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Sulphide ion reacts with solid sulphur ...

Sulphide ion reacts with solid sulphur
`S_((aq))^(2-) +S_((s)) hArr S_(2(aq))^(2-), k_1= 10`
`S_((aq))^(2-)+2S_((s)) hArr S_(3(aq))^(2-) k_2=130`
The equilibrium constant for the formation of `S_3^(2-) (aq)` from `S_2^(2-)(aq)` and sulphur is

A

10

B

13

C

130

D

1300

Text Solution

AI Generated Solution

The correct Answer is:
To find the equilibrium constant for the formation of \( S_3^{2-} (aq) \) from \( S_2^{2-} (aq) \) and solid sulfur, we will use the given reactions and their equilibrium constants. ### Given Reactions: 1. \( S^{2-} (aq) + S_{(s)} \rightleftharpoons S_2^{2-} (aq) \) with \( K_1 = 10 \) 2. \( S^{2-} (aq) + 2S_{(s)} \rightleftharpoons S_3^{2-} (aq) \) with \( K_2 = 130 \) ### Step 1: Reverse the first reaction To find the equilibrium constant for the formation of \( S_3^{2-} \) from \( S_2^{2-} \) and sulfur, we first need to reverse the first reaction: \[ S_2^{2-} (aq) \rightleftharpoons S^{2-} (aq) + S_{(s)} \] When we reverse a reaction, the equilibrium constant becomes the reciprocal: \[ K' = \frac{1}{K_1} = \frac{1}{10} \] ### Step 2: Write the second reaction The second reaction remains as is: \[ S^{2-} (aq) + 2S_{(s)} \rightleftharpoons S_3^{2-} (aq) \] ### Step 3: Add the two reactions Now, we can add the reversed first reaction and the second reaction: 1. \( S_2^{2-} (aq) \rightleftharpoons S^{2-} (aq) + S_{(s)} \) (from Step 1) 2. \( S^{2-} (aq) + 2S_{(s)} \rightleftharpoons S_3^{2-} (aq) \) (from Step 2) Adding these reactions gives: \[ S_2^{2-} (aq) + S^{2-} (aq) + 2S_{(s)} \rightleftharpoons S_3^{2-} (aq) + S_{(s)} \] ### Step 4: Simplify the equation Notice that the \( S^{2-} (aq) \) and one \( S_{(s)} \) cancels out: \[ S_2^{2-} (aq) + S_{(s)} \rightleftharpoons S_3^{2-} (aq) \] ### Step 5: Calculate the equilibrium constant for the overall reaction The overall equilibrium constant \( K \) for the combined reaction is the product of the equilibrium constants of the individual reactions: \[ K = K' \times K_2 = \left(\frac{1}{10}\right) \times 130 = \frac{130}{10} = 13 \] ### Final Answer: The equilibrium constant for the formation of \( S_3^{2-} (aq) \) from \( S_2^{2-} (aq) \) and sulfur is **13**. ---
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