Question

For the reaction shown here

8H2S(g) ---> <--- 8H2(g) + S8(g)

the equlibrium concentrations were found to be

[H2S] = 0.250 M
[H2] = 0.400 M
[S8] = 0.750 M

What is the equilibrium constant for this reaction?

Kc = ???

Answer 1

Answer:

Kc = 32.3

Explanation:

To calculate the Kc, it's the ratio of products over reactants (each raised to their respective stoichiometric coefficient). In this case:

$K_c = \frac{{[H_2]}^8[S_8]}{{[H_2S]}^8}\\K_c = \frac{0.4^8 \times 0.75}{0.25^8}\\K_c = \frac{4.91 \times 10^{-4}}{1.52 \times 10^{-5}}\\K_c = 32.3$

Answer 2

The equilibrium constant for the reaction is 32.2

The equilibrium constant (K꜀) for a given reaction is simply defined as the ratio of the concentration of the products raised to their coefficients to the concentration of the reactants raised to their coefficients

With the above information, we can obtain the equilibrium constant for the reaction given above as follow:

8H₂S(g) <=> 8H₂(g) + S₈(g)

Concentration of Hydrogen sulphide [H₂S] = 0.250 M

Concentration of Hydrogen [H₂] = 0.400 M

Concentration of sulphur [S₈] = 0.750 M

Equilibrium constant (K꜀) =?

$K_{C} = \frac{[H_{2}]^8[S_{8}]}{[H_{2}S]^{8}} \\\\K_{C} = \frac{0.4^{8} * 0.750}{0.25^{8}}$

K꜀ = 32.2

Therefore, the equilibrium constant for the reaction is 32.2

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