What are the units of the rate constant for first-order and zero-order reactions in terms of molarity, respectively?

A. [tex] s^{-1}, M \cdot s^{-1}[/tex]
B. [tex] s^{-1}, M[/tex]
C. [tex] M \cdot s^{-1}, s^{-1}[/tex]
D. [tex] M, s^{-1}[/tex]



Answer :

To determine the units of the rate constant for reactions of different orders, let's analyze the general form of the rate law equations for first-order and zero-order reactions.

### First-Order Reactions:
For a first-order reaction, the rate law is given by:
[tex]\[ \text{Rate} = k [A] \][/tex]

Here, [tex]\(\text{Rate}\)[/tex] is the reaction rate ([tex]\(\frac{M}{s}\)[/tex]), [tex]\([A]\)[/tex] is the concentration of the reactant ([tex]\(M\)[/tex]), and [tex]\(k\)[/tex] is the rate constant.

To find the unit of the rate constant [tex]\(k\)[/tex], we rearrange the equation to solve for [tex]\(k\)[/tex]:
[tex]\[ k = \frac{\text{Rate}}{[A]} \][/tex]

Substituting the units:
[tex]\[ \text{Unit of } k = \frac{\frac{M}{s}}{M} = s^{-1} \][/tex]

So, for a first-order reaction, the unit of the rate constant is [tex]\(s^{-1}\)[/tex].

### Zero-Order Reactions:
For a zero-order reaction, the rate law is given by:
[tex]\[ \text{Rate} = k \][/tex]

For zero-order reactions, the rate of reaction is independent of the concentration of reactants. Here, [tex]\(\text{Rate}\)[/tex] is again expressed in units of [tex]\(\frac{M}{s}\)[/tex].

Since the rate law is directly equal to [tex]\(k\)[/tex]:
[tex]\[ k = \text{Rate} \][/tex]

Thus, the unit of [tex]\(k\)[/tex] is the same as the unit of the reaction rate:
[tex]\[ \text{Unit of } k = \frac{M}{s} = M s^{-1} \][/tex]

So, for a zero-order reaction, the unit of the rate constant is [tex]\(M s^{-1}\)[/tex].

Comparing the answer choices with our derived units:
(a) [tex]\(s^{-1} \cdot M s^{-1}\)[/tex] — Combining units doesn’t make sense in this context.
(b) [tex]\(s^{-1}, M\)[/tex] — Incorrect because the zero-order unit is [tex]\(M s^{-1}\)[/tex], not [tex]\(M\)[/tex].
(c) [tex]\(M s^{-1} ; s ^{-1}\)[/tex] — Incorrect ordering, these represent zero and first-order rate constants but in reverse order.
(d) [tex]\(M, s^{-1}\)[/tex] — Incorrect because the zero-order unit is [tex]\(M s^{-1}\)[/tex], not [tex]\(M\)[/tex].

The correct answer is clearly based on the units derived:
```
('s^-1', 'M s^-1')
```

Thus, the correct answer is:

(b) [tex]\(s^{-1}, M\)[/tex] Since it matches the order and units for first-order and zero-order reactions respectively.