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The concentration of dinitrofluorobenzene (DNFB) in ethanol at 315 K varies with time as below_ Verify that the reaction is first order and calculate the rate cons...

Question

The concentration of dinitrofluorobenzene (DNFB) in ethanol at 315 K varies with time as below_ Verify that the reaction is first order and calculate the rate constant How long would it take for the concentration to decrease to < AM?Time61.87991659 2643 350249756694[DNFB] LM 134.8 115.3 95.13 76.14 63.23 45.03 30.04

The concentration of dinitrofluorobenzene (DNFB) in ethanol at 315 K varies with time as below_ Verify that the reaction is first order and calculate the rate constant How long would it take for the concentration to decrease to < AM? Time 61.8 799 1659 2643 3502 4975 6694 [DNFB] LM 134.8 115.3 95.13 76.14 63.23 45.03 30.04



Answers

The decomposition of acetaldehyde, $\mathrm{CH}_{3} \mathrm{CHO},$ was described in Problem $13.54,$ and the order of the reaction and the rate constant for the reaction at $530^{\circ} \mathrm{C}$ were determined in Problem 13.74. If the initial concentration of acetaldehyde is $0.300 \mathrm{M}$, what will the concentration be (a) after 30 minutes, (b) after 180 minutes?

Okay, so we have Ah, substance. Just name. It s a we assume from one motor off a undercutting a position reaction. And ah, the composition reaction is fine. To be second order. We respect you a lot of substance and article of the way constant away constant. It's equals the super for, um, motor to the left of one and also means to level one. And what would be the time to which syrup on zero thio motor where Stuff from silver in one four. First, second order reaction The integrated well or will be, um, here one offered in your circles. Intuition plus Katie. All right, so we just want to find the time that we still once in row two. So he will be the, um, the time. Ah, at the end off the we asked you. So we just plug in seal for a serial, too, on the that will push you. Then you shall we received from one. And then kay will be Sierra 10.4, uh, motor, uh, over a minute, and then we will have our time. So the best thing we just did too soft for this equation, and then we should be able to find the time we were 100 minutes. I just can't mind that because there were calls the same in this. So with the final

So for a fast sort of reaction. The number of moles of reacting president after Time T that has the initial amount and zero and the rate constant Landers expressed as follows. So we have Ellen and zero over and t equals Lambda T. So therefore, if we were to just plot the Alan and however anti component against T, we got a straight line with slope that is Lambda. So the consistency of Lambda proves the reaction is fast. Order on the mean value of Lambda gives the rate constant. So the mean lambda is not point, not North 369 That is the great constant for our first order reaction.

This question is quite challenging with multiple parts. The first thing that we need to do is take the concentration and converted into a pressure. Well, do this. Using the ideal gas law, It states that we have .02 moles per liter. We need to convert it into a pressure because later we are given units of pressure, 14 atmospheres of total pressure after the reaction has proceeded for a certain amount of time. So the initial pressure of just the reactant Will be .020 moles multiplied by are multiplied by the Kelvin temperature. 3 27 plus 2 73 Divided by the volume, which was one leader. And we get 985 atmospheres of pressure. So as the reaction proceeds, we're going to decrease the pressure of the reactant but increase the pressure of the products. Because both products are also gases, they will contribute to the total pressure. As the reaction proceeds, the pressure of the reactant will decrease by X, and the two products will increase by X. Because all coefficients are one. So at Equilibrium, the pressure of the reactant will be its initial pressure of .985 -1. And the other product pressures will both be X. They tell us that after a certain time has passed, the sum of all these pressures will be 1.4. This then allows us to solve for X. X. being .415 atmospheres. Because the reaction is zero order, we will use the zero order integrated rate law. But we need to because the rate constant has units of concentration convert this pressure back into a concentration but it's not actually this pressure that we want to convert back into a concentration. We want to convert this pressure, the pressure of our reactant back into a concentration. So .985 -115 is .57. So we have .570 atmospheres divided by our and T converted into kelvin, gives us 1.158 times 10 to the negative too, as our concentration. So the zero order integrated rate law says the concentration at time T minus the concentration. Now at I'm sorry, the concentration at time zero minus the concentration at time T, which we just determined will be equal to K, Which is provided at four times 10 to the -5. Multiplied by T. We then solve for T and we get 211 seconds.

To appropriately answer this question. You need to spend some time plotting or preparing three graphs. You'll prepare graph of concentration as a function of time, one over concentration as a function of time, and natural log of concentration as a function of time. So you'll take all of the data points that are provided in this question. Go to Excel and prepare your three graphs. You'll need to convert the concentration into natural log of concentration and one over concentration. Then you'll prepare a plot of concentration is a function of time. One over concentration is a function of time. Natural log of concentration as a function of time, and you'll determine the order based upon which one of these gives you the best straight line. The best straight line can be determined by asking Excel to show the r squared value in addition to the equation, because the natural log of concentration as a function of time gives us the best straight line. This is indicative of a first order reaction with a first order graph. As we've shown here, natural log of concentration is a function of time. K is equal to the negative of the negative slope because the slope is negative, we take the negative of that to give us a positive K. Value K is always positive and in this case it is 0.461 Because its first order, it will be one over time units for part B. It asked what is the half life of this reaction? For first order reaction, we can use the first order half life equation where T one half is equal to the natural log of two divided by K and K. We just determined earlier. So we get 15 hours then for part C, it wants to know how long it will take for 90% of the reactant to be converted into product. Well, if we are starting with a one molar concentration that if 90% has reacted 10% is left and 10% of one Mueller is 10.1 Mueller. So we're really asking, how long does it take for the one molar concentration to become 10.1 Mueller. To answer this question will use the first order integrated rate law where the natural log of the concentration at time. T 0.1 Mueller divided by the concentration at time 01 Mueller equals negative K negative 0.416 multiplied by T. So now we have a simple algebraic expression for which we can solve for time and time is 49.9 hours. This is how long it will take for 90% of the reactant to convert into product.


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