Question

A sample of solid ammonium chloride was placed in an evacuated chamber, and then heated causing it to decompose according to the following reaction: NH4Cl(s) NH3(g) + HCl(g) In a particular experiment the pressure of NH3(g) in the container was found to be 2.2 atm. Calculate the value of Kp for the decomposition of NH4Cl(s) at this temperature. Kp = SubmitShow Hints

Answer 1

The equilibrium constant (Kp) for the decomposition of ammonium chloride into ammonia and hydrogen chloride, given the pressure of NH3 as 2.2 atm, is calculated to be 4.84 atm^2.

Step-by-step explanation:

To calculate the equilibrium constant (Kp) for the decomposition of ammonium chloride (NH4Cl) into ammonia (NH3) and hydrogen chloride (HCl), we can use the fact that at equilibrium, the pressure of NH3 is the same as the pressure of HCl because they are produced in a 1:1 ratio according to the balanced equation:

NH4Cl(s) → NH3(g) + HCl(g)

Since the pressure of NH3 is given as 2.2 atm, we can infer that the pressure of HCl is also 2.2 atm at equilibrium. Kp is calculated using the partial pressures of the gaseous products:

Kp = (PNH3)(PHCl)

So the Kp for the decomposition at this temperature would be:

Kp = (2.2 atm) × (2.2 atm) = 4.84 atm2

Answer 2

Kp = 4.84

Step-by-step explanation:

Considering the ICE table for the equilibrium as:

NH₄Cl (s) ⇔ NH₃ (g) + HCl (g)

t = o

t = eq x x

--------------------------------------------- --------------------------

Moles at eq: x x

Given the pressure of NH₃ at equilibrium = 2.2 atm

x = 2.2 atm

The expression for the equilibrium constant is:

`K_p={P_(NH_3)P_([HCl)}`

So,

`K_p={x^2}`

x = 2.2 atm

Thus,

`K_p={2.2}^2`

Thus, Kp = 4.84