The activation energy of an uncatalyzed reaction is
95 kJmol. Th...

Question

The activation energy of an uncatalyzed reaction is
95 kJ>mol. The addition of a catalyst lowers the activation
energy to 55 kJ>mol. Assuming that the collision
factor remains the same, by what factor will the catalyst
increase the rate of the reaction at (a) 25 C

Accepted Answer

Hi everyone for this problem, it reads a catalyst increases a chemical reactions rate by lowering its activation energy. A particular chemical reaction has an activation energy of 125 kg per mole at 45°C. The activation energy is lowered by 55 kg joules per mole. When a catalyst is added to the reaction, calculate the factor by which the rate of the reaction is increased. Assume that the frequency factor remains the same. So our goal here is to calculate the factor by which the rate of the reaction is increased. To calculate the ratio by which the rate of the reaction increases. We can calculate the ratio of the rate constants and we can do that using the equation Ln of K is equal to negative activation energy over R times T plus Ln of a. Okay, so based off of the problem we have, this is what's given. So K one we can say is our uncapped realized reaction K two We can say is our catalyzed reaction. Okay, now our first activation energy is equal to 125 killer jewels Permal. And our second activation energy is going to equal that 1 25 kg joules per mole -55 killer jewels per mole, which gives us 70 killer jewels. Permal, our is our Constant, which is a value. We should know and that value is 8.314 jewels over Mole Times Kelvin And our temperature is 45°C. So let's just take a look at our units here to make sure that they all match. So for our we have our energy and jewels and our temperature and Kelvin. So that means we need to convert our activation energy to jewels and then our temperature to Kelvin. So to do that we can say in one kill a jewel, There's 1000 jewels and so our killer jewels will cancel and will be left with jewels Permal. So this gives us an activation energy of 125, jules Permal and the same thing for a second activation energy. So we have and one killer jewel, there's 1000 jules. So this gives us 70, jules per mole for a second activation energy. Now for a temperature to go from degrees Celsius to kelvin, we add 273.15. Okay, so this is going to give us a temperature in kelvin equal to 318.15 kelvin. Okay, so now what we can do to find the factor that we're looking for is we can set up our equation like this Ellen of K two minus Ln of K one is equal to our second activation energy minus the first activation energy. Okay, so negative activation energy to over R. T plus Ln of a minus our first activation energy. Okay, so we have everything that we need to plug in and solve. So we're just going to go ahead and do that now. So we have L N F K two minus Ln of K one is equal to negative 70, jules Permal over our times T Okay minus -1. jewels per mole over the same thing. Okay, so let's go ahead and clean this up a little bit and we can do that by factoring out. So we'll have negative 70,000 jules per mole plus 100 and 25,000 jules per mole. And this is going to be all over R times T. Okay, so by doing it that way we're going to get a ratio on the left side. So we'll have L N F K two over L. N of K one is equal to 55,000 jewels Permal over 2645 jewels per mole. Okay, so we're going to get LNFK two over LNFK one is equal to 20.7940. Now to get rid of the L. N. We're going to raise that 20.7940 to the E. Okay, so by doing that, the L. N. Are going to cancel and we'll just be left with K two over K one Is equal to 1. times 10 to the 9th. Okay and so what this means is the rate of the reaction Is increased by a factor of 1.1 times to the ninth. That's the end of this problem. And that's the answer to this problem. I hope this was helpful.

The activation energy of an uncatalyzed reaction is 95 kJ>mol. The addition of a catalyst lowers the activation energy to 55 kJ>mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) 25 C

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