For a reaction of order n, the unit of the rate constant is:

First, we recall the definition of the rate law for a reaction of overall order $$n$$. For a simple case where the rate depends on a single concentration $$[A]$$, we write

$$\text{Rate}=k\,[A]^n$$

Here, $$k$$ is the rate constant whose units we are asked to find, $$[A]$$ is the molar concentration of the reactant, and $$n$$ is the order of the reaction.

Now we analyse the units of each quantity one by one.

We know that the numerical value of rate is change in concentration divided by time. Therefore, in SI-compatible chemical units,

$$[\text{Rate}]=\dfrac{\text{mol L}^{-1}}{\text{s}}=\text{mol L}^{-1}\,\text{s}^{-1}$$

Next, the concentration $$[A]$$ itself has the unit $$\text{mol L}^{-1}$$. Raising this to the power $$n$$ gives

$$[A]^n=\bigl(\text{mol L}^{-1}\bigr)^n=\text{mol}^n\,\text{L}^{-n}$$

We now substitute these unit expressions into the rate law. Isolating the unit of $$k$$ from

$$[\text{Rate}]=[k]\,[A]^n$$

we have

$$[k]=\dfrac{[\text{Rate}]}{[A]^n}$$

Substituting, we obtain

$$[k]=\dfrac{\text{mol L}^{-1}\,\text{s}^{-1}}{\text{mol}^n\,\text{L}^{-n}}$$

To simplify, we divide the powers of identical units algebraically:

The exponent of $$\text{mol}$$ becomes $$1-n$$ because $$1-n=1-(n)$$.

The exponent of $$\text{L}$$ becomes $$-1-(-n)=n-1$$.

The exponent of $$\text{s}$$ remains $$-1$$ because no time unit appears in the denominator.

Hence, after simplification,

$$[k]=\text{mol}^{\,1-n}\,\text{L}^{\,n-1}\,\text{s}^{-1}$$

Comparing this with the options given, we find that this expression matches the unit written in Option C.

Hence, the correct answer is Option C.