The first-order rate constant for the decomposition of
N2O5, 2 N2...

Question

The first-order rate constant for the decomposition of
N2O5, 2 N2O51g2¡4 NO21g2 + O21g2, a t 70 C i s
6.82 * 10-3 s-1. Suppose we start with 0.0250 mol of
N2O51g2 in a volume of 2.0 L. (a) How many moles of
N2O5 will remain after 5.0 min?

Accepted Answer

Hey everyone today, we're being asked to find out how much of the reactant and molds will remain after six hours for the decomposition of chloral methane to produce methane and hydrogen chloride. Now there's a few things that we need to take note of. The first being that the reaction is stated to be first order. It is stated to be first order. So this means we're going to need to utilize the first order integrated rate law and that can be expressed as such as the natural log of the final concentration of the reactant is equal to negative K. The rate constant times time plus the natural log of the initial reactant concentration. And in the question were given a few of these values were given that K is equal to 1.6 times 10 to the negative 6th seconds. and eight of 1 were given that the time he's six hours Which can be converted to seconds. Since the rest of the quantities are also in seconds. So we have 60 men per one hour Times 60 seconds per one man, Which equals 21,600 seconds. These will cancel out. And finally we have the initial reacting concentration which is zero point 047457 moles Per 1.25 L which gives us 0.0366 Molar. So with all of this here, we can go ahead and start solving by substituting in the values that we have is equal to negative 1.6 times 10 to the negative sixth seconds. The native one Into 21, seconds Plus The Natural Log of 0.0366 Molar. So this gives us a value of negative 3.3426, solving for the concentration. Then it's E. To the power Of -3. - six. Just as the final value of 0. Molar. However, we're not done now, we have to calculate the number of moles of the reactant remaining. This is just the polarity. So we take 0.0353 molar, which is just simply molds for leaders. So let's rewrite that this is mold per liter, Multiplied by the volume, which is 1.25 L. So our leaders will cancel out and we will be left 0.0441 0.441 moles. Therefore, the amount of reactant in moles that will remain after six hours for the decomposition of chlorate will be 0.0441 moles. I hope this helps. And I look forward to seeing you all in the next one

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