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Consider the following reaction:

$2HgCl_{2(aq)}+C_2O_{4(aq)}^{2-}$→$2Cl_{(aq)}^-+2CO_{2(g)}+Hg_2Cl_{2(aq)}$

A rate study on this reaction was conducted and the following data gathered:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Run	\n Initial $[HgCl_{2(aq)}]$ \n (mol L^-1) \n	\n Initial $[C_2O_{4(aq)}^{2-}]$ \n (mol L^-1) \n	\n Initial Rate \n (mol L^-1 s^-1) \n
1	$\\var{b}$	$\\var{a}$	$\\var{c}$
2	$\\var{b}$	$\\var{b}$	$\\var{e}$
3	$\\var{a}$	$\\var{b}$	$\\var{d}$

", "advice": "

We determine the order with respect to each reactant using the following:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Effect on reaction rate	Order
No change in rate	Zero
Rate changes in same ratio as concentration	First
Rate changes by square of concentration change	Second
Rate changes by cube of concentration change	Third

Consider the data from the rate study:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Run	\n Initial $[HgCl_{2(aq)}]$ \n (mol L^-1) \n	\n Initial $[C_2O_{4(aq)}^{2-}]$ \n (mol L^-1) \n	\n Initial Rate \n (mol L^-1 s^-1) \n
1	$\\var{b}$	$\\var{a}$	$\\var{c}$
2	$\\var{b}$	$\\var{b}$	$\\var{e}$
3	$\\var{a}$	$\\var{b}$	$\\var{d}$

We will use runs 2&3 to investigate the order with respect to reactant $HgCl_{2}$ since its concentration varies, but that of $C_2O_{4}^{2-}$ remains constant.

We see that the rate of run 2 is double that of run 3. Note that this corresponds to the concentration of $HgCl_{2}$ doubling also, and so the reaction is first order for $HgCl_{2}$.

Consider the data from the rate study:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Run	\n Initial $[HgCl_{2(aq)}]$ \n (mol L^-1) \n	\n Initial $[C_2O_{4(aq)}^{2-}]$ \n (mol L^-1) \n	\n Initial Rate \n (mol L^-1 s^-1) \n
1	$\\var{b}$	$\\var{a}$	$\\var{c}$
2	$\\var{b}$	$\\var{b}$	$\\var{e}$
3	$\\var{a}$	$\\var{b}$	$\\var{d}$

We will use runs 1&2 to investigate the order with respect to reactant $C_2O_{4}^{2-}$ since its concentration varies, but that of $HgCl_{2}$ remains constant.

We see that the rate increases by a factor of 4 between runs 1&2. Note that this corresponds to the concentration of $C_2O_{4}^{2-}$ doubling also, and so the reaction is second order for $C_2O_{4}^{2-}$.

$Rate=k[A]^m[B]^n$

Where,

$k$ is the rate constant
$m$ is the order for reactant $A$
$n$ is the order for reactant $B$

In parts a) and b) we determined that the reaction is first order for $HgCl_{2}$ and second order for $C_2O_{4}^{2-}$.

Thus the rate law is given by,

$Rate=k[HgCl_{2}][C_2O_{4}^{2-}]^2$

What is the order for $HgCl_{2}$?

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What is the order for $C_2O_{4}^{2-}$?

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Which of the following correctly describes the rate law for the reaction?

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The rate law for a particular reaction is given by,

$Rate=k[X]^2$.

Experimentally, the initial rate of reaction is determined to be $\\var{a}$ mol L^-1 s^-1, where the initial concentration of reactant X is $\\var{b}$ mol L^-1.

", "advice": "

We are asked to find the value of $k$, given that the rate law is

$Rate=k[X]^2$ (1)

Rearranging equation (1) for $k$ gives,

$k=\\frac{Rate}{[X]^2}$ (2)

We are told that the initial rate when $[X]=\\var{b}$ mol L^-1 is $Rate=\\var{a}$ mol L^-1 s^-1. Plugging these values into equation (2) gives,

$k=\\frac{\\var{a}}{\\var{b}^2}=\\var{k}=\\var{k_rounded}$ (3 s.f.)

We can see from the rate law that the reaction is second order. The units for $k$ for a second order reaction are L mol^-1 s^-1.

Alternatively we can work out the units for $k$ as follows:

We are told $Rate=k[X]^2$ (1),

Rearranging for $k$ gives

$k=\\frac{Rate}{X^2}$ (2)

Note that the units for $Rate$ are mol L^-1 s^-1 and the units for concentration, $[X]$ are mol L^-1 .

Now, if we plug these units into equation (2) and cancel where appropriate we get,

$\\require{cancel}k=\\frac{molL^{-1}s^{-1}}{(molL^{-1})^2}=\\frac{molL^{-1}s^{-1}}{mol^2L^{-2}}=\\frac{molL^{-1}s^{-1}}{mol^2(L^{-1})^2}=\\frac{\\cancel{mol}\\cancel{L^{-1}}s^{-1}}{mol^{\\cancel{2}}(L^{-1})^\\cancel{2}}=Lmol^{-1}s^{-1}$

Calculate the rate constant, $k$, for the reaction.

$k$=[[0]]

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What are the units for $k$?

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Consider the following reaction,

$A_{(g)}+B_{(g)}$→$C_{(g)}+D_{(g)}$

The reaction is found to be second order with respect to $A$ and zero order with respect to $B$.

", "advice": "

We are told that the reaction is second order with respect to $A$ and zero order with respect to $B$. Thus, the rate law is given by,

$Rate=k[A]^2[B]^0=k[A]^2$

The rate law is independent of $[B]$, so altering $[B]$ has no effect on rate.

In the rate law, $[A]$ is raised to the power of $2$, thus increasing $[A]$ by a factor of $2$ leads to the rate increasing by a factor of $4$, i.e., it quadruples.

The rate law is independent of $[B]$, so altering $[B]$ has no effect on rate. We are told that $[A]$ is kept constant. Thus the rate remains the same.

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If $[A]$ is doubled and $[B]$ is halved, what happens to the rate?

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If $[A]$ stays the same and $[B]$ doubles, what happens to the rate?

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Consider the following reaction carried out at 298K:

$BrO_{3(aq)}^-+5Br^-_{(aq)}+6H^+_{(aq)}$→$3Br_{2(aq)}+3H_2O_{(l)}$

A rate study was conducted and the following data obtained:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Run	\n Initial $[BrO_{3(aq)}^-]$ \n (mol dm^-3) \n	\n Initial $[Br^-_{(aq)}]$ \n (mol dm^-3) \n	\n Initial $[H^+_{(aq)}]$ \n (mol dm^-3) \n	Initial rate (10^-3 mol dm^-3 s^-1)
1	$\\var{a}$	$\\var{a}$	$\\var{a}$	$\\var{d}$
2	$\\var{b}$	$\\var{a}$	$\\var{a}$	$\\var{x_1}$
3	$\\var{a}$	$\\var{c}$	$\\var{a}$	$\\var{f}$
4	$\\var{b}$	$\\var{a}$	$\\var{b}$	$\\var{g}$

", "advice": "

We determine the order of reaction with respect to each reactant according to the following:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Effect on reaction rate	Order
No change in rate	Zero
Rate changes in same ratio as concentration	First
Rate changes by square of concentration change	Second
Rate changes by cube of concentration change	Third

To determine the order of reaction with respect to $BrO_{3}^-$, we compare runs 1&2 since the concentrations of the other reactants remain constant and only $[BrO_{3(aq)}^-]$ varies.

We note that the concentration of $BrO_{3}^-$ increases by a factor of 2 (doubles) as does the rate.

So the reaction is first order with respect to $BrO_{3}^-$.

To determine the order of reaction with respect to $Br^-$, we compare runs 1&3 since the concentrations of the other reactants remain constant and only $[Br^-_{(aq)}]$ varies.

We note that the concentration of $Br^-$ increases by a factor of 3 as does the rate.

So the reaction is first order with respect to $Br^-$.

To determine the order of reaction with respect to $H^+$, we compare runs 2&4 since the concentrations of the other reactants remain constant and only $[H^+_{(aq)}]$ varies.

We note that the concentration of $H^+$ increases by a factor of 2 and the rate increases by a factor of 4 (2²).

So the reaction is second order with respect to $H^+$.

The overall reaction order is the sum of orders of each reactant.

The overall order of the reaction = 1 + 1 + 2 = 4

The rate equation is given by

$Rate=k[A]^m[B]^n$

Where,

$k$ = rate constant

$[A]$ = concentration of reactant $A$

$[B]$ = concentration of reactant $B$

$m$ = order with respect to reactant $A$

$n$ = order with respect to reactant $B$

The reaction is first order with respect to $BrO_{3}^-$, first order with respect to $Br^-$ and second order with respect to $H^+$.

Thus the rate equation is given by,

$Rate=k[BrO_3^-][Br^-][H^+]^2$

We determined the overall rate equation in part e), namely,

$Rate=k[BrO_3^-][Br^-][H^+]^2$ (1).

Rearranging equation (1) for $k$ gives,

$k=\\frac{Rate}{[BrO_3^-][Br^-][H^+]^2}$ (2).

Now we choose one line from the table, i.e., one run, and substitute values from this experiment into equation (2) to obtain a value for $k$.

Let's choose run 2, we have,

$k=\\frac{\\var{x_1}×10^{-3}}{\\var{b}×\\var{a}×\\var{a}^2}=\\var{k}=\\var{k_rounded}$ (3 s.f.)

Note: substituting the data from any of the 4 runs yields the same value for $k$, this is a helpful check!

We determined the overall rate equation in part e), namely,

$Rate=k[BrO_3^-][Br^-][H^+]^2$ (1).

Rearranging equation (1) for $k$ gives,

$k=\\frac{Rate}{[BrO_3^-][Br^-][H^+]^2}$ (2).

Now, substituting the units into equation (2) gives,

$k=\\require{cancel}\\frac{moldm^{-3}s^{-1}}{moldm^{-3}×moldm^{-3}×(moldm^{-3})^2}=\\frac{moldm^{-3}s^{-1}}{moldm^{-3}×moldm^{-3}×mol^2dm^{-6}}=\\frac{moldm^{-3}s^{-1}}{mol^4dm^{-12}}=\\frac{\\cancel{mol}\\cancel{dm^{-3}}s^{-1}}{mol^{\\cancel{4}3}dm^{\\cancel{-12}-9}}=dm^9mol^{-3}s^{-1}$

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What is the order of the reaction with respect to $BrO_{3}^-$?

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What is the order of the reaction with respect to $Br^-$?

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What is the order of the reaction with respect to $H^+$?

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What is the overall order of the reaction?

[[0]]

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Which of the following correctly represents the rate equation for this reaction?

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Find the value of the rate constant, $k$, at 298K.

[[0]]

Select the correct units for $k$.

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Consider the following reaction:

$CH_3COCH_{3(aq)}+Br_{2(aq)}+H^+_{(aq)}$→$CH_3COCH_2Br_{(aq)}+2H^+_{(aq)}+Br^-_{(aq)}$

A rate study for the reaction was conducted and the following results obtained:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

\n Run \n	\n Initial $[CH_3COCH_3]$ \n (mol dm^-3) \n	\n Initial $[Br_2]$ \n (mol dm^-3) \n	\n Initial $[H^+]$ \n (mol dm^-3) \n	\n Initial rate of reaction \n (mol dm^-3 s^-1) \n
1	$\\var{a}$	$\\var{a}$	$\\var{a}$	$\\var{c}$
2	$\\var{b}$	$\\var{a}$	$\\var{a}$	$\\var{d}$
3	$\\var{b}$	$\\var{b}$	$\\var{a}$	$\\var{d}$
4	$\\var{a}$	$\\var{a}$	$\\var{b}$	$\\var{d}$

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In order to deduce the rate equation for the reaction we must determine the order of the reaction with respect to each reactant.

We determine the order with respect to each reactant using the following:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Effect on reaction rate	Order
No change in rate	Zero
Rate changes in same ratio as concentration	First
Rate changes by square of concentration change	Second
Rate changes by cube of concentration change	Third

Consider the data from the rate study:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

\n Run \n	\n Initial $[CH_3COCH_3]$ \n (mol dm^-3) \n	\n Initial $[Br_2]$ \n (mol dm^-3) \n	\n Initial $[H^+]$ \n (mol dm^-3) \n	\n Initial rate of reaction \n (mol dm^-3 s^-1) \n
1	$\\var{a}$	$\\var{a}$	$\\var{a}$	$\\var{c}$
2	$\\var{b}$	$\\var{a}$	$\\var{a}$	$\\var{d}$
3	$\\var{b}$	$\\var{b}$	$\\var{a}$	$\\var{d}$
4	$\\var{a}$	$\\var{a}$	$\\var{b}$	$\\var{d}$

We will use runs 1&2 to investigate the order with respect to reactant $CH_3COCH_3$ since its concentration varies, but the concentrations of the other reactants remain constant.

We see that the rate of run 2 is double that of run 1. Note that this corresponds to the concentration of $CH_3COCH_3$ doubling also, and so the reaction is first order for $CH_3COCH_3$.

We will use runs 2&4 to investigate the order with respect to reactant $Br_2$ since its concentration varies, but the concentrations of the other reactants remain constant.

We see that the rates of runs 2&3 are the same despite the concentration of $Br_2$ doubling also, and so the reaction is zero order for $Br_2$.

We will use runs 1&4 to investigate the order with respect to reactant $H^+$ since its concentration varies, but the concentrations of the other reactants remain constant.

We see that the rate of run 4 is double that of run 1. Note that this corresponds to the concentration of $H^+$ doubling also, and so the reaction is first order for $H^+$.

The general form for a rate equation is given by,

$Rate=k[A]^m[B]^n$

Where,

$k$ is the rate constant
$m$ is the order for reactant $A$
$n$ is the order for reactant $B$

In part a) we determined that the reaction is first order for $CH_3COCH_3$ and $H^+$ and zero order for $Br_2$ .

Thus the rate law is given by,

$Rate=k[CH_3COCH_3][H^+]$

We determined the overall rate equation in part a), namely,

$Rate=k[CH_3COCH_3][H^+]$ (1).

Rearranging equation (1) for $k$ gives,

$k=\\frac{Rate}{[CH_3COCH_3][H^+]}$ (2).

Now we choose one line from the table, i.e., one run, and substitute values from this experiment into equation (2) to obtain a value for $k$.

Let's choose run 1, we have,

$k=\\frac{\\var{c}}{\\var{a}×\\var{a}}=\\var{ans}=\\var{ans_rounded}$ (3 s.f.)

Note: substituting the data from any of the 4 runs yields the same value for $k$, this is a helpful check!

The overall reaction order is the sum of orders of each reactant.

The overall order of the reaction = 1 + 1 + 0 = 2

So overall, the reaction is second order.

We determined the overall rate equation in part b), namely,

$Rate=k[CH_3COCH_3][H^+]$ (1).

Rearranging equation (1) for $k$ gives,

$k=\\frac{Rate}{[CH_3COCH_3][H^+]}$ (2).

Now, substituting the units into equation (2) gives,

$\\require{cancel}k=\\frac{moldm^{-3}s^{-1}}{moldm^{-3}×moldm^{-3}}=\\frac{moldm^{-3}s^{-1}}{mol^2dm^{-6}}=\\frac{\\cancel{mol}dm^{\\cancel{{-3}}}{^3} s^{-1}}{mol^{\\cancel{2}1}\\cancel{dm^{-6}}}=dm^3mol^{-1}s^{-1}$

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Which correctly represents the rate equation for the reaction?

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Determine the value of the rate constant for the reaction.

$k=$[[0]]

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What is the overall order of the reaction?

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Select the correct units for the rate constant, $k$.

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We can determine the overall order of a reaction using the units of the rate constant, $k$.

Recall that,

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Overall order	Units for $k$
Zero	mol L^-1 s^-1
First	s^-1
Second	L mol^-1 s^-1
Third	L² mol^-2 s^-1

Though this table is handy, it is useful to be able to calculate the units for $k$ given the overall reaction order.

The general rate equation for a zero order reaction is

$Rate=k[X]^0$, which simplifies to

$Rate=k$, i.e., the units for the rate constant, $k$, are the same as those for rate: $molL^{-1}s^{-1}$.

The general rate equation for a first order reaction is

$Rate=k[X]$, which can be rearranged for $k$, we have

$k=\\frac{Rate}{[X]}$

Now substituting in the units for rate and concentration gives,

$\\require{cancel}k=\\frac{{molL^{-1}}s^{-1}}{molL^{-1}}=\\frac{\\cancel{molL^{-1}}s^{-1}}{\\cancel{molL^{-1}}}=s^{-1}$

The general rate equation for a second order reaction is

$Rate=k[X]^2$, which can be rearranged for $k$, we have

$k=\\frac{Rate}{[X]^2}$.

Now substituting in the units for rate and concentration gives,

$\\require{cancel}k=\\frac{molL^{-1}s^{-1}}{(molL^{-1})^2}=\\frac{molL^{-1}s^{-1}}{mol^2L^{-2}}=\\frac{\\cancel{mol}\\cancel{L^{-1}}s^{-1}}{mol^\\cancel{2}L^{\\cancel{-2}1}}=Lmol^{-1}s^{-1}$

The general rate equation for a third order reaction is

$Rate=k[X]^3$, which can be rearranged for $k$, we have

$k=\\frac{Rate}{[X]^3}$.

Now substituting in the units for rate and concentration gives,

$\\require{cancel}k=\\frac{molL^{-1}s^{-1}}{(molL^{-1})^3}=\\frac{molL^{-1}s^{-1}}{mol^3L^{-3}}=\\frac{\\cancel{mol}\\cancel{L^{-1}}s^{-1}}{mol^{\\cancel{3}2}L^{\\cancel{-3}-2}}=L^2mol^{-2}s^{-1}$

We are told that the rate constant, k, has units s^-1.

Thus the reaction is first order overall.

We are told that the rate constant, k, has units L² mol^-2 s^-1.

Thus the reaction is third order overall.

We are told that the rate constant, k, has units L mol^-1 s^-1.

Thus the reaction is seoncd order overall.

We are told that the rate constant, k, has units s^-1.

Thus the reaction is first order overall.

A reaction for which the units of $k$ are L⁴ mol^-4 s^-1 is fifth order overall.

An overall fifth order reaction has general rate equation

$Rate=k[X]^5$, which can be rearranged for $k$ giving

$k=\\frac{Rate}{[X]^5}$

Now substituting in the units for rate and concentration gives,

$\\require{cancel}k=\\frac{molL^{-1}s^{-1}}{(molL^{-1})^5}=\\frac{molL^{-1}s^{-1}}{mol^5L^{-5}}=\\frac{\\cancel{mol}\\cancel{L^{-1}}s^{-1}}{mol^{\\cancel{5}4}L^{\\cancel{-5}-4}}=L^4mol^{-4}s^{-1}$

The following equation describes the gas phase decomposition of dinitrogen pentoxide:

$N_2O_5$→$NO_2^{•}+NO_3^{•}$

The rate constant is given by $k=2.53×10^{-2}$ s^-1, what is the overall order of the reaction?

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Consider the following complex reaction:

$2NO_{(g)}+2H_{2(g)}$→$N_{2(g)}+2H_2O_{(g)}$

The rate constant is given by $k=1.25$ L² mol^-2 s^-1, what is the overall order of the reaction?

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The decomposition of difluorine oxide is described by the following balanced equation:

$2F_2O_{(g)}$→$2F_{2(g)}+O_{2(g)}$

The reaction is found to have rate constant, $k=0.35$ L mol^-1 s^-1. What is the overall order of the reaction?

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Consider the following reaction of cyclobutane decomposing to ethane:

$C_4H_{8(g)}$→$2C_2H_{4(g)}$

The rate constant is given by $k=0.00014$ s^-1, what is the overall order of the reaction?

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Suppose the rate constant, $k$, for a reaction is found to have units: L⁴ mol^-4 s^-1. What would the overall order of reaction be?

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