Chemical Reactions

• The following table summarizes the common Driving forces that tend to make reactions go in the direction of the arrow:

Reactants Products

• When two or more chemicals are brought together, if any of the above can occur, a chemical reaction is likely to take place!

Formation of a solidFormation of waterFormation of gasTransfer of electrons

CHEMISTRYCHEMISTRY - science of properties and transformations of matter

CHEMICAL REACTIONS - central to CHEMSITRY

CLASSIFICATION of REACTIONS

Precipitation

Oxidation-reduction

Acid-base

Gas forming

Precipitation ReactionsSolubility Rules

• There are simple guidelines that allow one to predict whether or not a reaction will produce an insoluble compound: these are called the Solubility Rules

Ionic Equations:Net Ionic

• The ions that appear on both sides of the equation in identical forms. These ions are not involved directly in the reaction and are called spectator ions.

• In the net ionic equation, the spectator ions are omitted leaving only the ions directly involved in the reaction.

PbPb2+2+(aq)(aq) + 2NO + 2NO33--(aq) + 2K(aq) + 2K++(aq) + (aq) + 22II--(aq)(aq) PbIPbI22 (s) (s) + 2K + 2K++(aq) + (aq) +

2NO2NO33--(aq)(aq)

spectator ionsspectator ions

PbPb2+2+(aq)(aq) + 2 + 2II--(aq)(aq) PbIPbI22 (s) (s)

Ionic Equations:Summary

• Summary of the 3 different chemical equations:– Molecular equation

– Complete ionic equation

– Net ionic equation

Pb(NOPb(NO33))22(aq) + 2KI(aq) (aq) + 2KI(aq) PbI PbI22 (s) +2KNO (s) +2KNO33(aq)(aq)

PbPb2+2+(aq)(aq) + 2NO + 2NO33--(aq) + 2K(aq) + 2K++(aq) + (aq) + 22II--(aq)(aq) PbIPbI22 (s) (s) + 2K + 2K++(aq) + (aq) +

2NO2NO33--(aq)(aq)

PbPb2+2+(aq)(aq) + 2 + 2II--(aq)(aq) PbIPbI22 (s) (s)

Acid-Base Reactions:Neutralization

• Consider the reaction between Mg(OH)2 [milk of magnesia] and HCl (stomach acid)

• The net reaction is:

)(2)()()()(2 222 lOHaqMgClaqOHMgaqHCl

)()()( 2 lOHaqOHaqH

The acid has been neutralized

و اکسایش عدد دارای یونهایمتفاوت کاهش

Half-Reaction Method OfBalancing Redox Equations

• Assign oxidation numbers to each atom and identify the species undergoing oxidation and reduction. Write two skeleton half-reactions.

• oxidation # increase => oxidation; oxidation # decrease => reduction

• (1) Balance the numbers of atoms in each half-reaction.• (a) Balance all atoms except O and H.• (b) Balance O atoms by adding H2O molecules.• (c) Balance H atoms by adding H+ ions. • (2) Balance electric charge by adding electrons to more positive

side.• (3) Combine two half-reactions to obtain balanced equation.• (a) Multiply each half-reaction by factors that make the

number of • electrons in two half-reactions equal.• (b) Combine the adjusted half-equations into an overall

equation.• Simplify the balanced equation if necessary.• Verify that the equation is balanced.

• Balance the following redox equation in acidic solution:• Fe2+ (aq) + Cr2O7

2- (aq) Fe3+ (aq) + Cr3+ (aq)• Fe2+ (aq) Fe3+ (aq)+e• Cr2O7

2- (aq) +e Cr3+ (aq)

شیمیایی سینتیک تعریف

خود ترمودینامیک• پیرامون که است علمی شیمیایییا فیزیکی تغییر یک نبودن بخود خود یا بودن بخود

انرژی شیمیایی نظر نماید از می مطالعهمطالعه سینتیک• تندی شیمیایی یا های سرعت واکنش

. یک سرعت تعیین در کمی عوامل است شیمیایی . کمک ما به عوامل این مطالعه دارند دخالت واکنشمواد تبدیل نحوه پیرامون نخهایی سر تا کند میشیمیایی واکنش در ها فراورده به دهنده واکنش . محصول به اولیه مواد تبدیل مسیر آوریم بدست

. دارد نام واکنش مکانیسم

واکنش سرعت تعریفReaction rate: changes in a concentration of a product or a reactant per unit time.

[ ] concentrationReaction rate = ——

t

[ ]

t

[ ]

t

change

Define reaction rate and explainAverage reaction rateInstantaneous reaction rate(2 tangents shown)Initial reaction rate

واکنش نسبی سرعتFor a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation.

Note the expression and reasons for their relations for the reaction

2 NO + O2 (g) = 2 NO2 (g)

[O2] 1 [NO] 1 [NO2] Reaction rate = – ——— = – — ———— = — ———

t a t b t

Make sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for

2 N2O5 = 4 NO2 + O2

How can the rate expression be unique and universal?

Calculating reaction rateThe concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 s after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for

2 N2O5 = 4 NO2 + O2

Solution: (0.93 – 1.24)e-2 – 0.31e-2 MDecomposition rate of N2O5 = – ———————— = – ——————

1200 – 600 600 s

= 5.2e-6 M s-1.

Note however, rate of formation of NO2 = 1.02e-5 M s-1.rate of formation of O2 = 2.6e-6 M s-1.

The reaction rates are expressed in 3 forms

Be able to do this type problems

قانون یا واکنش کلی سرعت Dependence of reaction rate on the concentrations of reactants isسرعت

called the rate law, which is unique for each reaction.

For a general reaction, a A + b B + c C products

the rate law has the general form

order wrt A, B, and C, determined experimentally reaction rate = k [A]X [B]Y [C]Z

the rate constant

For example, the rate law is

rate = k [Br-] [BrO3-] [H+] for

5 Br- + BrO3- + 6 H+ 3Br2 + 3 H2O

The reaction is 1st order wrt all three reactants, total order 3.

Use differentials to express rates

غلظت- 1 عبارتهای توانهای مجموعمرتبه سرعت قانون معادله در

شود می نامیده واکنشمرتبه- 2 صورت دو به واکنش مرتبه

گردد می بیان جزئی مرتبه یا کلیواکنش- 3 یک جزئی یا کلی مرتبه

توان می آزمایش طریق از تنها راآورد بدست

واکنش مرتبه

مرتبه با واکنش برای سرعت تغییرات نمودارمختلف های

First order, rate = k [A]

k = rate, 0th order

[A]

rate

2nd order, rate = k [A]2

The variation of reaction rates as functions of concentration for various order is interesting.

Mathematical analysis is an important scientific tool, worth noticing.

[A] = ___?

مثالبدست را واکنشگرها تک تک به نسبت واکنش مرتبه زیر واکنش در

. در نمایید محاسبه را واکنش سرعت ثابت مقدار سپس آوردهواکنش سرعت باشد موالر یک واکنشگرها تمام غلظت که صورتی

است؟ چقدر

H2O2 + 3 I- + 2 H+ I3- + 2 H2O

Exprmt [H2O2] [I-] [H+] Initial rate M s-1

1 0.010 0.010 0.0050 1.15e-6 2 0.020 0.010 0.0050 2.30e-6 3 0.010 0.020 0.0050 2.30e-6

4 0.010 0.010 0.0100 1.15e-6

Solution next

R = k [H2O2]1 [I-]y [H+]z

R = k [H2O2]x [I-]y [H+]z

R = k [H2O2]1 [I-]1 [H+]z

R = k [H2O2]1 [I-]1 [H+]0

R = 1.15e-6 = k (0.010)(0.010)

k =0.0115 M-1 s-1

R = k [H2O2] [I-]=0.0115(1)(1)=0.0115

Differential Rate Law determination - continue

From the following reaction rates observed in 4 experiments, derive the rate law for the reaction A + B + C products where reaction rates are measured as soon as the reactants are mixed.

Expt 1 2 3 4 [A]o 0.100 0.200 0.200 0.100[B]o 0.100 0.100 0.300 0.100 [C]o 0.100 0.100 0.100 0.400 rate 0.100 0.800 7.200 0.400

Solution nextThis example illustrates the strategy to determine, and a reliable method to solve rate-law experimentally.

Differential Rate Law determination - continue

From the following reaction rates, derive the rate law for the reaction A + B + C products where reaction rates are measured as soon as the reactants are mixed.

Expt 1 2 3 4 order[A]o 0.100 0.200 0.200 0.100 3 from expt 1 & 2

[B]o 0.100 0.100 0.300 0.100 2 expt 1, 2 & 3 [C]o 0.100 0.100 0.100 0.400 1 expt 1 & 4 rate 0.100 0.800 7.200 0.400

0.800 k 0.2x 0.1y 0.1z

----- = ----------------------0.100 k 0.1x 0.1y 0.1z

Therefore 8 = 2x

log 8 = x log 2x = log 8 / log 2 = 3

Assume rate = k [A]x[B]y[C]z

سرعت قانون انتگرالی و مشتقی نمایش

مشتقی نمایش انتگرالی نمایش

R= – d[A] / dt = k [A] = [A]o – k t

d[A]R= – —— = k [A] [A] = [A]o e – k t or ln [A] = ln [A]o – k t

d t

d[A] 1 1 [A] conc at t R= – —— = k [A]2 —— – —— = k t

d t [A] [A]o [A]o conc at t = 0

Describe, derive and apply the integrated rate laws

Learn the strategy to determine rate-law

سرعت در غلظت تغییرات نموداراول مرتبه

Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at various times.

[A]

t

ln[A]

t

[A] = [A]o e – k t ln [A] = ln [A]o – k t

t½

دوم مرتبه سرعت در غلظت تغییرات نمودار

t = 1 5 10 15 30 35

[B] = 1.67 1.0 0.67 0.50 0.29 0.25[B] 1

— [B]

t t

1 1——= k t + ——[B] [B]o

[B]o

[B] = ——————[B]o k t + 1

What kind of plot is linear for 1st and 2nd reactions?

عمر نیمهThe time required for half of A to decompose is called half life t1/2.

Since [A] = [A]o e – k t or ln [A] = ln [A]o – k t

When t = t1/2, [A] = ½ [A]o

Thus ln ½ [A]o = ln [A]o – k t1/2

– ln 2 = – k t1/2

k t1/2 = ln 2 = 0.693 relationship between k and t1/2

Radioactive decay usually follow 1st order kinetics, and half life of an isotope is used to indicate its stability.

Evaluate t½ from k or k from t½

1st order reaction calculationN2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.

Solution next

If the rate-law is known, what are the key parameters?

1st order reaction calculationN2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.

Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or 0.9 = 1.0 e – k t apply [A]o = [A] e– k t ln 0.9 = ln 1.0 – k 30 s– 0.1054 = 0 – k * 30 k = 0.00351 s – 1

t½ = 0.693 / k = 197 s apply k t ½ = ln 2

[A] = 1.0 e – 0.00351*500 = 0.173Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 %

After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed.After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed.

Apply integrated rate law to solve problems

Typical Problem wrt 1st Order Reaction

The decomposition of A is first order, and [A] is monitored. The following data are recorded:

t / min 0 2 4 8[A]/[M] 0.100 0.0905 0.0819 0.0670

Calculate k (What is the rate constant? k = 0.0499)Calculate the half life (What is the half life? Half life = 13.89)Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?)Calculate t when [A] = 0.0100. (Estimate the time required for 90% of A to decompose.)

Work out all the answers

های مرتبه با واکنش مقایسهمختلف

صفر مرتبه یک مرتبه دو مرتبه

سرعت قانون

سرعت ثابت واحد

راست خط معادله

از عرض و شیبمبدا

عمر نیمه

The Arrhenius Equation

1903 Nobel Prize citation” …in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation”

The temperature dependence of the rate constant k is best described by the Arrhenius equation:

k = A e – Ea / R T

or ln k = ln A – Ea / R T

If k1 and k2 are the rate constants at T1 and T2 respectively, then

k1 Ea 1 1 ln —— = – — — – — k2 R T1 T2

How does temperature affect reaction rates?

Derive and apply these relationship to solve problems, and recall the Clausius-Clapeyron equation.

Energy in chemical reactionsPotential energy

Progress of reaction

R + A

P + DH exothermic

Ea Ea for reverse reaction

Endothermic rxn

RA-PDactivated complex

Explain the various terms and energy changes in a chemical reaction

تفسیر برای موجود های نظریهسینتیکی های پدیده

برخورد -1 نظریهCollision theory

گذار -2 حالت نظریهTransition state theory

Ea

A---B---C

A + BC

AB + CReactants

Products

Potential Energy

Reaction Progress

The height of the barrier is called the

activation energy, Ea.

The configuration of atoms at the

maximum in the P.E. profile is called

the transition state.

A2(g) + B2(g) 2AB(g)

فعال کمپلکس تشکیل فرضیه

موثر برخورد

نظر از واکنشها انواعمکانیسمی

بنیادی - 1 مرحله واکنشهای یک از تنها که هستند واکنشهاییضرایب مجموع برابر واکنش مرتبه آنها در و اند شده تشکیل

. است واکنش استوکیومتری

O3 O2 + O rate = k [O3]

NO2 + NO2 NO3 + NO rate = k [NO2]2

بنیادی - 2 غیر مراحل واکنشهای آنها در که واکنشهاییدر گویند را دارد جود و محصول به اولیه مواد تبدیل برای مختلفی

نمی برابر استوکیومتری ضرایب با واکنش مرتبه واکنشها گونه این. باشند می نوع این از واکنشها همه تقریبا باشد

کمک به و بوده فرضی واقعی، واکنش یک در بنیادی واکنشهایواکنش سرعت قانون و واسطها حد تشخیص توان مکانیسم، می

برد پی آنها صحت به

NO2 + CO NO + CO2

Step 1 NO2 + NO2 NO3 + NO (an elementary reaction)

Step 2 NO3 + CO NO2 + CO2 (an elementary reaction)

مولکوالریته

را بنیادی واکنش یک در واکنش کل گویند مولکوالریتهمرتبه

را باشد یک واکنش مرتبه آن در که که بنیادی تک واکنشهاییمولکولی

را باشد دو واکنش مرتبه آن در که که بنیادی دو واکنشهاییمولکولی

را باشد سه واکنش مرتبه آن در که که بنیادی واکنشهایی سه وگویندمولکولی

Molecularity of elementary reactions - Example

Some elementary reactions for the reaction between CH4 and Cl2 are

Cl2 2 Cl

2 Cl Cl2

2Cl + CH4 Cl2 + CH4*

Cl + CH4 HCl + CH3

CH3 + Cl CH3Cl

CH3 + CH3 CH3-CH3

CH3Cl + Cl HCl + CH2Cl

CH2Cl + Cl CH2Cl2

* * * (more)

Write down the rate laws and describe them as uni- bi- or ter-molecular steps yourself, please.

Rate Laws and MechanismsA mechanism is a collection of elementary steps devise to explain the the reaction in view of the observed rate law. You need the skill to derive a rate law from a mechanism, but proposing a mechanism is task after you have learned more chemistry

For the reaction, 2 NO2 (g) + F2 (g) 2 NO2F (g), the rate law is,

rate = k [NO2] [F2] .

Can the elementary reaction be the same as the overall reaction?

If they were the same the rate law would have been

rate = k [NO2]2 [F2],

Therefore, they the overall reaction is not an elementary reaction. Its mechanism is proposed next.

Rate-determining Step in a MechanismThe rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction.

The (determined) rate law is, rate = k [NO2] [F2],for the reaction, 2 NO2 (g) + F2 (g) 2 NO2F (g), and a two-step mechanism is proposed:

i NO2 (g) + F2 (g) NO2F (g) + F (g)ii NO2 (g) + F (g) NO2F (g)

Which is the rate determining step?

Answer:The rate for step i is rate = k [NO2] [F2], which is the rate law, this suggests that step i is the rate-determining or the s-l-o-w step.

Explain rate determining step in a mechanism and use it to derive the rate law.

2مثال

The decomposition of H2O2 in the presence of I– follow this mechanism,

i H2O2 + I– k1 H2O + IO– slow ii H2O2 + IO– k2 H2O + O2 + I– fast

What is the rate law? Energy

Solve the problem

Eai

Eaii

reaction

The decomposition of H2O2 in the presence of I– follow this mechanism,

i H2O2 + I– k1 H2O + IO– slow ii H2O2 + IO– k2 H2O + O2 + I– fast

What is the rate law?

Solution

The slow step determines the rate, and the rate law is:

rate = k1 [H2O2] [I –]

Since both [H2O2] and [I –] are measurable in the system, this is the rate law.

پایان 24تمرین ( 3مثال فصل)

صورت به سرعت معادله زیر واکنش درN2O5+NO 3NO2

است زیر صورت به پیشنهادی مکانیسم است

پایا حالت تقریب به توجه با و بوده واکنش کند مرحله سوم مرحلهکند می تایید را سرعت معادله همان فوق مکانیسم که دهید نشان

که صورتی مینماید k2<<k3در تغییری چه سرعت معادله باشد

][][

]][[

322

523152 NOkNOk

NOONkkR ON

23

3

522

32

321

52

2NONONO

ONNONO

NONOON

k

k

k

حل راهمشخص- 1 مسئله در چون واکنش سرعت معادله نوشتن

این بنابر است واکنش کند مرحله مرحله کدام که نشدهنظر در سرعت کننده تعیین مرحله عنوان به آخر مرحله

شود می گرفتهواکنش = k3[NO][NO3]سرعت

پایا -2 حالت که: تقریب شود می فرض تقریب این درو آن شدن تولید سرعت و است ثابت واسط حد غلظت

پس باشد برابر شدنش مصرف سرعت[NO3] مصرف تولید [NO3]= سرعت سرعت

K1[N2O5]=[NO3] تولید سرعت

K2[NO2][NO3]+k3[[NO][NO3]=[NO3] مصرف سرعت

واسط -3 حد مصرف و تولید سرعت دادن قرار برابر باداشت خواهیم

سرعت رابطه در واسط حد غلظت کردن باجانشینداشت خواهیم

که داشت k2<<k1 درصورتی خواهیم

][][

][][

322

5213 NOkNOk

ONkNO

][][

]][[

322

523152 NOkNOk

NOONkkR ON

][][ 52523

3152

ONkONk

kkR ON

Deriving rate laws from mechanisms – steady-state approximation

The steady-state approximation is a general method for deriving rate laws when the relative speed cannot be identified.It is based on the assumption that the concentration of the intermediate is constant.

Rate of producing the intermediate, Rprod, is the same as its rate of consumption, Rcons.

Rprod = Rcons

[Intermediate]

time

Rprod < RconsRprod > Rcons

Be able to apply the steady-state approximation to derive rate laws

Steady-state approximatio

n - 2

Let’s assume the mechanism for the reaction. H2 + I2 2 HI

as follows.Step (1) I2 —k1 2 IStep (1) 2 I —k-1 I2

Step (2) H2 + 2 I —k2 2 HIDerive the rate law.Derivation:

rate = k2 [H2] [I] 2 (‘cause this step gives products)but I is an intermediate, this is not a rate law yet.Since k1 [I2] (= rate of producing I)

= k-1 [I]2 + k2 [H2] [I]2 (= rate of consuming I)

Thus, k1 [I2] [I]2 = ——————

k-1 + k2 [H2]

rate = k1 k2 [H2] [I2] / {k-1 + k2 [H2] }

Steady state

Steady-state approximation - 3From the previous result:

k1 k2 [H2] [I2]rate = ———————

{k-1 + k2 [H2] }

Discussion:

(i) If k-1 << k2 [H2] then {k-1 + k2 [H2]} = k2 [H2] ,then rate = k1 k2 [H2] [I2] / {k2 [H2] } = k1 [I2] (pseudo 1st order wrt I2)using large concentration of H2 or step 2 is fast (will meet this condition).

(ii) If step (2) is slow, then k2 << k1, and if [H2] is not large, we have {k-1 + k2 [H2]} = k-1 and rate = k1 k2 [H2] [I2] / k1 = k2 [H2] [I2]

Steady-state approximation - 4In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O8

2- + SO32- + 2 OH- 3 SO4

2- + H2O.The following mechanism has been proposed:

i S2O82- + SO3

2- —k1 S2O72- + SO4

2- ii S2O7

2- + H2O —k2 2 SO42- + 2 H+

iii H+ + OH- —k3 H2O (fast equilibrium to be discussed)

Steady-state approximation follows these steps:

What is or are the intermediates I?Use which step to give the rate law that may involve [I]? Express the rates of producing and consuming intermediate(s)Express [I] of intermediate(s) in terms of [Reactants] Derive the rate law in terms of [Reactants]

Discuss See page 607 PHH Text

Catalysis

A catalyst is a substance that changes the rate of a reaction by lowing the activation energy, Ea. It participates a reaction in forming an intermediate, but is regenerated.

Enzymes are marvelously selective catalysts.

A catalyzed reaction,NO (catalyst)

2 SO2 (g) + O2 — 2 SO3 (g)via the mechanism i 2 NO + O2 2 NO2 (3rd order) ii NO2 + SO2 SO3 + NO

Uncatalyzed rxn

Catalyzed rxn

rxn

Energy

Catalyzed decomposition of ozoneR.J. Plunkett in DuPont discovered carbon fluorine chlorine compounds.

The CFC decomposes in the atmosphere:CFCl3 CFCl2 + ClCF2Cl3 CF2Cl + Cl.

The Cl catalyzes the reaction via the mechanism: i O3 + h v O + O2, ii ClO + O Cl + O2

iii O + O3 O2 + O2.The net result or reaction is

2 O3 3 O2

Scientists sound the alarm, and the CFC is banned now.

Homogenous vs. heterogeneous catalysts

A catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst. It effective, but recovery is difficult.

When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Chemisorption, absorption, and adsorption cause reactions to take place via different pathways.

Platinum is often used to catalyze hydrogenation

Catalytic converters reduce CO and NO emission.

Heterogeneous catalystsCeryx's vision is to design, produce, and commercialize advanced systems that balance Cost, Performance, Emissions Reduction, and Fuel Penalty to make the economics of pollution control viable.

We explore new ways to look at the air quality challenges faced by industry and search for potential solutions by combining proven technologies with state-of-the-art science.

Catalyzed reactions:

CO + O2 CO2

2 NO N2 + O2

Enzymes – selective catalystsEnzymes are a long protein molecules that fold into balls. They often have a metal coordinated to the O and N sites.

Molecules catalyzed by enzymes are called substrates. They are held by various sites (together called the active site) of the enzyme molecules and just before and during the reaction. After having reacted, the products P1 & P2 are released.

Enzyme + Substrate ES (activated complex)

ES P1 + P2 + E

Enzymes are biological catalysts for biological systems.

                                                                                      

      

X-ray 3-D structure of fumarate reductase. It reduces fumerate, an important role in the metabolism of anaerobic bacteria, from Max Planck Inst.

Chemical Kinetics - SummaryExplain how the various factors affect reaction rates.

Define reaction rates, average rates, initial rates and rate constants.Evaluate rate law from experiments

Properly apply 1st and 2nd differential rate laws and integrated rate laws.

Interpret elementary reactions and mechanisms. Derive rate laws from a given mechanism. Apply the steady-state method to derive the rate law of a given mechanism, and discuss the results.

Explain the action of catalysts in terms of chemistry and in terms of energy of activation.

Fig. 13.21

Fig. 13.23

Fig. 13.26

A) What is the Rate Law For This Reaction? B) What is the order in respect to each reactant?C) What is the overall reaction order? D) What is k, the rate constant?