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 $2.26\times 10^{-3}M$

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 $NH_4Cl=\frac{0.564}{1}=0.564M$

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:

$K_c=5.10\times 10^{-6}$

Putting values in above equation, we get:

$5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M$

Negative sign is neglected because concentration cannot be negative.

So, $[HCl]=2.26\times 10^{-3}M$

Hence, the equilibrium concentration of HCl is $2.26\times 10^{-3}M$