This reaction is first order in N₂O₅: N₂O₅(g) → NO₃(g) + NO₂(g) The rate constant for the reaction at a certain temperature is 0.053/s. b. What would the rate of the reaction be at the concentration indicated in part a if the reaction were second order? Zero order? (Assume the same numerical value for the rate constant with the appropriate units.)

Video transcript

Welcome back everyone to another video. The dissociation of the nitrogen pide is first order we're given the reaction, the nitrogen pide gas dissociates into nitrogen trioxide gas and nitrogen dioxide gas at a certain temperature. The rate constant for the reaction is 7.91 multiplied by its sensitive power of negative second per second. Determine the rate of the reaction at the concentration of 0.890 molar. If the reaction were second order zeroth order assume the same numerical value for the rate constant with the appropriate units. So let's begin solving this problem. First of all, we understand that the reaction is the first or the reaction, which also corresponds to the units, right? Because the value of K is given in the units of molar to the power of one minus the reaction order O second to the negative first. OK. So if O is one, then we don't have any molar present, we just have second is the power of negative first. What we're going to do is calculate the rate starting with the zeroth order. And we essentially can say that if we have the zeroth order reaction, the rate is equal to K multiplied by the concentration of N 25 to the power of zero, which is just the weight constant K. And in this case, that'd be 7.91 multiplied by 10 to the power of negative seconds molar per second, right? We are changing our units. The order is zero, one minus zero, gives us one. So we have molar per second. Now, what about the first order? Well, essentially in this case, we have the weight constant K multiplied by N 205 to the power of first because now we have the first order reaction. So we have K multiplied by N 205, we substitute the rate constant 7.901 multiplied by 10, the power of negative second second to the power or second to the power of negative first. And we multiply by the concentration of N 205 which is 0.8 and 90 molar. And in this case, we end up with the rate of 7.0 multiplied by sensitive power of negative second molar per second. And finally, the rate assuming that this is the second or the reaction would be equal to K multiplied by N 205 to the power of second. So in this case, we're taking our rate constant 7.91 multiplied by its sensitive power of negative second. What would be the units? One minus two is negative one? So we have molar to the negative 1st 2nd to the negative first multiplied by the concentration of 0.890 molar squared. And in this case, we get 6.3 multiplied by 10, the power of negative seconds molar per second. So now we have our answers. Let's label them. Now, if the reaction was the zeroth order, we would have a rate of 7.91 multiplied by 10. The power of negative second molar per second. If the reaction is the first order, we get 7.0 multiplied by 10, the power of negative second molar per second. And for the second order reaction, we have a rate of 6.3 multiplied by its senses of the power of negative second molar per second. That would be it for today. And thank you for watching.