Question

Chemical Kinetics !

Answer 1

@Zale101

Answer 2

What is the order of this rate?

Answer 3

1

Answer 4

If i increase the concentration of both a and b, what would happen?

Answer 5

reaction shifts to forward direction

Answer 6

She means, what would happen to the \(rate\), not to the equilibrium concentrations. After all this is kinetics, right?

Also whats the order of [B] in the reaction?

Answer 7

well the rate of reaction is of first order

Answer 8

how can we determine order of any reactant ?

Answer 9

yes, the overall rate is to the first order. The rate of each reactant in the reaction is indicated by the exponents - shown in the rate law.

Answer 10

okay,so B is of zero rate

Answer 11

but B has no order

Answer 12

no its to the zeroth order. it's rate is independent of its concentration.

Answer 13

correct !

Answer 14

by it's rate, i meant the rate of the reaction

Answer 15

so, what graph illustrates this?

Answer 16

option A

Answer 17

@aaronq

Answer 18

@Rushwr

Answer 19

so you are saying that option A is correct ?

Answer 20

sorry to say,answer is wrong

Answer 21

but i'm getting option C

Answer 22

Oh well that's actually possible no i was taking the integral law so that confused me a little bit.
Actualy the slope of the graph shows the rate right ? But rate we know that is zero ! Then the slope should be zero which makes it the third answer ! :)

Answer 23

So first off it states that initially the concentrations are both a, I'll put a little o to mean it's initial concentrations, when \(t=0\), here: \([A]_0 = [B]_0 = a\).

Don't worry about this next calculus thing but we can write the rate as:

\[rate = - \frac{ d [A]}{dt}\]
The negative sign is a matter of convention, whether you're looking from the perspective of the products or reactants. If you don't know any calculus don't worry if you do well enjoy pluging this in:

\[- \frac{d[A]}{dt} = k [A]\]

(I omitted [B] because \([B]^0 =1\) )

We now can rearrange this separable differential equation and then integrate. The limits of integration mean we're going from the start t=0 to our current time t and we're going from the initial concentration \([A]_0\) to the current concentration \([A]\).

\[\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = \int_0^t -k dt\]
\[\ln \frac{[A]}{[A]_0} = -kt\]
\[[A] = [A]_0 e^{-kt}\]

Alright but we really wanted information about B not A but of course they're related since for every 1 amount of A we need 2 amounts of B. All this really amounts to is that it decays away twice as fast. Main thing to note is that the concentration never really goes to 0, so that really leaves just either 3 or 4 as your possible answer since it requires infinite time to reach the steady state for [B] so it's gotta be (4) since this is the only one that works.

I haven't thought about this in years so there could be flaws in my reasoning and you should doubt it but if you ask some questions (I realize I probably did wayyyy too much calculus sorry!) I'll try to help sort this out and I could use refreshing my memory on this stuff lol

Answer 24

so what do you think what is the answer ?

Answer 25

@Photon336

Answer 26

@Cuanchi

Answer 27

if you notice something

\[r = k[A][B]^{0}\]

The reaction is zero order with respect to b

and first order with respect to a.

so what we can do is add both the orders together to find the total order of reaction. 0+1 = 1

so the reaction is first order.


you know that a is going to decrease by some amount. originally you're going to have all a and it's going to decrease. now the next part is just figuring out how the graph will look.

Answer 28

but question asks about concentration of B

Answer 29

@Photon336

Answer 30

the reaction is first order with respect to A. so from my knowledge the concentration of B doesn't affect the rate of reaction.

Answer 31

so what do you think what is the answer ?

Answer 32

@Kainui