# Homework Help: Quick Chemical kinetics question

1. Feb 12, 2005

The rate constant for a certain reaction has the units concentration/time. What is the order of the raction?

A. 0
B. 1
C.2
D.3

My reasoning:

K = [A]^m x ^n

Since the concentration/time is to the power of one, both m and n must be one i think which give an overall order of one?.... since if they were zero there would be no reaction?

2. Feb 12, 2005

### Sirus

There is a formula for this.

$$k=\frac{\left(\frac{L}{mol}\right)^{\mbox{order}-1}{\mbox{units of time}}$$

EDIT: for some reason the latex is not working.

k=[(L/mol)^{order - 1}]/(units of time)

3. Feb 12, 2005

Do you mean that in order to have the unit as concentration/time, it should be:

K = C^(2-1)/t

which gives the answer C?

4. Feb 12, 2005

### Sirus

Use the formula. If K must be in units of concentration over time, what must the order be in the formula?

5. Feb 12, 2005

### Sirus

Here's the fixed latex:

$$k=\frac{\left(\frac{L}{mol}\right)^{\mbox{order}-1}}{\mbox{units of time}}$$

6. Feb 13, 2005

So its C i suppose

Thank you for the help!

7. Feb 13, 2005

### Sirus

How did you get C? That's incorrect.

8. Feb 13, 2005

if the unit is concentration/time it should be:

(Concentration^1) = K
-------------------
time

So i think it's :

[Concentration]^[2] - 1
------------------------ =
time

(Concentration^1)
------------------- = K
time

the order should be: [2]

Thats how i reasoned.

9. Feb 13, 2005

### Sirus

Review the formula. What are the units for concentration? What are the units in the formula?

10. Feb 13, 2005

### Gokul43201

Staff Emeritus

From your reasoning above where would you get the dimensions of time from ? The error is in that your equation is for the rate, not the rate constant.

You solve this problem, you can either remember the formula provided by Sirus, or simply work it out from first principles. The order n, of the reaction determines the rate law :

$$rate = -\frac {dX}{dt} = kX^n$$ where X is a concentration.

So $$-k = X^{-n} \frac {dX}{dt}$$

So the dimensions of k will be $$\frac {[X^{1-n}]}{[T]}$$

In this particular case, the fact that k has the same units as the rate, makes it all the more easy and negates the need to go beyond the first step.

11. Feb 16, 2005

### GCT

Just focus on the equation rate=kX^n, the units for the rate will always be in terms of concentration/T. The right side of the equation, besides k, will be in terms of concentration, concentration squared, etc...depending on the order number (you should be able to see this). Thus if we have k in terms of concentration/time (M/s), and rate=concentration/time (M/s), dividing the rate by k will give you 1, meaning zero order.