### Homework Question

Posted:

**Thu Feb 09, 2017 5:06 pm**Hi,

can anyone explain how to do problem 13 in the Chapter 15 homework?

Thanks

can anyone explain how to do problem 13 in the Chapter 15 homework?

Thanks

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=147&t=19103

Page **1** of **1**

Posted: **Thu Feb 09, 2017 5:06 pm**

Hi,

can anyone explain how to do problem 13 in the Chapter 15 homework?

Thanks

can anyone explain how to do problem 13 in the Chapter 15 homework?

Thanks

Posted: **Fri Feb 10, 2017 10:43 pm**

The problem describes the reaction H_{2} (g) + I_{2} (g) -> 2HI (g) and reveals that it behaves as a second order elementary reaction (first order with respect to each reactant). I_{2} is a solid at room temperature but sublimates readily. The problem gives us mass and volume so we will solve for gaseous concentrations of each reactant.

Begin by taking the definition of the rate of a second order elementary reaction: rate = k[A][B]. Here, A and B are [H_{2}] (g) and [I_{2}].

Plug in the concentrations of the reactants and the given rate constant and you can solve for initial rate.

If we double the initial concentration of H_{2}, we can write the rate equation as rate = 2k[A][B] which means the rate will double.

Begin by taking the definition of the rate of a second order elementary reaction: rate = k[A][B]. Here, A and B are [H

Plug in the concentrations of the reactants and the given rate constant and you can solve for initial rate.

If we double the initial concentration of H

Posted: **Sun Feb 19, 2017 2:56 pm**

Hi!

For this question, I'm confused as to why we don't square H_{2} when it asks for the reaction rate with doubling the concentration of H_{2}. Doesn't the equation for the rate law involve k[H_{2}]^{a}, in which a= the coefficient, which would be 2 since we doubled it? Or am I mixing up concepts?

Thanks!

For this question, I'm confused as to why we don't square H

Thanks!

Posted: **Sun Feb 19, 2017 3:59 pm**

Zoe Robertson 2H wrote:Hi!

For this question, I'm confused as to why we don't square H_{2}when it asks for the reaction rate with doubling the concentration of H_{2}. Doesn't the equation for the rate law involve k[H_{2}]^{a}, in which a= the coefficient, which would be 2 since we doubled it? Or am I mixing up concepts?

Thanks!

We do not put a=2 because it is found that [H

Posted: **Sun Feb 19, 2017 11:42 pm**

So does the fact that it is a second order elementary reaction mean that you have added up the powers to get an overall reaction order of two?