If a chiral molecule reacts with an achiral molecule will the product be chiral or achiral?
The key here is to look at the transition state through which the reaction proceeds. 

The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.

The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:

$$\ce{($S$)-A + B -&gt; [($S$,$R$)-AB]^{\ddagger} + [($S$,$S$)-AB}]^{\ddagger}$$

These are diastereomeric and the rates through these two transition states is different.

But $\ce{($R$)-A}$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:


$\ce{[($S$,$R$)-AB]^{\ddagger}}$
$\ce{[($S$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$R$)-AB]^{\ddagger}}$


Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.

So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.