease since there are fewer products available to form HCl (g).
- Kp, on the other hand, will remain the same as it is a constant at a given temperature.
2. If more NH4Cl is added without changing the temperature (T):
- The increase in NH4Cl concentration will shift the equilibrium to the right side, favoring the formation of NH3 (g) and HCl (g).
- The partial pressure of HCl (g) will increase as more NH4Cl decomposes and produces HCl (g).
- Kp may change depending on the increase in the partial pressures of NH3 and HCl relative to the partial pressure of NH4Cl. Without specific values, it's difficult to determine the exact effect on Kp.
3. If the volume of the box is reduced by 50% at the same temperature (T):
- The decrease in volume will increase the pressure inside the box.
- According to Le Chatelier's principle, the equilibrium will shift in the direction that reduces the total gas moles.
- Since the forward reaction produces two moles of gas (NH3 and HCl), the equilibrium will shift to the right, favoring the formation of NH3 (g) and HCl (g).
- The partial pressure of HCl (g) will increase due to the shift in equilibrium.
- Kp may change depending on the changes in partial pressures, but without specific values, it's difficult to determine the exact effect on Kp.
In summary, changes in the concentration or pressure of reactants/products and changes in volume can affect the equilibrium position and the partial pressures of the gases involved. However, the effect on Kp will depend on the specific changes and the relationship between the partial pressures of each component in the equilibrium expression.