Question

Consider the reaction for the decomposition of hydrogen disulfide:

2H₂S(g)⇌2H₂(g)+S₂(g),  
Kc = 1.67×10^-7 at 800∘C

The reaction is carried out at the same temperature with the following initial concentrations:

[H₂S]=3.20×10^-4M

[H₂]=0.00M

[S₂]=0.00M

Find the equilibrium concentration of S₂.

Express the concentration to three significant figures and include the appropriate units.

Answer 1

Equilibrium reaction is

2 H₂S (g)  $\rightarrow$ H₂ (g) + S₂ (g)

Now ICE table for the reaction is

	[H2S]	[H2]	[S2]
initial	3.20*10-4	0.00	0.00
change	-2x	+2x	+x
equilibrium	(3.20*10-4 - 2x)	2x	x

Now, Equilibrium constant (Kc ) = $\frac{[H_{2}]^{2}[S]}{[H_{2}S]^{2}}$

now, putting the value of concentration of H₂S, H₂ and S₂ at equilibrium

Kc =

(22.)2 x (3.20 × 10-4-22)2

Solving Cubic equation with this small data is very tough.

As, Equilibrium constant is very less, the reaction will not proceed very far towards right then x will be also very small

Therefore, (3.20*10^-4 - 2x) $\approx$ 3.20*10^-4

now,

$\frac{(2x)^{2}\times x}{(3.20\times 10^{-4})^{2}}$ = Kc

or , $\frac{(2x)^{2}\times x}{(3.20\times 10^{-4})^{2}}$ = 1.67*10^-7

or, 4x³ =  1.67*10^-7 * (3.20*10^-4)²

or, 4x³ = 17.10*10^-15

or, x³ = (17.10/4) * 10^-15

or, x³ = 4.27*10^-15

or, x = 1.62*10^-5 M

therefore

at equilibrium, [H₂] = 2*x = 2*1.62*10^-5 = 3.24*10^-5 M

[S₂] = x = 1.62*10^-5 M

[H₂S] = 3.20*10^-4 - (2*1.62*10^-5 ) M = 2.88*10^-4 M

now, K_c = $\frac{[H_{2}]^{2}[S]}{[H_{2}S]^{2}}$ =  $\frac{(3.24*10^{-5})^{2}*1.62*10^{-5}}{(2.88*10^{-4})^{2}}$

= 2 *10^-7  

The calculated Kc value is nearly close to the given value of Kc , hence answer was valid.