Iodide ion is oxidized by acidified dichromate ions as shown in this equation
Cr2O7^2- +9I- +14H+ ---> 2Cr3+ +3I3- + 7H2O
These data were obtained when the reaction was studied at a constant pH.
Experiment [Cr2O7^2-]M [I-]M Rate, M x s^-1
1 0.0040 0.010 0.00050
2 0.0080 0.010 0.0010
3 0.0120 0.020 0.0060
What is the order of the reaction with respect to Cr2O7^2- and I-?
Cr2O7^2- +9I- +14H+ ---> 2Cr3+ +3I3- + 7H2O
These data were obtained when the reaction was studied at a constant pH.
Experiment [Cr2O7^2-]M [I-]M Rate, M x s^-1
1 0.0040 0.010 0.00050
2 0.0080 0.010 0.0010
3 0.0120 0.020 0.0060
What is the order of the reaction with respect to Cr2O7^2- and I-?
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Hi Que!
To find the reaction order, you simply select two experiments in the chart and try to cancel out one of the orders of the reactant so that you can solve for the other. For instance, you should pick two experiments where one of the reactants has a constant concentration and the other one has its concentration changed. This will allow us to solve for the reaction order of that reactant and then we can use that value to solve for the other reactant’s reaction order.
Divide all the values of experiment 2 by the values of experiment 1 because as you can see, the concentration of Iodide remains constants (0.010) and so then we can solve for the reaction order of the dichromate.
(0.0080/0.0040)^x * (0.010/0.010)^y = (0.0010/0.00050). Where x represents the reaction order of dichromate and y represents the reaction order of iodide. We don’t need to know the value of y first because it is 1 in this equation. Solve for x and you should get 1. Therefore, the reaction order of dichromate is 1.
Now that you know the reaction order of dichromate, use it to find the reaction order of iodide. Divide the values of experiment 3 by the values of experiment 1 and then solve for y.
(0.0120/0.0040)^1 * (0.020/0.010)^y = (0.0060/0.00050).
You should get 2 and so the reaction order of iodide is 2. Just as additional information, the total reaction order would then be 3 because the sum of all the reactant’s reaction orders equals 3.
I hope this helped and good luck with chemistry! :)
To find the reaction order, you simply select two experiments in the chart and try to cancel out one of the orders of the reactant so that you can solve for the other. For instance, you should pick two experiments where one of the reactants has a constant concentration and the other one has its concentration changed. This will allow us to solve for the reaction order of that reactant and then we can use that value to solve for the other reactant’s reaction order.
Divide all the values of experiment 2 by the values of experiment 1 because as you can see, the concentration of Iodide remains constants (0.010) and so then we can solve for the reaction order of the dichromate.
(0.0080/0.0040)^x * (0.010/0.010)^y = (0.0010/0.00050). Where x represents the reaction order of dichromate and y represents the reaction order of iodide. We don’t need to know the value of y first because it is 1 in this equation. Solve for x and you should get 1. Therefore, the reaction order of dichromate is 1.
Now that you know the reaction order of dichromate, use it to find the reaction order of iodide. Divide the values of experiment 3 by the values of experiment 1 and then solve for y.
(0.0120/0.0040)^1 * (0.020/0.010)^y = (0.0060/0.00050).
You should get 2 and so the reaction order of iodide is 2. Just as additional information, the total reaction order would then be 3 because the sum of all the reactant’s reaction orders equals 3.
I hope this helped and good luck with chemistry! :)