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Mashutka [201]
3 years ago
12

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

rium constant Kfinal for the following reaction? S2−(aq)+2H+(aq)⇌H2S(aq) Enter your answer numerically.
Chemistry
1 answer:
Dvinal [7]3 years ago
5 0

<u>Answer:</u> The value of K_c for the final reaction is 7.16\times 10^{25}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1

<u>Equation 2:</u>  HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2

The net equation follows:

S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=\frac{1}{K_1}\times \frac{1}{K_2}

We are given:

K_1=9.57\times 10^{-8}

K_2=1.46\times 10^{-19}

Putting values in above equation, we get:

K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}

Hence, the value of K_c for the final reaction is 7.16\times 10^{25}

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