Because the rate of a chemical reaction is always in . The  value must always cancel with other units to produce these units. In a zero order reaction there is nothing to cancel, so the units are simply .

Because the rate of a chemical reaction is always in . The k value must always cancel with other units to produce these units. In a first order reaction the rate equation is 

If we replace the variables above with their units, the equation will look like this:

Therefore, in order to match,  must simply be .

Because the rate of a chemical reaction is always in . The k value must always cancel with other units to produce these units. In a second order reaction the rate equation is 

If we replace the variables above with their units, the equation will look like this:

Therefore, in order to match,  must be have the units .

Because the rate of a chemical reaction is always in . The  value must always cancel with other units to produce these units. In a third order reaction the rate equation is 

If we replace the variables above with their units, the equation will look like this:

Therefore, in order to match,  must be have the units .

Because B is in the rate determining step (the slow step) then we know that it is not zero order with respect to the concentration of B.

However, because there is only one molecule of B reacting in the rate determining step, we know that this reaction is first order with respect to the concentration of B.

The rate law is determined by the slowest step in a reaction mechanism. Because B is not a reactant in the slow step, then it will not appear in the reaction mechanism.

Therefore, the reaction is zeroth order in relation to the concentration of B.

For this question, we need to determine a true statement with regard to reactions that are catalyzed by enzymes.

To answer this, it's important to distinguish between the thermodynamics of a reaction and the kinetics. When adding an enzyme, the activation energy for the reaction is lowered. What this means is that it becomes easier for the reactants to achieve the high-energy transition state on their way to becoming product. Thus, enzymes affect the kinetics of a reaction by increasing both the forward rate and the reverse rate.

But if we look at the thermodynamics of an enzyme-catalyzed reaction, there is no change. In other words, the difference in free energy between the reactants and the products remains the same, regardless of the presence of enzyme. Thus, an enzyme will not cause a reaction to change its equilibrium position; it cannot shift a reaction towards the left or towards the right.

Because the thermodynamics of the reaction remain unchanged, the reaction will not become more exergonic, nor will it become more endergonic. Furthermore, the equilibrium of the reaction will not shift.

For this question, we're asked to select a true statement regarding reactions that are influenced by enzymes.

When thinking about how enzymes affect reactions, its important to draw a distinction between kinetics and thermodynamics. Enzymes increase a reaction's rate, but they don't change the equilibrium of a reaction. In other words, they make reactions go faster, but they don't make reactions go.

Since enzymes don't affect equilibrium, we can rule out two of the answer choices. Moreover, we know that enzymes increase the reaction rate by lower the activation energy. Consequently, this raises the rate constant for the reaction.