Question

The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibrium concentration of HCl when 0.564 moles of NH4Cl(s) are introduced into a 1.00 L vessel at 548 K.[HCl] = _____ M

Answer 1

Answer: The equilibrium concentration of HCl is
`2.26* 10^(-3)M`

Step-by-step explanation:

We are given:

Moles of
`NH_4Cl(s)` = 0.564 moles

Volume of vessel = 1.00 L

Molarity is calculated by using the equation:

`\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}`

Molarity of
`NH_4Cl=(0.564)/(1)=0.564M`

The given chemical equation follows:

`NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)`

Initial: 0.564

At eqllm: 0.564-x x x

The expression of
`K_c` for above equation follows:

`K_c=[NH_3][HCl]`

The concentration of pure solid and pure liquid is taken as 1.

We are given:

`K_c=5.10* 10^(-6)`

Putting values in above equation, we get:

`5.10* 10^(-6)=x* x\\\\x=2.26* 10^(-3)M,-2.26* 10^(-3)M`

Negative sign is neglected because concentration cannot be negative.

So,
`[HCl]=2.26* 10^(-3)M`

Hence, the equilibrium concentration of HCl is
`2.26* 10^(-3)M`