At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Certainly! Let's solve this problem step by step.
We start with the dissociation reaction:
[tex]\[X \leftrightarrow Y + 2Z.\][/tex]
Let the initial molar solubility of [tex]\(X\)[/tex] be [tex]\(s\)[/tex].
At equilibrium:
- The concentration of [tex]\(X\)[/tex] is [tex]\([X] = s\)[/tex].
- The concentration of [tex]\(Y\)[/tex] is [tex]\([Y] = s\)[/tex].
- The concentration of [tex]\(Z\)[/tex] is [tex]\([Z] = 2s\)[/tex].
The equilibrium constant [tex]\(K_s\)[/tex] is expressed as:
[tex]\[K_s = [Y][Z]^2.\][/tex]
Substituting the equilibrium concentrations, we get:
[tex]\[K_s = (s) \cdot (2s)^2.\][/tex]
Simplifying, we have:
[tex]\[K_s = s \cdot 4s^2 = 4s^3.\][/tex]
Now, let's consider the condition where the molar solubility of [tex]\(X\)[/tex] is multiplied by 2. Therefore, the new molar solubility is:
[tex]\[s_{\text{new}} = 2s.\][/tex]
We now need to find the new equilibrium constant [tex]\(K_s\)[/tex] when the solubility is [tex]\(s_{\text{new}}\)[/tex].
At this new solubility:
- [tex]\([X] = s_{\text{new}} = 2s\)[/tex].
- [tex]\([Y] = s_{\text{new}} = 2s\)[/tex].
- [tex]\([Z] = 2s_{\text{new}} = 2(2s) = 4s\)[/tex].
Substituting these new values into the expression for [tex]\(K_s\)[/tex], we get:
[tex]\[K_s' = [Y][Z]^2 = (2s) \cdot (4s)^2.\][/tex]
Simplifying this, we have:
[tex]\[K_s' = 2s \cdot 16s^2 = 32s^3.\][/tex]
Thus, the new equilibrium constant [tex]\(K_s\)[/tex] is:
[tex]\[K_s' = 32s^3.\][/tex]
We need to determine the factor by which [tex]\(K_s\)[/tex] is multiplied. Initially, we had:
[tex]\[K_s = 4s^3.\][/tex]
Now, with the increased solubility:
[tex]\[K_s' = 32s^3.\][/tex]
The multiplication factor is:
[tex]\[\frac{K_s'}{K_s} = \frac{32s^3}{4s^3} = 8.\][/tex]
Therefore, the value of [tex]\(K_s\)[/tex] is multiplied by [tex]\(8\)[/tex].
The correct answer is [tex]\(8\)[/tex].
We start with the dissociation reaction:
[tex]\[X \leftrightarrow Y + 2Z.\][/tex]
Let the initial molar solubility of [tex]\(X\)[/tex] be [tex]\(s\)[/tex].
At equilibrium:
- The concentration of [tex]\(X\)[/tex] is [tex]\([X] = s\)[/tex].
- The concentration of [tex]\(Y\)[/tex] is [tex]\([Y] = s\)[/tex].
- The concentration of [tex]\(Z\)[/tex] is [tex]\([Z] = 2s\)[/tex].
The equilibrium constant [tex]\(K_s\)[/tex] is expressed as:
[tex]\[K_s = [Y][Z]^2.\][/tex]
Substituting the equilibrium concentrations, we get:
[tex]\[K_s = (s) \cdot (2s)^2.\][/tex]
Simplifying, we have:
[tex]\[K_s = s \cdot 4s^2 = 4s^3.\][/tex]
Now, let's consider the condition where the molar solubility of [tex]\(X\)[/tex] is multiplied by 2. Therefore, the new molar solubility is:
[tex]\[s_{\text{new}} = 2s.\][/tex]
We now need to find the new equilibrium constant [tex]\(K_s\)[/tex] when the solubility is [tex]\(s_{\text{new}}\)[/tex].
At this new solubility:
- [tex]\([X] = s_{\text{new}} = 2s\)[/tex].
- [tex]\([Y] = s_{\text{new}} = 2s\)[/tex].
- [tex]\([Z] = 2s_{\text{new}} = 2(2s) = 4s\)[/tex].
Substituting these new values into the expression for [tex]\(K_s\)[/tex], we get:
[tex]\[K_s' = [Y][Z]^2 = (2s) \cdot (4s)^2.\][/tex]
Simplifying this, we have:
[tex]\[K_s' = 2s \cdot 16s^2 = 32s^3.\][/tex]
Thus, the new equilibrium constant [tex]\(K_s\)[/tex] is:
[tex]\[K_s' = 32s^3.\][/tex]
We need to determine the factor by which [tex]\(K_s\)[/tex] is multiplied. Initially, we had:
[tex]\[K_s = 4s^3.\][/tex]
Now, with the increased solubility:
[tex]\[K_s' = 32s^3.\][/tex]
The multiplication factor is:
[tex]\[\frac{K_s'}{K_s} = \frac{32s^3}{4s^3} = 8.\][/tex]
Therefore, the value of [tex]\(K_s\)[/tex] is multiplied by [tex]\(8\)[/tex].
The correct answer is [tex]\(8\)[/tex].
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
