Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. 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.
### Given:
- The rate of change of iodide ion [tex]\([- \Delta [I^{-}] / \Delta t = 4.8 \times 10^{-4} \, \text{M/s}\] ### 1. Value of \(\Delta[I_3^-] / \Delta t\)[/tex]
To determine the value of [tex]\(\Delta[I_3^-] / \Delta t\)[/tex], we need to understand the stoichiometry of the balanced chemical equation. Assuming a 1:1 stoichiometric ratio between [tex]\(I^{-}\)[/tex] and [tex]\(I_3^{-}\)[/tex], the rate at which [tex]\(I_3^{-}\)[/tex] is produced will be equal to the rate at which [tex]\(I^{-}\)[/tex] is consumed, because the change in concentration of [tex]\(I_3^{-}\)[/tex] directly depends on the loss of [tex]\(I^{-}\)[/tex].
Therefore:
[tex]\[-\Delta [I^{-}] / \Delta t = \Delta [I_3^{-}] / \Delta t\][/tex]
Hence:
[tex]\[\Delta [I_3^{-}] / \Delta t = 4.8 \times 10^{-4} \, \text{M/s}\][/tex]
### 2. Average Rate of Consumption of [tex]\(H^{+}\)[/tex]
To find the average rate of consumption of [tex]\(H^{+}\)[/tex], we need to know the stoichiometric ratio between [tex]\(I^{-}\)[/tex] and [tex]\(H^{+}\)[/tex] in the balanced chemical reaction. Assuming a hypothetical balanced equation shows that for every mole of [tex]\(I^{-}\)[/tex] consumed, 3 moles of [tex]\(H^{+}\)[/tex] are also consumed (this is a typical stoichiometric ratio in such reactions):
Thus, the rate of consumption of [tex]\(H^{+}\)[/tex] will be three times the rate of consumption of [tex]\(I^{-}\)[/tex]:
[tex]\[\text{Rate of consumption of } H^{+} = 3 \left(\text{Rate of consumption of } I^{-}\right) \][/tex]
[tex]\[ \text{Rate of consumption of } H^{+} = 3 \times 4.8 \times 10^{-4} \, \text{M/s}\][/tex]
[tex]\[ \text{Rate of consumption of } H^{+} = 1.44 \times 10^{-3} \, \text{M/s} \][/tex]
Thus, the average rate of consumption of [tex]\(H^{+}\)[/tex] during that time interval is [tex]\(1.44 \times 10^{-3} \, \text{M/s}\)[/tex].
### Summary:
1. The value of [tex]\(\Delta[I_3^-] / \Delta t\)[/tex] is [tex]\(4.8 \times 10^{-4} \, \text{M/s}\)[/tex].
2. The average rate of consumption of [tex]\(H^{+}\)[/tex] is [tex]\(1.44 \times 10^{-3} \, \text{M/s}\)[/tex].
### Given:
- The rate of change of iodide ion [tex]\([- \Delta [I^{-}] / \Delta t = 4.8 \times 10^{-4} \, \text{M/s}\] ### 1. Value of \(\Delta[I_3^-] / \Delta t\)[/tex]
To determine the value of [tex]\(\Delta[I_3^-] / \Delta t\)[/tex], we need to understand the stoichiometry of the balanced chemical equation. Assuming a 1:1 stoichiometric ratio between [tex]\(I^{-}\)[/tex] and [tex]\(I_3^{-}\)[/tex], the rate at which [tex]\(I_3^{-}\)[/tex] is produced will be equal to the rate at which [tex]\(I^{-}\)[/tex] is consumed, because the change in concentration of [tex]\(I_3^{-}\)[/tex] directly depends on the loss of [tex]\(I^{-}\)[/tex].
Therefore:
[tex]\[-\Delta [I^{-}] / \Delta t = \Delta [I_3^{-}] / \Delta t\][/tex]
Hence:
[tex]\[\Delta [I_3^{-}] / \Delta t = 4.8 \times 10^{-4} \, \text{M/s}\][/tex]
### 2. Average Rate of Consumption of [tex]\(H^{+}\)[/tex]
To find the average rate of consumption of [tex]\(H^{+}\)[/tex], we need to know the stoichiometric ratio between [tex]\(I^{-}\)[/tex] and [tex]\(H^{+}\)[/tex] in the balanced chemical reaction. Assuming a hypothetical balanced equation shows that for every mole of [tex]\(I^{-}\)[/tex] consumed, 3 moles of [tex]\(H^{+}\)[/tex] are also consumed (this is a typical stoichiometric ratio in such reactions):
Thus, the rate of consumption of [tex]\(H^{+}\)[/tex] will be three times the rate of consumption of [tex]\(I^{-}\)[/tex]:
[tex]\[\text{Rate of consumption of } H^{+} = 3 \left(\text{Rate of consumption of } I^{-}\right) \][/tex]
[tex]\[ \text{Rate of consumption of } H^{+} = 3 \times 4.8 \times 10^{-4} \, \text{M/s}\][/tex]
[tex]\[ \text{Rate of consumption of } H^{+} = 1.44 \times 10^{-3} \, \text{M/s} \][/tex]
Thus, the average rate of consumption of [tex]\(H^{+}\)[/tex] during that time interval is [tex]\(1.44 \times 10^{-3} \, \text{M/s}\)[/tex].
### Summary:
1. The value of [tex]\(\Delta[I_3^-] / \Delta t\)[/tex] is [tex]\(4.8 \times 10^{-4} \, \text{M/s}\)[/tex].
2. The average rate of consumption of [tex]\(H^{+}\)[/tex] is [tex]\(1.44 \times 10^{-3} \, \text{M/s}\)[/tex].
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
