## Homework Question

$\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$

Elnaz Guivatchian 3L
Posts: 11
Joined: Wed Sep 21, 2016 2:55 pm

### Homework Question

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

Chem_Mod
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### Re: Homework Question

The problem describes the reaction H2 (g) + I2 (g) -> 2HI (g) and reveals that it behaves as a second order elementary reaction (first order with respect to each reactant). I2 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 [H2] (g) and [I2].

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 H2, we can write the rate equation as rate = 2k[A][B] which means the rate will double.

Zoe Robertson 2H
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

### Re: Homework Question

Hi!

For this question, I'm confused as to why we don't square H2 when it asks for the reaction rate with doubling the concentration of H2. Doesn't the equation for the rate law involve k[H2]a, in which a= the coefficient, which would be 2 since we doubled it? Or am I mixing up concepts?

Thanks!

Desiree Martin 2A
Posts: 25
Joined: Fri Jul 22, 2016 3:00 am

### Re: Homework Question

Zoe Robertson 2H wrote:Hi!

For this question, I'm confused as to why we don't square H2 when it asks for the reaction rate with doubling the concentration of H2. Doesn't the equation for the rate law involve k[H2]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 [H2] is first order, this means that a=1. The other reactant is first order as well where [I2] has a=1. Therefore, the overall order is second order, since you add the sums of the orders of the reactants being 1+1=2. This is why we don't raise [H2] to the second power. It is asked that we double the concentration, which means multiply by two, not raise to the second power. If it is first order we understand that whatever we do to one concentration has a proportional effect on the rate. If we triple the concentration of H2, then the rate would be 3*k*[H2] *[I2].

Suzanne Reyes-Ingwersen 3A
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Joined: Wed Sep 21, 2016 2:56 pm

### Re: Homework Question

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?