CdnraueMrUnder some conditions, the rate Iaw for the reaction NO(g) = O3lg)-> NO2lg) 02(g)Fale0.06 M-2 $ 1 "INoj?o3] When the concentraton of NO 1.30 x 10-3...

Use the data in Table 13.2 to calculate the rate of the reaction at the time when $\left[\mathrm{F}_{2}\right]=0.010 \mathrm{M}$ and $\left[\mathrm{ClO}_{2}\right]$ $=0.020 M$.

We can calculate the rate of the reaction at different concentration by first calculating the rate constant. You know that rate is equal to the rate constant multiplied by the concentration raised to the order. The order, we're told, is one. So rearranging this we get K. The rate constant is equal to rate divided by the concentration, or 1.3 times. Send the negative four, divided by 40.4 gives us 0.3 to 51 over seconds for the rate constant. So going back to this equation here, the new rate is going to be equal to the rate constant multiplied by the new concentration 0.25 Moeller, which is 8.1 times 10 to the negative five Moeller per second.

According to question. Mm It was, yeah. Mhm. Oh my God. Yeah The rate constant K one for a certain reaction is equal to 1.4 into 10 days. To the power -5. Mhm Permal per minute. At temperature the one Equals to 483 Calvin. And activism energy E. Is equal to 2.11 in two. 10 days to the power three jule per mole. We have to find rate constant for the reaction at temperature 611 Kelvin. Let yeah. Okay. To be there great constant. Yeah. Mhm. Mhm. At temperature. Okay, excuse me. He noted by T two Which is equal to 611. Kill me from our genius equation. What what you know? Yeah. Ellen K two by Cuban is equal to activism and as the upon art into the tu minus Stephen divided by The 1- two. Mm. Okay, thank you. For K two is alone, Kevin is equal to 1.4 in two, 10 days. To the power -5 formal for a minute. Here is equal to 2.11 and took 10 days. three power 3 Joel Permal R. is the gas constant, which is which is equal to 8.31 jewel caramel pour carving The two is equal to 611 curving Stephen is equal to 483 Kelvin. Yeah. Mhm. You can. Mhm. They get Ellen K two to fight by 1.4 into changes to the par minus fight Moeller universe. Minute universe is equal to. We get this value is equal to zero 1101. We know that Ellen has a base E. So the videos look at me logarithmic property here, K two, divide by 1.4 into 10 days. To the power -5. Model universe minute inwards is equal to eat. To the power 0.1101. Mhm. Just during cross multiplication, Wicked 1.4 in to 10 days to the power -5 Modern universe minute inverse into The value of a raise to the power 0.1101 Is equal to 1.12. So we get the value of great constant. K two is equal to 1.6 to 10 days. To the power minus five Moeller universe minute inverse. This mhm Mhm required mhm. Great constant. You just forget it. K two Is equal to 1.6 into 10 days to the ball -5 Mueller universe, minute in first. This is our final answer for this problem.

In this question, we have a reaction of CH three cl Reacting with three moles of cl two producing CCL four and three moles of hcl. We are given for data for four experiments where concentration of the reactant are provided and the initial rate is provided and were asked to determine the order of the reaction with respect to each reactant so that we can write the rate law for the reaction and then calculate the K value. To do this, we need to determine, we need to determine the order with respect to each reactant We noticed in experiments one and two, the concentration of cl two is unchanging or staying constant, But the concentration of CH three cl is being doubled. If we double the concentration of CH three cl, we see that the rate doubles from 30.142 point 0 to 9. It's not an exact doubling, but it's probably because of rounding issues. So, a doubling of the concentration, resulting in a doubling in the rate is indicative of first order. So the reaction is first order with respect to CH three cl But then if we look at experiments two and three Where the CH three cl concentration is staying constant And we are doubling the cl two concentration, We're going from .052.1. We have something that is not double nor quadruple the rate we go from .0292.041. If it were first order doubling, the concentration would double the rate And the rate would be more like .058. If it were second order, then doubling the concentration would quadruple the rate that didn't happen. If it was zero order, then doubling the concentration would result in the rate staying constant. So it's not one of these very common orders. 0 1st or second, it must be a fraction order to calculate the order. That is a fraction. We need to set up an algebraic expression where we will take the rate law and plug in the data for experiment to where the rate is 0.0 to nine. Set that equal decay multiplied by the concentration of CH three cl raised to the first power because we determined that up here multiplied by the concentration of cl two raised to an unknown power. Well then divide that by the differential rate law with the data. For experiment too. Where the rate is .041. Set that equal to K. Multiplied by the concentration of CH three cl again raised at the first power, Multiplied by the concentration of Cl two raised to an unknown power. We set up a ratio of rate laws so that the case will cancel. You'll notice that the .1 concentration staying constant also cancel. So we'll take the a ratio of these two numbers which gives us .707. And then the ratio of these two numbers gives us five. But remember it's raised to the X. So we can now take the log of both sides. We take the log of both sides. The exponent now becomes a multiplier. We can then divide both sides by little log of 50 And X. is the log of .77 divided by the log of .50 which is .5. So the order with respect to the L two is .5 or 1/2 order. The overall order would then be first order plus 1/2 order or 1.5 order. So to calculate the K value. We can take data for any experiment, plug it into a single rate law. I'm choosing the experiment one where the rate is .014. Okay. I will set that equal to K. Which I don't know. The concentration of both of them is 0.5. So it'll be five. Race to the first power for CH three cl and then five raised to the one half power for all to I can then solve for K. and I get 1.25 With units of one over polarity to the one half seconds because the overall order is 1.5. A better K value could be calculated by doing the same thing for all four experiments, Summing them up and divided by 4 to get an average K value. I'm just going to go with the K. Value. I got for experiment one because it should be pretty close to the true value. So then I can say that the rate law is simply equal to the rate constant, which I calculated to be 1.251 over, Mueller to the one half seconds, multiplied by the concentration of CH three cl raised to the first power, multiplied by the concentration of cl two raised to the one half.

We know that when the temperature increased by 10 degree Celsius, the rate of reaction would double. And since we know that the degrees of Celsius and degrees of Calvin are just units on a different scale, So the increase of 10 degree Kelvin will also lead to the increase. I started a rate doubling. So now if at 300 k we have a rate of zero going to to most of Eno to permanent at 3 20 k, there is a 20 K increase in temperature, which means that the rate would double twice, which is four times. Therefore the rate the final rate would be through a point a most of n 02 per minute at 3 20 k.