AP Chemistry - Rate Laws

If a reaction is second order with respect to , which of the following quantities will produce a straight line when graphed against (time)?

$\frac{1}{[H]}$

$\frac{1}{ln[H]}$

$\frac{1}{[H]^2}$

Explanation

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

If the reaction:

$A_2+2B \rightarrow 2AB$

Follows the mechanism:

$i. A_2 \rightarrow 2A$ (Slow)

$ii. 2A+2B \rightarrow 2AB$ (Fast)

What is the rate order with respect to ?

Explanation

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.

If the reaction:

$A_2+2B \rightarrow 2AB$

Follows the mechanism:

$i. A_2+B \rightarrow A+AB$ (Slow)

$ii. A+B \rightarrow AB$ (Fast)

What is the rate order with respect to ?

Explanation

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 reaction

$2NO_2+F_2\rightarrow 2NO_2F$

Follows the mechanism:

$i. NO_2+F_2\rightarrow NO_2F+F$ (slow)

$ii. NO_2+F\rightarrow NO_2F$ (fast)

Determine the rate law for this reaction.

Explanation

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

$-\frac{1}{2}\frac{d[NO_2]}{dt}=k[NO_2][F_2]$

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

What are the units for the rate constant of a third order reaction?

$\frac{1}{M^2\cdot s}$

$\frac{1}{M\cdot s}$

$\frac{1}{s}$

$\frac{M}{s}$

Explanation

Because the rate of a chemical reaction is always in $\frac{M}{s}$ . 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:

$\frac{M}{s}=k*M^3$

Therefore, in order to match, must be have the units $\frac{1}{M^2\cdot s}$ .

If a reaction is zero order with respect to , which of the following quantities will produce a straight line when graphed against (time)?

$\frac{1}{ln([H])}$

$\frac{1}{[H]}$

Explanation

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

What are the units for the rate constant for a second order reaction?

$\frac{1}{M\cdot s}$

$\frac{1}{M^2\cdot s}$

$\frac{1}{s}$

$\frac{M}{s}$

$\frac{M^2}{s}$

Explanation

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

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

$\frac{M}{s}=k*M^2$

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

What are the units for the rate constant for a first order reaction?

$\frac{1}{s}$

$\frac{M}{s}$

$\frac{M^2}{s}$

$\frac{1}{M\cdot s}$

$\frac{1}{M^2\cdot s}$

Explanation

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

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

$\frac{M}{s}=k*M$

Therefore, in order to match, must simply be $\frac{1}{s}$ .

What are the correct units for the rate constant of a zero order reaction?

$\frac{M}{s}$

$\frac{1}{s}$

$\frac{1}{M\cdot s}$

$\frac{M^2}{s}$

Explanation

Because the rate of a chemical reaction is always in $\frac{M}{s}$ . 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 $\frac{M}{s}$ .