Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide.
NH4I(s) Picture NH3(g) + HI(g)
At 400°C, Kp = 0.215. Calculate the partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C.
To answer this question, we need to use the expression for the equilibrium constant (Kp) and the given value to calculate the partial pressure of ammonia at equilibrium.
The equation for the dissociation of ammonium iodide is:
NH4I(s) ⇌ NH3(g) + HI(g)
The equilibrium constant expression for this reaction in terms of partial pressures is:
Kp = (P(NH3) * P(HI))/(P(NH4I))
where P(NH3), P(HI), and P(NH4I) are the partial pressures of ammonia, hydrogen iodide, and ammonium iodide, respectively.
Given that Kp = 0.215 and the solid ammonium iodide is in excess, we can assume that the concentration of NH4I(s) is sufficiently high and will not affect the equilibrium concentration of ammonia.
Since NH4I is a solid, its partial pressure is considered constant and does not contribute to changes in the equilibrium partial pressures of ammonia and hydrogen iodide. Therefore, we can rewrite the equilibrium constant expression as follows:
Kp = P(NH3) * P(HI)
We are given the value of Kp as 0.215. We need to find the partial pressure of ammonia at equilibrium.
Let's assume the partial pressure of ammonia at equilibrium is P(NH3). Since the partial pressure of hydrogen iodide is not given, we will denote it as P(HI).
Now, we can rewrite the expression for Kp:
0.215 = P(NH3) * P(HI)
To calculate the partial pressure of ammonia at equilibrium, we rearrange the equation:
P(NH3) = 0.215 / P(HI)
We can conclude that the partial pressure of ammonia at equilibrium is 0.215 divided by the partial pressure of hydrogen iodide (P(HI)). However, since we do not have the partial pressure of hydrogen iodide, we cannot determine the exact value of P(NH3). Additional information about the system or a balanced equilibrium constant expression would be required to calculate P(NH3) accurately.
What's with the picture stuff? Just put an arrow there.
NH4I(s) ==> NH3(g) + HI(g)
.............p..........p
Kp = pNH3*pHI = 0.215 (what? atm?)
Substitute into Kp and solve for p.