When the particles of a gas decrease in energy what happens?

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

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