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The rate constant for this second-order reaction 0.550 M-!at 300 'C.productsHow long; seconds, would it take for the concentration of to decrense from 0.980 M ...

Question

The rate constant for this second-order reaction 0.550 M-!at 300 'C.productsHow long; seconds, would it take for the concentration of to decrense from 0.980 M to 0.350 M?

The rate constant for this second-order reaction 0.550 M-! at 300 'C. products How long; seconds, would it take for the concentration of to decrense from 0.980 M to 0.350 M?



Answers

For the reaction $A \longrightarrow$ products, the following data give $[\mathrm{A}]$ as a function of time: $t=0 \mathrm{s},[\mathrm{A}]=0.600 \mathrm{M}$ $100 \mathrm{s}, 0.497 \mathrm{M} ; 200 \mathrm{s}, 0.413 \mathrm{M} ; 300 \mathrm{s}, 0.344 \mathrm{M} ; 400 \mathrm{s}$ $0.285 \mathrm{M} ; 600 \mathrm{s}, 0.198 \mathrm{M} ; 1000 \mathrm{s}, 0.094 \mathrm{M}.$ (a) Show that the reaction is first order. (b) What is the value of the rate constant, $k ?$ (c) What is $[\mathrm{A}]$ at $t=750 \mathrm{s} ?$

This question will take a little bit of work because the best way to justify the order of a reaction in this case First order is when you just have time and concentration data is to plot the concentration as a function of time, the natural log of the concentration as a function of time and one over the concentration as a function of time. Whichever one of these graphs gives you a straight line that will tell you the order of the reaction. Yes, concentration is a function of time gives you a straight line. It's zero order natural log of concentration as a function of time gives you a straight line. Its first order and one over concentration as a function of time gives you a straight line, then its second order. So it suggests this reaction is first order, so to verify that it's first order, we need to go into excel and plot the natural log of the concentration as a function of time. So you'll need to put in all the time values you'll need to put in all the concentrations as provided in the question, and then calculate the natural log of the concentration at all the time values and then plot the natural log of the concentration as a function of time to validate that you have a straight line in excel. You can ask it to show the equation. The equation will be the best fit, straight line and the better fit you have. The closer the R squared value will be to one. So because we plotted the natural log of the concentration as a function of time and we got in r squared value, that is essentially one, this validates that this reaction follows first order kinetics, which is what the question wanted you to do when it asks, Show that the reaction is first order. Now that we have plotted the data as a function of the natural log of the concentration as a function of time and obtained are straight line and its equation, then the slope allows us to identify the K value or the rate constant for the reaction if its first order will always get a negative slope. So Kay is going to be equal to the negative of the negative slope because K, the rate constant is always positive. So in this case, the rate constant is 0.200185 positive because the negative of negative 0.185 is positive 0.185 Now that we know the rate constant and we have validated that it is first order, we can use first order kinetics in order to calculate the concentration of a at time T Time T in this case is going to be 750 seconds. So it should have a concentration somewhere between 0.1 point 198 and 0.94 because the time is somewhere between here. So we'll set up our first order. Integrated rate like equation. Natural log of concentration at time t over concentration at time. Zero equals negative. K, which was just determined, is 0.185 one over seconds, multiplied by seconds. The seconds is the 7 15. If you remember, your log rules natural log of a value over another value is the natural log of the numerator minus the natural log of the denominator. And then that will be equal to the product of this right here. Well, then, add natural log of 0.6 to both sides and we get natural. Log of X is equal to negative 1.898 Well, then do e to the both sides in order to get out of the natural log in into X and so e to the negative. 1.898 gives us our X value of 0.150 which is definitely between these two values here, as we predicted.

Hello. So today we're going to be looking at this reaction right here. So this is a second order reaction as we conceive from our rate constant that we've been given. And we know that we're starting off with 0.62 mil Arat, ease off nitrogen dioxide and we want to know what what time will have 0.28 mil? Araji. Well, remember what's unique about a second order reaction? Well, if you plot the inverse of its reacted concentration in this case, nitrogen dioxide over time you'll get a straight line on the slope of this line is the reactant. Our case. Nitrogen dioxide at some time is equal to the rate constant times, time plus one over the initial concentration. And you'll see well, we know what we're starting with. We know what we end up with, and we know the rate constant. All right, What we want is the time. So why don't we just plug in? So let's plug in. You have 1/0 0.28 is equal to 0.54 times time, plus one over 0.62 And so let's do some math. So one divided by 10.62 is 1.6 and we take one divided, right, 2.28 and then some. Track that, then divide by 0.54 and we will see that our answer is three 0.6 seconds. So in 3.6 seconds, we go from 0.62 Miller T 20.28 mil. Araji, There we have it.

For this question, we have a generic reaction of a plus B goes to C plus D. They tell us that it is second order in a and zero order in B that then give us the rate constant to be 0.103 one over Moller minutes. It then asks, What is the rate of this reaction when we have a given a concentration and a given be concentration? To answer this question, we're going to use the differential rate law or just the rate law itself. The rate is equal to K multiplied by a raised to its order which is second to multiplied by be raised to its order, which is zero because it's zero order. Anything raised to the zero is one. So it essentially cancels from our rate law. So right then will be equal to K, which is provided at 0.0 mhm one throw three and then multiplied by the concentration which is provided at 0.116 Moller and we're going to square that this will give us a rate of 1.39 times 10 to the negative four Mueller per minute


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