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

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = $K_c=5.10\times 10^{-6}$

Moles of ammonium chloride = 0.573 mol

Concentration of ammonium chloride = $\frac{0.573 mol}{1.00 L}=0.573 M$

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

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:

$K_c=\frac{[HCl][NH_3]}{[NH_4Cl]}$

Putting values in above equation, we get:

$5.10\times 10^{-6}=\frac{x\times x}{(0.573-x)}$

$2.9223\times 10^{-6}-5.10x\times 10^{-6}=x^2$

$x^2-2.9223\times 10^{-6}+(5.10\times 10^{-6})x=0$

On solving this quadratic equation we get:

x = 0.01707 M

The equilibrium concentration of HCl is 0.01707 M.