A sample of only solid ammonium chloride was heated in a 1 L container at 500 C. At equilibrium, the pressure of NH3 (g) was found to be 1.75 atm. what is the equilibrium constant, Kc, for the decomposition at this temperature?
R= .08206 L atm/K Mol
NH4CL (s) <---> NH3(g) + HCL (g)
Kc = (NH3)(HCl)
Use PV = nRT to calculate n for NH3 from the pressure in the problem. Then (NH3) = moles/L. (HCl) is the same. Substitute and solve for Kc.
To find the equilibrium constant, Kc, for the decomposition of solid ammonium chloride (NH4Cl) at 500°C, we need to use the expression for Kc and the given information.
The equilibrium constant expression for the reaction is written as:
Kc = [NH3] * [HCl]
We can use the ideal gas law to find the concentrations of NH3 and HCl in the expression.
According to the ideal gas law, PV = nRT, where:
P is the pressure,
V is the volume,
n is the number of moles,
R is the gas constant (0.08206 L atm/K mol), and
T is the temperature in Kelvin.
In this case, we are given the pressure of NH3 (g) as 1.75 atm. However, we need to convert it to moles per liter (M) before we can use it in the equilibrium expression.
To convert pressure to concentration, we can use the equation:
P = n/V
Since we have a 1 L container, the volume (V) is 1 L. Therefore, the concentration, [NH3], is:
[NH3] = P / (R * T)
Let's calculate [NH3]:
[NH3] = 1.75 atm / (0.08206 L atm/K mol * (500 + 273) K)
[NH3] = 0.0193 M
Similarly, we can find the concentration of HCl using the same approach. However, since we don't have the pressure for HCl, we can assume it is negligible compared to NH3. This assumption is valid because HCl is a strong acid, and at equilibrium, most of it will dissociate.
Therefore, we can consider [HCl] ≈ 0.
Now, we can plug the concentrations into the equilibrium expression:
Kc = [NH3] * [HCl]
Kc = 0.0193 M * 0
Kc = 0
Thus, the equilibrium constant, Kc, for the decomposition of ammonium chloride at 500°C is 0.