Answer to Question #258589 in General Chemistry for emman

Question

1. Hydrogen sulfide decomposes according to the following reaction, for which Kc = 0.93 at 700 °C:

2H2S(g) <---> 2H2(g) + S2(g)

If 0.75 mol of H2S is placed in a 3.0-L container, what are the equilibrium concentrations of all the species?

Answer 1

The balanced reaction equation is: 2H2S(g)⇋2H2(g)+S2(g)

The equilibrium constant expression is: Kc=9.30×10^-8

^=$\frac{2H2×S2}{H2S}$

Initially, we only have the reactant:

$[ H 2 S ] = 0.45 m o l ÷ 3.0 L = 0.15 M$

$[ H 2 S ] e q ≈ 0.15 M$

$H 2 ] e q = 2 x$

$[ S 2 ] e q = x$

$9.30 × 10 ^{− 8} = \frac{( 2 x )^ 2 ( x )}{ ( 0.15 M ) ^2}$

$[H 2] e q = 2 × 8.06 × 10^{ − 4}M$

$= 1.61 × 10 ^{− 3} M$