Increasing the amount of reactant increases the rate of reaction only if the concentration of each reactant increases and if the reactants are dissolved in a solution. This is not the case for all reactions. Heating the reaction and introducing a catalyst enable the reactants to reach the activation energy needed to proceed into the reaction. Increasing the surface area between the reactants enables more of each reactant to interact with one another. 

Remember, the rate law is always determined by the rate-determining step. This is ALWAYS the "slow" step in the reaction. In this case, the slow step is step (i).

The rate for this step alone is

\(\displaystyle -\frac{1}{2}\frac{d[NO_2]}{dt}=k[NO_2][F_2]\)

And because both of the original reactants are in this rate law, it is also the rate law for the overall reaction.

The correct answer is \(\displaystyle [H]\) because the differential rate law for a zero order reaction is \(\displaystyle \frac{-dX}{dt}=k\). Therefore, the plot has a linear relationship to the concentration of hydrogen.

Because the differential rate law for a first order reaction is \(\displaystyle \frac{-dH}{dt}=k[H]\) the graph will have a linear relationship to \(\displaystyle ln[H]\).

Because the differential rate law for a second order equation is \(\displaystyle \frac{-dH}{dt}=k[H]^2\) the graph will have a linear relationship to \(\displaystyle \frac{1}{[H]}\).

In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to \(\displaystyle [A]\). When comparing experiments 1 and 3, we see that the concentration of B was doubled, and this resulted in a doubling of the rate. Therefore, the reaction is first order with respect to \(\displaystyle [B]\). Then, we plug in all of this information into the rate law, and we get the correct answer of \(\displaystyle rate=k[A]^2[B]\).

In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to \(\displaystyle [A]\). When comparing experiments 1 and 3, we see that the concentration of B was doubled, and this resulted in a doubling of the rate. Therefore, the reaction is first order with respect to \(\displaystyle [B]\). Then, we plug in all of this information into the rate law, and we get the correct answer of \(\displaystyle rate=k[A]^2[B]\).

Finally, we take the values from one of the rows of the experimental data. It does not matter which one, they will all result in the same answer. In this case, I chose experiment 1.

\(\displaystyle 5=k(.5)^2(.5)\)

\(\displaystyle k=\frac{5}{(.5)^2(.5)}\)

\(\displaystyle k=40\)

Because the values from the table all only have one significant figure, our answer can also only have one significant figure.

Because the rate of a chemical reaction is always in \(\displaystyle \frac{M}{s}\). The \(\displaystyle k\) 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 \(\displaystyle \frac{M}{s}\).

Because the rate of a chemical reaction is always in \(\displaystyle \frac{M}{s}\). The k value must always cancel with other units to produce these units. In a first order reaction the rate equation is 

\(\displaystyle rate=k[A]\)

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

\(\displaystyle \frac{M}{s}=k*M\)

Therefore, in order to match, \(\displaystyle k\) must simply be \(\displaystyle \frac{1}{s}\).

Because the rate of a chemical reaction is always in \(\displaystyle \frac{M}{s}\). The k value must always cancel with other units to produce these units. In a second order reaction the rate equation is 

\(\displaystyle rate=k[A]^2\)

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

\(\displaystyle \frac{M}{s}=k*M^2\)

Therefore, in order to match, \(\displaystyle k\) must be have the units \(\displaystyle \frac{1}{M\cdot s}\).