Question

Hydrogen sulfide decomposes according to the following reaction, for whichK_c = 9.30 × 10⁻⁸ at 700°C: 2 H₂S(g) ⇌ 2 H₂(g) + S₂(g)

If 0.27 mol of H₂S is placed in a 3.0−L container, what is the equilibrium concentration of H₂(g) at 700°C?

____M

Answer 1

The equilibrium reaction is

2H₂S (g) 2H₂ ( g) + S₂ (g)

Inital concentration of H₂S(g) = 0.27/3 = 0.09 M

ICE table for the reaction

	[H2S]	[ H2]	[S2]
Initial	0.09	0	0
Change	-2x	+2x	+x
Equilibrium	(0.09-2x)	2x	x

Now, equilibrium constant , Kc = (1)

Putting the values in above equation

Kc = (2x)²×( x) / ( 0.09-2x)²

Given, value of Kc is very small, hence the reaction at equilibrium does not procced to far in the right side , therefore x is very small , therefore we can approximate (0.09-2x) (0.09)

Therefore, Kc = 4x³/(0.09)²

Or, 4x³ = 9.30*10^-8×(0.09)²

Or, 4x³ = 0.075×10^-8

or, x³ = 1.88×10^-10

Or, x = 5.736×10^-4 M

Now,

[ S₂] = 5.736 ×10^-4 M

[ H₂S] =2×5.736 ×10^-3 M = 11.47×10^-4 M

and, [ H₂S] = (0.09 -2×5.736×10^-4) = 0.0888 M

Now, putting above calculated values in Eq.1

Calculated Kc = (11.47×10^-4)²×(5.736×10^-4)/(0.0888)²

= 9.56×10^-8

Again

x*100÷[ intial concentration of H₂S ]

=( 5.736×10^-4×100)÷(0.09)

= 0.6376

So, as calculated Kc is nearly equal to given value of Kc and degree of dissociation(x) is less than 5% of initial concentration of H₂S, then the approximation was valid.

Therefore, concentration of H₂(g) at equilibrium (700⁰c)

= 11.47×10^-4 M

or, 1.147×10^-3 M.