Order and rate law of a reactionTo understand how elementary steps make up a mechanism and howthe rate law for an elementary step can be determined.The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular.Furthermore, the rate law can be determined directly from the number of each type of molecule in an elementary step. For example, the rate law for step 1 isVery often, a reaction does not tell us the whole story. For instance,the reactionrate =k[NO2]?NO2(9) + CO(9)¬NO(g)+CO2(g)The exponent "2" is used because the reaction involves two NO2 molecules. The rate law for step 2 isdoes not involve a collision between an NO2 molecule and aCO molecule. Based on experimental data at moderate temperatures,this reaction is thought to occur in the following two steps:rate = k[NO3]"[CO]' = k[NO3][CO]because the reaction involves only one molecule of each reactant the exponents are omitted.1. NO2 (g) + NO2(9)→NO3(9)+NO(g)2. NO3 (9) + CO(9)→CO2(9)+NO2(9)Analyzing a new reactionEach individual step is called an elementary step. Together, theseelementary steps are called the reaction mechanism.Consider the following elementary steps that make up the mechanism of a certain reaction:Overall, the resulting reaction is1. 2X→Y+Z2. Y+ 2L→M+ZNO2 (g) + CO(g)→NO(g)+CO2(g)Notice that in the elementary steps NO3 appears both as a productand then as a reactant; therefore it cancels out of the final chemicalequation. NO3 is called a reaction intermediate. Also notice that 2molecules of NO2 appear in the reactants of the first step and 1molecule of NO2 appears as product of the second step, the neteffect leaves only 1 molecule of NO2 as a reactant in the netequation.Part AWhat is the overall reaction?Express your answer as a chemical equation.• View Available Hint(s)Molecularity is the proper term for "how the molecules collide" in areaction. For example, step 1 is bimolecular because it involves thecollision of two molecules. Step 2 is also bimolecular for the samereason. Unimolecular reactions involve only one molecule in thereactants. Though rare, collisions among three molecules can occur.ΑΣφSuch a reaction would be called termolecular. To understand how elementary steps make up a mechanism and howthe rate law for an elementary step can be determined.Part CVery often, a reaction does not tell us the whole story. For instance,the reactionWhat is the rate law for step 1 of this reaction?NO2 (g) + CO(g)→NO(g)+CO2(g)Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C]³ type k*[A]*[c]^3.does not involve a collision between an NO2 molecule and aCO molecule. Based on experimental data at moderate temperatures,this reaction is thought to occur in the following two steps:• View Available Hint(s)Rate =1. NO2 (g) + NO2(9)→NO3(9)+NO(g)2. NO3 (g) + CO(9)→CO2(g) +NO2 (9)SubmitEach individual step is called an elementary step. Together, theseelementary steps are called the reaction mechanism.Overall, the resulting reaction isPart DNO2(9) + CO(g)→NO(g)+CO2(g)Notice that in the elementary steps NO3 appears both as a productWhat is the rate law for step 2 of this reaction?and then as a reactant; therefore it cancels out of the final chemicalequation. NO3 is called a reaction intermediate. Also notice that 2molecules of NO2 appear in the reactants of the first step and 1molecule of NO2 appears as product of the second step, the neteffect leaves only 1 molecule of NO2 as a reactant in the netequation.Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C]3 type k*[A]*[c]^3.• View Available Hint(s)Rate =Molecularity is the proper term for "how the molecules collide" in areaction. For example, step 1 is bimolecular because it involves thecollision of two molecules. Step 2 is also bimolecular for the samereason. Unimolecular reactions involve only one molecule in thereactants. Though rare, collisions among three molecules can occur.SubmitSuch a reaction would be called termolecular.

The two elementary step give are as follow 1) 2X → Y + Z 2) Y + 2L → M + Z Part A : 1) For…

Analyzing a new reaction Consider the following elementary steps that make up the mechanism of a certain reaction: 1. 2X→Y+Z 2. Y+ 2L¬M+Z Part A What is the overall reaction? Express your answer as a chemical equation. > View Available Hint(s)

Analyzing a new reaction Consider the following elementary steps that make up the mechanism of a certain reaction: 1. 2X→Y+Z 2. Y+ 2L¬M+Z Part A What is the overall reaction? Express your answer as a chemical equation. > View Available Hint(s)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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67) A reaction is carried out in a container. A pure compound X is shown at the start of the experiment. Compound X undergoes a first order reaction to form compound Y, with no reverse reaction. Which of the following might represent the container at 20 minutes ? The diagram shows the number of molecules of each sample X and Y. t=0 mins 1-10 mins X 15 X=30 Y=0 Y = 15 t = 20 mins X = 15 Y = 15 Thaum Whic A) X = 7 Y = 23 C) D) X = 23 Y = 7 X = 0 Y = 30
I need help determining the orders please. Thank you
PLEASE ANSWER EVERYTHING ON THE PICTURES, THANK YOU
1) NO(g) + O2 (g) >< NO3 (g) 2) NO3(g) + NO(g) > < 2NO2(g0 for the reaction mechanism given, what is the rate law for each elementary equation? What is the rate law for the overall reaction if the mechanism is correct?
Nitrous oxide, NO, reacts with hydrogen to produce nitrogen and water. The initial rate data obtained for this reaction are given below: NO(g) + H2(g) –→ N2(g) + H2O(l) Using this data, determine the experimental rate law and calculate the rate constant. Experiment [NO] / mol L-1 [H2] / mol L-1 Initial Rate / mol L-1 s-1 1 0.420 0.122 0.136 0.210 0.122 0.0339 3 0.210 0.244 0.0678
1) Determine the order of the reaction and the rate constant. 2) What is the half-life of this reaction? 3) What is the concentration of the reactant after 2.00 hours of the experiment?
19. The initial rate of reaction between Substance A and Substance B was measured at a series of concentrations of each reactant with the following results. Solve for the rate law expression of this reaction. Note: Include all rough notes, show all calculations, follow the rules of significant figures, and box your final answer. ( A + B → AB Trial Number Trial #1 Trial #2 Trial #3 [A] 0.10 M 0.30 M 0.30 M [B] 0.10 M 0.10 M 0.20 M Initial rate 5.0 X 104 1.5 X 103 6.0 X 103 (mol/L·s)
The data for the reaction given NOCI (g) The following experimental data was measured for the reaction. Use the Method of Initial Rates to determine the rate law equation for this reaction 1 2 3 NO (g) + 1/2 Cl2₂ (g) [NO] [Cl₂] Initial Rate Experiment (mol.L-¹) (mol.L-¹) (mol.L-¹ s¯¹) 1.4x10-6 2.8x10-6 5.6x10-6 0.250 0.250 0.500 0.250 0.500 0.250 → 1. Determine the order of the reaction with respect to NO oxide 2. Determine the order of the reaction with respect to Cl₂ 3. What is the overall order of this reaction? 5. Calculate the value of the rate constant, k → 4. Select the correct rate law (rate equation) for this reaction: Rate =
A student wishes to experimentally determine the rate law of the following reaction. A+B+C+D The following tables demonstrate four students' plans for how to mix reactants and measure the rate of the reaction. Which can be used to determine the order of the reaction with respect to A and B? I. Trial 1 [A]initial 1M [B]initial 1M Trial 1 [A]initial 1 M [B]initial 3 M 2 1 M 1M III. 2 2 M 3 M 3 1 M 1 M 3 2 M 2 M Trial II. 1 1 M 2 1 M 1M 3 4 M 1 M [A]initial 2 M [B]initial IV. Trial [A]initial 1 1M [B]initial 1 M 2 1 M 1 M 3 2 M 2 M Table I only Table II only Table I & IV Table III only ○ Table II & III I don't know
Mechanism Suppose a reaction occurs with the following mechanism. All are elementary steps. Step 1: 0₂(9) 20(g) fast K₂ Step 2: O(g) + N₂(g) Step 3: O(g) + N(g) گی 3 NO(g) + N(g) slow very fast K3, NO(g) $ 4 % 5 6 What is the rate law for the rate-determining step? Rate =k,[0₂]² Rate = kr[O]² Rate-k,[O][N] Rate = kf[0₂]/k,[0]² Rate = k₂[O][N₂] Rate = K[0₂] & 7 What is the rate law for the forward reaction in step 1? Rate = kr[O]² Rate = kr[0₂] Rate =k+[0₂1² Rate = kr[0₂1² Rate = kf[0]² Rate =k+[0₂] What is the overall rate law for this reaction? Rate = k[N₂][0₂][N][0] MacBook Pro 8 ( 9 ☆ 88
The following initial rate data are for the reaction of ICI with hydrogen: 2IC1 + H₂ → I₂ + 2HCl Experiment [IC]o, M [H₂], M 0.198 0.141 0.396 0.141 0.198 0.281 0.396 0.281 1 2 3 4 Complete the rate law for this reaction in the box below. Use the form k[A] [B]", where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n. Rate = k = -2 M-². Initial Rate, M. s-¹ S 0.00657 0.0131 0.0261 0.0522 -1 S