Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Given the same reactant concentrations, the reaction$$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)$$at $250^{\circ} \mathrm{C}$ is $1.50 \times 10^{3}$ times as fast as the same reaction at $150^{\circ} \mathrm{C}$. Calculate the activation energy for this reaction. Assume that the frequency factor is constant.

$$135 \mathrm{kJ} / \mathrm{mol}$$

Chemistry 102

Chapter 6

Chemical Kinetics

Kinetics

University of Central Florida

Rice University

University of Maryland - University College

Brown University

Lectures

22:42

In probability theory, the conditional probability of an event A given that another event B has occurred is defined as the probability of A given B, written as P(A|B). It is a function of the probability of B, the probability of A given B, and the probability of B.

04:55

In chemistry, kinetics is the study of the rates of chemical reactions. The rate of a reaction is the change in concentration of a reactant over time. The rate of reaction is dependent on the concentration of the reactants, temperature, and the activation energy of the reaction.

02:37

Consider the second-order …

03:04

For the reaction$$\mat…

02:57

Calculate the activation e…

01:23

The rate constant of a rea…

01:44

03:07

Consider the first-order r…

Hello said that I was going to be looking at a situation where the rate constant at 250 degrees Celsius is ah 1.50 temps into the third times as great larger than the right constant on 150 degrees Celsius. And now we want to find the activation energy. So for this reaction, so remember now the natural log of K over one over tea the slope If you graph it is equal to the negative of activation energy over our, which is 8.3 when four jewels over Calvin and moles. So talk about we try to find the soap given what we have here. So first. So the natural log so are final point will be here. So it will be the smaller K which in this case is the K at 150 Celsius. Subtracting natural log okay, at 250 degrees Celsius and that will be over one over 150 which converted to Calvin. We just had 273. That would be 423. It's attracting. So now we convert 250 degrees Celsius to Calvin. So there we go. And when you should notice is that we can convert the numerator here. You can convert it into the natural log off. Okay, 150 three Celsius over the K of 250 degrees Celsius so we can use the ratio here. So we know that the ratio is 1/1 0.50 times 10 to the third. So let's take the natural log event and then let's divide it by then. Verse of the temperature. I'm denominator here, and we're going to get negative. 16,000 179 is equal to negative activation under G over our, which is a 0.314 So let's times both sides, by negative 8.314 we will see that the activation energy is under it. And 30 5000 jewels promote well, which can also be written as 135 killer jewels. Permal. So there we go. We found activation energy

View More Answers From This Book

Find Another Textbook

01:59

Give an example of an ion or molecule containing Al that (a) obeys the octet…

02:55

In general, atomic radius and ionization energy have opposite periodic trend…

03:18

In $2009,$ thousands of babies in China became ill from drinking contaminate…

01:26

What is the osmotic pressure (in atm) of a $1.36 \mathrm{M}$ aqueous solutio…

04:26

As mentioned in the chapter, the Lewis structure for $\mathrm{O}_{2}$ is…

01:07

Explain, in terms of their electron configurations, why $\mathrm{Fe}^{2+}$ i…

00:38

Calculate the osmotic pressure of a $0.0500 \mathrm{M}$ $\left.\mathrm{MgSO}…

01:46

Under the same conditions of temperature and density, which of the following…

03:19

An aqueous solution of a $0.10 \mathrm{M}$ monoprotic acid HA has an osmotic…

06:03

A student titrates an unknown monoprotic acid with a NaOH solution from a bu…