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Ch.14 - Chemical Kinetics

Chapter 14, Problem 33c

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl- + I- → OI- + Cl- . This rapid reaction gives the following rate data:

[OCl4-] (M) [I-] (M) Initial Rate (M,s)

1.5 * 10-3 1.5 * 10-3

1.36 * 10-4 3.0 * 10-3 1.5 * 10-3 2.72 * 10-4

1.5 * 10-3 3.0 * 10-3 2.72 * 10-4 (c) Calculate the rate when [OCl-] = 2.0 * 10-3 M and [I-] = 5.0 * 10 - 4 M.

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Hello, everyone. Today we have the following problem. The reaction of AC with B is as follows, given the following data showing the rate of the reaction. What is the rate when the concentration of AC is equal to 0.0037 and the concentration of B is 0.0047 moller. So first, you want to recognize what our rate law would be and that would be that the rate is equal to some constant K times the concentration of A C. And we're going to raise that with an exponent of X times the concentration of B and we're gonna raise that exponent to Y. And so first, you want to calculate the order with respect to A C and this is how we do. So we have to look at the data where the concentration of A sea changes while the concentration of B remains the same. And so that's going to be between the first and the third of columns. So we're going to take their rates, 0.00094, divide that by 0.00, And then equal that to the same ratio for the concentrations. So we have 0.008 divided by 0.004. And if we simplify this, we will get two is equal to two raised to the power of X X is going to have to equal one. And so we know this is going to be first order, then we're gonna look at the concentration with respect to be, we're gonna follow a similar process. We're gonna find where be changes, but A C remains the same. And so that's going to be between our 2nd and 3rd column. We're gonna have our 0.94 Divided by 0.00047. And we're gonna equal that to our 0.008 divided by 0.004. So we're gonna raise that to the power of why? Of course, we're going to get two is equal to two raised to the power of Y Y is going to equal one. And as before, this is going to representative of a first order reaction. So our rate law from before is going to be archaic, constant terms of concentration of A C raised to the power of one times the concentration of B raised to the power of one, which is going to simplify to some constant K times the concentration of A C times the concentration of B. And so we're gonna use our first row to calculate for K. So we're going to rearrange this equation to say that K is equal to the rate divided by the concentration of A C times the concentration of B business for the first row. So our rate for the first row is 0.00094. And our concentration of AC is 0.008 In our concentration to be 0.004. And so when we simplify this, We were going to get 0.38, Moeller raised to the power of negative one times second, raise the power of negative one. And so we have this rate now for our first experiment. So then we have to calculate when A C and B are equal to the following concentrations given to us in the question stem. So we're gonna do five up here. It's our last step. It's going to be the rate is equal to our constant K which was 29.38 Moeller race to the negative one seconds race and negative one times our concentration of what A C would be, which is 0.37 times are 0.47 for B. And this is overall going to give us a rate of 0. Moeller per second as our final answer overall, I hope this helped. And until next time.
Related Practice
Textbook Question

The react ion between ethyl bromide 1C2H5Br2 and hydroxide ion in ethyl alcohol at 330 K, C2H5Br1alc2 + OH- 1alc2¡ C2H5OH1l2 + Br - 1alc2, is first order each in ethyl bromide and hydroxide ion. When 3C2H5Br4 is 0.0477 M and 3OH- 4 is 0.100 M, the rate of disappearance of ethyl bromide is 1.7 * 10-7 M>s. (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

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Textbook Question

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl- + I- → OI- + Cl- . This rapid reaction gives the following rate data:

[OCl4-] (M) [I-] (M) Initial Rate (M,s)

1.5 * 10-3 1.5 * 10-3

1.36 * 10-4 3.0 * 10-3 1.5 * 10-3 2.72 * 10-4

1.5 * 10-3 3.0 * 10-3 2.72 * 10-4

(a) Write the rate law for this reaction.

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Textbook Question

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl - + I - ¡OI - + Cl - . This rapid reaction gives the following rate data:

[OCl4-] (M) [I-] (M) Initial Rate (M,s)

1.5 * 10-3 1.5 * 10-3

1.36 * 10-4 3.0 * 10-3 1.5 * 10-3 2.72 * 10-4

1.5 * 10-3 3.0 * 10-3 2.72 * 10-4

(b) Calculate the rate constant with proper units.

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Textbook Question

The following data were measured for the reaction 

(b) What is the overall order of the reaction?

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Textbook Question

The following data were measured for the reaction 

(c) Calculate the rate constant with proper units?

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Textbook Question

Consider the gas-phase reaction between nitric oxide and bromine at 273 C: 2 NO1g2 + Br21g2¡2 NOBr1g2. The following data for the initial rate of appearance of NOBr were obtained: Experiment 3no4 1M 2 3br2 4 1M 2 Initial Rate 1M,s2 1 0.10 0.20 24 2 0.25 0.20 150 3 0.10 0.50 60 4 0.35 0.50 735 (b) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets.

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