Question

The equilibrium constant, K, for the following reaction is 5.10X10 at 548 K. NH_CH(s) 2 NH3(E) + HC1(2) Calculate the equilibrium concentration of HCl when 0.573 moles of NH CI(s) are introduced into a 1.00 L vessel at 548 K. (HCI) =

Answer 1

The equilibrium concentration of HCl is 0.01707 M.

Step-by-step explanation:

Equilibrium constant of the reaction =
`K_c=5.10* 10^(-6)`

Moles of ammonium chloride = 0.573 mol

Concentration of ammonium chloride =
`(0.573 mol)/(1.00 L)=0.573 M`

`NH_4HCl(s)\rightleftharpoons 2 NH_3(g) + HCl(g)`

Initial: 0.573 0 0

At eq'm: (0.573-x) x x

We are given:

`[NH_4Cl]_(eq)=(0.573-x)`

`[HCl]_(eq)=x`

`[NH_3]_(eq)=x`

Calculating for 'x'. we get:

The expression of
`K_(c)` for above reaction follows:

`K_c=([HCl][NH_3])/([NH_4Cl])`

Putting values in above equation, we get:

`5.10* 10^(-6)=(x* x)/((0.573-x))`

`2.9223* 10^(-6)-5.10x* 10^(-6)=x^2`

`x^2-2.9223* 10^(-6)+(5.10* 10^(-6))x=0`

On solving this quadratic equation we get:

x = 0.01707 M

The equilibrium concentration of HCl is 0.01707 M.