The rate law for the reaction below is given by the expression k[A][B]
$$A + B \to$$ Product
If the concentration of B is increased from 0.1 to 0.3 mol, keeping the value of A at 0.1 mol, the rate constant will be:

We are told that the experimentally determined rate law for the reaction $$A + B \rightarrow \text{Product}$$ is

$$r = k[A][B]$$

Here $$r$$ is the rate of the reaction, $$[A]$$ and $$[B]$$ are the molar concentrations of the reactants, and $$k$$ is the rate constant. By definition, the rate constant $$k$$ depends only on temperature (and the presence of a catalyst, if any). It does not change when we merely alter the concentrations of the reactants.

Nevertheless, to see this algebraically, let us write the rate twice—once with the original concentrations and once after changing the concentration of $$B$$.

Originally we have

$$[A]_1 = 0.1\ \text{mol L}^{-1}, \qquad [B]_1 = 0.1\ \text{mol L}^{-1}.$$

Substituting these values into the rate law, the initial rate becomes

$$r_1 = k [A]_1 [B]_1 = k(0.1)(0.1) = 0.01k.$$

Now the concentration of $$B$$ is increased to $$[B]_2 = 0.3\ \text{mol L}^{-1}$$ while $$[A]$$ is kept the same, so $$[A]_2 = 0.1\ \text{mol L}^{-1}$$. The new rate is therefore

$$r_2 = k [A]_2 [B]_2 = k(0.1)(0.3) = 0.03k.$$

Dividing the new rate by the original rate gives

$$\frac{r_2}{r_1} = \frac{0.03k}{0.01k} = 3.$$

The factor of 3 tells us that the rate triples, but notice that the factor $$k$$ cancels out in the ratio:

$$\frac{0.03\cancel{k}}{0.01\cancel{k}} = 3.$$

Since $$k$$ cancels, it is clear that changing the concentration of $$B$$ does not and cannot alter the numerical value of the rate constant. The only quantities affected are the rate itself and any directly concentration-dependent terms.

Hence the rate constant remains exactly the same, $$k$$.

Hence, the correct answer is Option D.