Question

Can I get some Chemistry help?

Answer 1

@AZ

Answer 2

sure, post your question and I can take a look at it

Answer 3

@az

Answer 4

so we're looking for the reaction order for H2

so look at the reaction rate

we want the amount of [P4] to stay the same but the [H2] to change so we can determine the reaction order for H2

Answer 5

so look at the ratio

when we keep [P4] the same

H2 goes from 0.0075 to 0.0150

and the reaction rate goes from 3.20*10^-4 to 6.40*10^-4

Answer 6

Yes

Answer 7

So would it be 2?

Answer 8

so remember we're trying to find the reaction order

for this equation, we would write it as

\(\Large\text{rate = k [P}_4]^x\text{ [H}_2]^y\)

Answer 9

we're trying to find what y is

so we kept P4 the same

and so that means we're only looking at H2

Answer 10

and so you divided the rate and the H2 levels and you get

2 = 2^y

what would 'y' be ?

Answer 11

0.0150?

Answer 12

so we divided the two numbers so we could figure out what power H2 is at

so now

2 to the power of what will give you 2

Answer 13

1

Answer 14

so that would be your reaction order for \(\text{H}_2\)

Answer 15

Does that make sense? How you're supposed to solve questions like this?

When they're asking about H2, you want to keep the other reactant constant so you see that when we chose 0.0075 and 0.0150 values for H2

the P4 values remain the same at 0.0110

Answer 16

I believe so

Answer 17

So with respect to H2 the answer would be 1.

Answer 18

ohhh now I remember your username, you're the one taking college chemistry xD

Answer 19

yes

Answer 20

so when we have a reaction

A + B --> Products

the rate law is written as

Rate = k [A]^x ^y

Answer 21

yes xD

Answer 22

the second one should be 1 as well right?

Answer 23

k is a constant
and when we keep one of the reactants constant

we can then determine the ratio between the rate and the other reactant to determine the order (or the exponent of it)

Answer 24

\(\color{#0cbb34}{\text{Originally Posted by}}\) @kamachavis
the second one should be 1 as well right?
\(\color{#0cbb34}{\text{End of Quote}}\)

no no so be careful here

Answer 25

so

P4 goes from 0.011 to 0.022

and the rate remains the same

Answer 26

so

\( \dfrac{6.40\times 10^{-4}}{6.40\times 10^{-4}} = \left(\dfrac{0.022}{0.011}\right)^{\large{x}}\)

Answer 27

P4 's reaction order would be the value of 'x'

Answer 28

It would be 0

Answer 29

exactly!

Answer 30

that's why the rate stays the same- because the reaction order is 0 for P4

Answer 31

thank you that helped me out a lot

Answer 32

big smart smh

Answer 33

i will probably post a few more questions today but you can get back to me whenever i am just practicing for an exam tomorrow.

Answer 34

Of course! If it's multiple questions and I'm not online, feel free to close the question to post another one and just tag me so I can take a look at it

Answer 35

so you need to know the equations to solve this and you're going to have to memorize them

If it was a first order reaction:
\(ln[A_t] = -kt + ln[A_0] \)

and half life
\(t_{1/2} = \dfrac{0.693}{k}\)



For second order reactions:
\( \dfrac{1}{[A_t]} = kt + \dfrac{1}{[A_0]}\)

and half life
\(t_{1/2} = \dfrac{1}{k[A_0]} \)

Answer 36

since the equation is a second order, just plug in the numbers they gave you

\( [A_t]\) = 0.1
\( [A_0]\) = 1
k = 0.02

and you solve for t

so plug them in

\( \dfrac{1}{[A_t]} = kt + \dfrac{1}{[A_0]}\)

Answer 37

For #9 (if you need help with that as well), it's also something you need to understand and then figure it out or you can memorize it

reaction rate has units of M/s but the units for the constant, k, will depend on the order of the reaction

so if was a zero order
rate = k * [A]^0
rate = k
so k has the same units as the rate so it would be M/s
(that's Molar which is moles/liter, just in case you weren't aware)

if it was first order, now you have

rate = k * [A]^1

we know the units for rate are M/s and [A] has units of M
so the units for K has to be just 1/s

and similarly for second order

rate = k * [A]^2
M/s = ?? * M^2

and if you divide M^2 on both sides to determine the units of k in a second order reaction, you'll see that the units are 1/(M*s)

so it's just something you need to memorize or if you understand it, something you can just derive and figure out

Answer 38

remember that M = moles of liter

so you can replace that but be careful when dealing with the fractions