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kotegsom [21]
3 years ago
7

☆BRAINLIEST to correct answer☆​

Chemistry
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

Explanation:

K2+H2O--------->KOH

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2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 0.47 moles of H2S are placed in a 3.0 L container and the system is allowed t
Anon25 [30]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 1.648\times 10^{-3}M

<u>Explanation:</u>

We are given:

Initial moles of hydrogen sulfide gas = 0.47 moles

Volume of the container = 3.0 L

The molarity of solution is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}

So, \text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M

The given chemical equation follows:

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

<u>Initial:</u>                  0.1567

<u>At eqllm:</u>           0.1567-2x       2x         x

The expression of K_c for above equation follows:

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

We are given:

K_c=9.3\times 10^{-8}

Putting values in above equation, we get:

9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}

So, equilibrium concentration of hydrogen gas = 2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M

Hence, the concentration of hydrogen gas at equilibrium is 1.648\times 10^{-3}M

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