How do you solve this:

Consider the decomposition of ammonium chloride at a cerain temperature

NH4Cl(s) <- NH3(g) + HCl(g)
->

Calculate the equilibrium constant Kp if the total pressure is 2.2 atm at that temperature.

I think I would do the following:

NH4Cl(s) ==> NH3(g) + HCl(g)

Ptotal = PNH3 + PHCl
2.2 atm = PNH3 + PHCl
From the equation, PNH3 = PHCl; therefore,
2.2 = 2PNH3 and you can go from there.
Then substitute partial pressures into
Kp = PNH3*PHCl to find Kp.

thank you. so what did you get as a final answer for this problem then? and what did u do after doing 2.2=2PHN3

and after subsituting the partial pressure??

Thanks

The remainder of the problem is up to you. We try to HELP you do the work but we don't do it. You can substitute and calculate Kp; it's just multiplying two numbers together.

1.2

25.00 atm kp

To calculate the equilibrium constant, Kp, for the given reaction, you need to know the partial pressures of each gas at equilibrium. The equilibrium constant for a reaction is a ratio of the concentrations (or in this case, partial pressures) of the products and reactants at equilibrium.

In this case, the reaction is:

NH4Cl(s) ⇌ NH3(g) + HCl(g)

You are given that the total pressure at equilibrium is 2.2 atm. However, you still need to determine the partial pressure of NH3 and HCl at equilibrium.

To solve this, you need additional information such as the initial amount of NH4Cl and other equilibrium constants. If you have the initial amount or concentration of NH4Cl and the equilibrium constant expression (Kc or Kp) for this reaction, you can proceed with the calculations.

If you do not have that information, it is not possible to determine the equilibrium constant (Kp) just from the given total pressure. More information would be needed, such as the initial amounts or concentrations of the reactants or the equilibrium constant expression.

In summary, without additional information, it is not possible to calculate the equilibrium constant (Kp) for the given reaction with just the total pressure.