Initial rate data at 25 °C are listed in the table for the reaction NH4 +1aq2 + NO2 -1aq2S N21g2 + 2 H2O1l2 (b) What is the value of the rate constant?

Video transcript

Hello. In this problem we are told the initial rate data for the reaction between ammonium ions and hydroxide ions to form ammonia and water at 25 degrees Celsius are shown in the table below. We are asked to calculate the rate constant of the reaction. Let's begin by writing the general form of the reaction rate law. So the rate is equal to the reaction rate constant times the concentration of ammonium eyes to some order X times the concentration of hydroxide ions to some order Y. So we'll use the experiments in the table to determine the orders X and Y. Recall that to determine the orders. We have to hold one of our reactant concentrations constant while the other is changing. So we can compare experiments one and two where the amount of money mine is changing but the amount of hydroxide in is being held constant. We can then compare experiment two and 3 where the amount of money mine is being held constant in the amount of hydroxide iron is changing. So let's begin by comparing experiment one and experiment to. So we have the rate of experiment one over that of experiment too. Then is equal to our reaction rate constant times the concentration of Ammonia mines. For experiment one which is .500 to the X. Power times the concentration of hydroxide ions which is 0.400 to the white power. And then for the second experiment we have our reaction rate constant times the concentration of the minimize to the X power times the concentration of our hydroxide ions to the white Power. This then is equal to the rate for experiment one which is 5.40 times 10 to the minus five Over that for experiment two which is 2.70 times 10 to the -5. So our case counsel as do our concentrations for the hydroxide island. So then we get 0.500 divided by 0.250 to the X. is equal to two. And on the left hand side this simplifies two to the X. Power is equal to two. That means then the X. Is equal to one. So it's first order with regard to the ammonia mines. Now we're going to compare experiments two and 3 in a similar way. So we're gonna put rate three over rate too. And so we get then rate for experiment three is our reaction rate constant times the concentration of pneumonia mines which is .250 to the X. Which is now, you know, is to be one times the concentration of our hydroxide ions which is 0.600 to the Y power. And then for right to we have a reaction rate constant times the concentration of ammonia mines which again is 0.250 to the first order divided by then the concentration of hydroxide ions which is 0.400 to order Y. And then taking the rates for both of these experiments. The rate for experiment three is 4.5 times 10 negative five. And that four, experiment two is 2.70 times 10 to the negative five. So our reaction rate constants cancel and our concentration of ammonia mine cancels. So we're left with 0.600 over 0.400 to the white power is equal to then 3/2ves on the left hand side. This also works out to 3/2 to the Y power is equal to three halves. That implies then that Y is equal to one. So the reaction is also first order with regard to hydroxide. So our reaction rate law then our rate is equal to our reaction rate constant times the concentration of ammonia mines. To the first order times the concentration of hydroxide ions. Also to the first order. And now we are going to find our reaction rate constant K. So taking our rate law expression, we are going to rearrange it so that we solve for a reaction rate constant. It's okay then will be equal to the rate divided by the concentration of ammonia mines And the concentration of hydroxide ions. We can pick any one of the three experiments will just pick the first one. The rate for the first experiment was 5.40 times 10 to the -5 Concentration of Ammonia Mines was 0.500 Molar. Let me get my units on my rate which is concentration per second. And the concentration of the hydroxide was .400 moller. So our reaction rate constant works out to 2.70 times 10 to the -4. Looking at our units, we see that one of the units of polarity cancels and we're left with units of malaria, T and seconds in the denominator. So we can also rewrite our reaction rate constant as 2.70 times 10 to the - per concentration per second. So the reaction rate constant for our reaction between ammonium ions and hydroxide ions is equal to 2.70 times 10 to minus four per morality per second. Thanks for watching. Hope. This helped.