AP Chemistry Unit 5
What graph will be linear for a second order reaction?
1/[X]
Na2C37H34N2S3O9 + OCl- (blue) --> products (colorless) Blue food coloring can be oxidized by household bleach (which contains OCl-) to form colorless products, as represented by the equation above. A student used a spectrophotometer set at a wavelength of 635 nm to study the absorbance of the food coloring over time during the bleach process. In the study, bleach is present in large excess the concentration of OCl- is essentially constant throughout the reaction. The student used data from the study to generate the graphs on slide 6. a. Based on the graph above, what is the order of the reaction with respect to the blue food coloring?
1st order
X + 2Y ---> z + 3Q For the reaction represented above, the initial rate of decrease in [X] was 2.8x10^-3 mol L-1s-1. What was the initial rate of decrease in [Y]?
5.6x10^-3 mol
X + 2Y ---> Z + 3Q For the reaction represented above, the initial rate of decrease in [X] was 2.8x10^-3 mol L-1s-1. What was the initial rate of decrease in [Y]?
5.6x10^-3 mol L^-1 s^-1
Reaction A:O+O→O2 Reaction B:C2H4+C2H4→C4H8 Reaction C:CO+O2→CO2+O Reaction D:CH3I+Br−→CH3Br+I− The equations shown above represent four elementary reactions. Which of the following identifies the reaction in which the number of successful collisions and reaction rate are independent of the orientation of the reactants and explains why? A Reaction A, because the electron clouds of the O atoms are distributed symmetrically. B Reaction B, because each C2H4 molecule has a double bond. C Reaction C, because both CO and O2 are linear molecules. D Reaction D, because the reactant Br− and the product I− are negatively charged.
A Reaction A, because the electron clouds of the O atoms are distributed symmetrically.
Homogenous Catalysis
A system by which both the catalyst and the reactants are in the same phase or state.
Heterogeneous Catalysis
A system by which the catalyst and the reactants are in a different phase or state.
The diagram on slide 20 shows the distribution of molecular collision energies for equimolar samples of a reactant at different temperatures. Based on the diagram, at which temperature will the reactant be consumed at the fastest rate, and why?
At T4, because a larger fraction of the molecules have an energy that is equal to or greater than the activation energy.
2NO(g) + 2H2(g) ---> N2(g) + 2H2O(g) The information in the data table on slide 16 represents two different trials for an experiment to study the rate of the reaction between NO(g) and H2(g), as represented by the balanced equation above the table. Which of the following statements provides the correct explanation for why the initial rate of formation of N2 is greater in trial 2 than in trial 1? Assume that each trial is carried out at the same constant temperature. A The activation energy of the reaction is smaller in trial 2 than it is in trial 1. B The frequency of collisions between reactant molecules is greater in trial 2 than it is in trial 1. C The value of the rate constant for the reaction is smaller in trial 2 than it is in trial 1. D The value of the rate constant for the reaction is greater in trial 2 than it is in trial 1.
B The frequency of collisions between reactant molecules is greater in trial 2 than it is in trial 1.
What happens at the transition state?
Bonds are partially formed/broken
X ---> products The rate constant (k) for the decay of the radioactive isotope X, according to the equation above, is 2.3x10^-2 days-1. The slope of which of the graphs on slide 8 is correct for the decay and could be used to confirm the value of k?
C
C12H22O11(aq) + H2O(l) → 2C6H12O6(aq) The chemical equation shown above represents the hydrolysis of sucrose. Under certain conditions, the rate is directly proportional to the concentration of sucrose. Which statement supports how a change in conditions can increase the rate of this reaction? A Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis. B Decreasing the temperature will increase the frequency of the collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis. C Increasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules. D Decreasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.
C Increasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.
CH3I+NaOH→CH3OH+NaI The rate of the reaction represented by the chemical equation shown above is expressed as rate=k[CH3I][NaOH]. Based on this information, which of the following claims is correct? A The reaction will proceed at a slower rate with increasing temperature. B The rate of the reaction will double when the concentrations of both CH3I and NaOH are doubled. C The rate of the reaction will double if the concentration of CH3I is doubled while keeping the concentration of NaOH constant. D A larger amount of CH3OH will be produced if the concentrations of CH3I and NaOH are halved.
C The rate of the reaction will double if the concentration of CH3I is doubled while keeping the concentration of NaOH constant.
Step 1:H2+IBr→HI+HBr(slow) Step 2:HI+IBr→I2+HBr(fast) A proposed mechanism for the reaction H2+2IBr→I2+2HBr is shown above. Two experiments were performed at the same temperature but with different initial concentrations. Based on this information, which of the following statements is correct? A The rate of the reaction will undergo a 4-fold increase in the experiment in which the initial concentrations of both HIHI and IBr were doubled. B The rate of the reaction will undergo a 2-fold increase in the experiment in which the initial concentrations of both HI and IBr were doubled. C The rate of the reaction will undergo a 4-fold increase in the experiment in which the initial concentrations of both H2 and IBr were doubled. D The rate of the reaction will undergo a 8-fold increase in the experiment in which the initial concentrations of both H2 and IBr were doubled.
C The rate of the reaction will undergo a 4-fold increase in the experiment in which the initial concentrations of both H2 and IBr were doubled.
2X + Y2 ---> X2Y2 A chemist is studying the reaction between the gaseous chemical species X and Y2, represented by the equation above. Initial rates of reaction are measured at various concentrations of reactants. The results are recorded on slide 15. A second chemist repeated the three experiments and observed that the reaction rates were considerably greater than those measured by the first chemist, even though the concentrations of the reactants and the temperature in the laboratory were the same as they were for the first chemist. Which of the following is the best pairing of a claim about a most likely cause for the greater rates measured by the second chemist and a valid justification for that claim? A The pressures of the gases used by the second chemist must have been lower than those used by the first chemist, thus the collisions between reacting particles were less frequent than they were in the first chemist's experiments. B The pressures of the gases used by the second chemist must have been lower than those used by the first chemist, thus the number of collisions with sufficient energy to cause reaction was lower than it was in the first chemist's experiments. C The second chemist must have added a catalyst for the reaction, thus providing a different reaction pathway for the reactant particles to react with an activation energy that was lower than that of the uncatalyzed reaction in the first chemist's experiments. D The second chemist must have added a catalyst for the reaction, thus providing energy to reactant particles to increase their rate of reaction compared to their rate of reaction in the first chemist's experiments.
C The second chemist must have added a catalyst for the reaction, thus providing a different reaction pathway for the reactant particles to react with an activation energy that was lower than that of the uncatalyzed reaction in the first chemist's experiments.
Zn(s)+2 HCl(aq)→ZnCl2(aq)+H2(g) Zn(s) reacts with HCl(aq) according to the equation shown above. In trial 1 of a kinetics experiment, a 5.0g piece of Zn(s) is added to 100mL of 0.10MHCl(aq). The rate of reaction between Zn(s) and HCl(aq) is determined by measuring the volume of H2(g) produced over time. In trial 2 of the experiment, 5.0g of powdered Zn(s) is added to 100mL of 0.10MHCl(aq). Which trial will have a faster initial rate of reaction and why? A Trial 1, because there is a higher concentration of Zn(s) in the reaction mixture. B Trial 1, because the sample of Zn(s) has less surface area for the reaction to take place. C Trial 2, because there is a higher concentration of HCl(aq) in the reaction mixture. D Trial 2, because the sample of Zn(s) has a greater surface area for the reaction to take place.
D Trial 2, because the sample of Zn(s) has a greater surface area for the reaction to take place.
The two diagrams on slide 19 represent collisions that take place at the same temperature between a CO molecule and an NO2NO2 molecule. The products are CO2 and NO. Which diagram most likely represents an effective collision, and why?
Diagram 1 represents an effective collision because the two molecules have the proper orientation to form a new C−O bond as long as they possess enough energy to overcome the activation energy barrier.
The mechanisms on slide 14 are proposed for the gas phase decomposition of ozone, O3. A student claims that the rate of the first mechanism is faster than the rate of the second mechanism because the first mechanism has fewer steps. Do you agree or disagree? Justify your claim.
Disagree. Mechanism 1 has an intermediate (O) and leads to the correct balanced equation. Mechanism 2 has intermediates (NO2 and O) and catalyst (NO). Catalysts increase the rate of reaction; therefore Mechanism 2 has a lower activation energy and is likely the faster reaction.
How do you find enthalpy?
E(products) - E(reactants)
D + 3E ---> 2F When the chemical reaction above is carried out under certain conditions, the rate of disappearance of D is 2.5x10^-2 Ms-1. What is the rate of disappearance of E and the rate of appearance of F under the same conditions?
E: 7.5x10^-2 Ms-1 F: 5.0x10^-2 Ms-1
How do you find activation energy?
Ea = E(activated complex) - E(reactants)
What is the overall reaction of the elementary steps on slide 11.
H2O2(aq) + 2I- (aq) + 2H+(aq) ---> I2(aq) + 2H2O(l)
Step 1:HCOOH+H2SO4→HCOOH2++HSO4− Step 2:HCOOH2+→HCO+++H2O Step 3:HCO++HSO4−→CO+H2SO4 The elementary steps in a proposed mechanism for the decomposition of HCOOH are represented above. Identify the catalyst in the overall reaction and justify your choice.
H2SO4, because it is consumed in the first step of the mechanism and regenerated in a later step.
Rate Constant Units in second order
L mol^-1 s^-1 (or M^-1 s^-1)
Rate Constant Units in third order
L^2 mol^2 s^-1 (or M^-2 s^-1)
What are the units for reaction rate?
Molarity/time
The proposed mechanism for a reaction involves the three elementary steps represented by the particle models shown on slide 22. On the basis of this information, what is an intermediate in the overall reaction?
N2O
The diagram on slide 23 shows the reaction energy profiles for a reaction with and without a catalyst. Identify the reaction energy profile for the catalyzed reaction and justify.
Profile Y, because it introduces a different reaction path that reduces the activation energy.
2 NH3(g)→N2(g)+3H2(g) The catalyzed decomposition of NH3(g) at high temperature is represented by the equation above. The rate of disappearance of NH3(g) was measured over time for two different initial concentrations of NH3(g) at a constant temperature. The data are plotted in the graph on slide 18. On the basis of the data in the graph, what best represents the rate law for the catalyzed decomposition of NH3(g)?
Rate = k
H2(g)+I2(g)→2HI(g) For the reaction between H2 and I2, shown above, the following two-step reaction mechanism is proposed. Step 1:I2⇄2I(fast equilibrium) Step 2:H2+2I→2HI(slow) What is the rate law expression for this reaction if the second step is rate determining?
Rate = k[H2][I2]
Nitrogen dioxide, NO2(g), is produced as a by-product of the combustion of fossil fuels in internal combustion engines. At elevated temperatures NO2(g) decomposes according to the equation below. 2NO2(g) ---> 2NO(g) + O2(g) The concentration of a sample of NO2(g) is monitored as it decomposes and is recorded on the graph on slide 4. The two graphs that follow it are derived from the original data. Write the rate law for the decomposition of NO2(g).
Rate = k[NO2]2
Step 1:2NO2(g)→NO3(g)+NO(g)(slow) Step 2:NO3(g)+CO(g)→NO2(g)+CO2(g)(fast) A proposed two-step mechanism for the chemical reaction NO2(g)+CO(g)→NO(g)+CO2(g) is shown above. What is the overall rate law?
Rate = k[NO2]2
O3(g)+O(g)→2O2(g) The decomposition of O3 occurs according to the balanced equation above. In the presence of NO, the decomposition proceeds in two elementary steps, as represented by the following mechanism. Step 1:O3+NO→NO2+O2(slow) Step 2:NO2+O→NO+O2(fast) Based on the information, which of the following is the rate law?
Rate = k[O3][NO]
2X + Y2 ---> X2Y2 A chemist is studying the reaction between the gaseous chemical species X and Y2, represented by the equation above. Initial rates of reaction are measured at various concentrations of reactants. The results are recorded on slide 15. Given the information in the table, what is the experimental rate law? Determine the initial rate of disappearance of X in experiment 1.
Rate = k[X][Y2] 64 M/s
What things must happen for a reaction to occur?
Reactant particles must: 1. collide 2. have proper orientation 3. have sufficient energy
The diagram on slide 21 shows the progress of the chemical reaction for the synthesis of ammonia from its elements. The adsorption of the N2 molecules on the surface of Ru weakens the triple bond between the two N atoms. Based on the diagram, what is the role of Ru in this process?
Ru is a catalyst
Step 1: ? (slow) Step 2: NO2(g)+F(g)→NO2F(g)(fast) Overall:2 NO2(g)+F2(g)→2NO2F(g) The overall reaction represented above is proposed to take place through two elementary steps. What is the chemical equation for step 1 and the rate law for the overall reaction?
The chemical equation for step 1 is NO2(g)+F2(g)→NO2F(g)+F(g), and the rate law for the overall reaction is rate=k[NO2][F2].
Step 1: H2O2+I−→IO−+H2O Step 2: H2O2+IO−→H2O+O2+I− The mechanism for a chemical reaction is shown above. What is the overall reaction and the rate laws of elementary step 1?
The chemical equation for the overall reaction is 2H2O2→2 H2O+O2, and the rate law for elementary step 1 is rate=k[H2O2][I−].
Step 1:N2O5→NO2+NO3(slow) Step 2:NO2+NO3→NO2+NO+O2(fast) Step 3:NO+N2O5→3NO2(fast) A proposed reaction mechanism for the decomposition of N2O5(g) is shown above. Based on the proposed mechanism, what is the chemical equation and the rate law for the overall reaction.
The chemical equation for the overall reaction is 2N2O5 ---> 4NO2 + O2 and the rate law is rate = k[N2O5]
Two samples of Mg(s) of equal mass were placed in equal amounts of HCl(aq) contained in two separate reaction vessels. Particle representations of the mixing of Mg(s) and HCl(aq) in the two reaction vessels are shown in Figure 1 and Figure 2 on slide 2. Water molecules are not included in the particle representations. Which of the reactions will initially proceed faster, and why?
The reaction in Figure 2, because more Mg atoms are exposed to HCl(aq) in Figure 2 than in Figure 1.
3S8 + 8OH- ---> 8S3- + 4HOOH In an experiment, a student studies the kinetics of the reaction represented above and obtains the data shown in the table on slide 10. Use the data in the table to determine the order of the reaction with respect to S8 and determine the value of [OH-] that was used in trial 3, considering that the reaction is first order with respect to OH-. Justify your answer. The next day the student conducts trial 4 using the same concentrations of S8 and OH- as in trial 1, but the reaction occurs at a much slower rate than the reaction in trial 1. The student observes that the temperature in the lab is lower than it was the day before. Using particle-level reasoning, provide TWO explanations that help to account for the fact that the reaction rate is slower in trial 4.
When the initial [S8] is tripled, while [OH-] remains constant, the rate is also tripled, therefore 1st order with respect to [S8]. if the reaction is 1st order with respect to [OH-], then when the rate doubles, while [S8] remains constant, the [OH-] must have doubled. Therefore the [OH-] for trial 3 was 0.0200 M. 1. The temperature was lower on the second day, so the average kinetic energy of the reactant particles was lower. Therefore, there were fewer collisions between particles with sufficient energy to react. 2. Since the temperature was lower, the kinetic energy was lower and the average speed of the particles was lower. At lower speeds, the reactant particles collide less frequently.
X(g) + 2Y(g) --> XY2(g) In order to determine the order of the reaction represented above, the initial rate of formation of XY2 is measured using different initial values of [X] and [Y]. The results of the experiment are shown in the table on slide 1. In trial 2 which of the reactants would be consumed more rapidly, and why?
Y, because the rate of disappearance will be double that of X.
The rate law expression reaction of hydrogen peroxide and iodide in an acidic solutions is found to be first order with respect to hydrogen peroxide and first order with respect to iodide. Does the proposed mechanism on slide 12 support this rate law expression?
Yes, experiments lead to 1st order with respect to H2O2 and I- because the slow step in this mechanism is first order in hydrogen peroxide and first order in iodide, in agreement with the experiment.
What graph will be linear for a zero order reaction?
[X]
Elementary reaction
a process in a chemical reaction that occurs in a single event or step
A reaction occurs according to the proposed three-step mechanism below Step 1: A + A ---> C Step 2: C + B --> D + F Step 3: D + B ---> E + F Overall: ? a. Write the rate law for each elementary step. b. What is the overall chemical equation for the reaction? c. It is the unlikely that this overall reaction in a single elementary step, explain why.
a. Step 1 : rate = k[A]2 Step 2: rate = k[C][B] Step 3: rate = k[D][B] b. 2A + 2B ---> E + 2F c. It is highly unlikely that 4 molecules would all collide at the exact same moment, with the proper orientation and sufficient energy to overcome the activation energy barrier.
A two-step reaction mechanism is proposed for a reaction, as represented below: Step 1: NO(g) + Cl2(g) ---> NOCl2(g) Step 2: NOCl2(g) + NO(g) ---> 2NOCl(g) a. Write the rate law for elementary step 1. b. Identify the intermediate and justify your choice. c. What is the overall chemical equation for the reaction above?
a. rate = k[NO][Cl2] b. NOCl2(g) is the intermediate, it is produced in step 1 and consumed in step 2. c. 2NO(g) + Cl2(g) ---> 2NOCl(g)
Which of the following statements best explains why an increase in temperature of 5-10 degrees Celsius can substantially increase the rate of a chemical reaction? a. The activation energy for the reaction is lowered. b. The number of effective collisions between reactant particles is increased. c. The rate of the reverse reaction is increased. d. DeltaH for the reaction is lowered. e. DeltaG for the reaction becomes more positive.
b. The number of effective collisions between reactant particles is increased.
2H2O2(aq) ---> 2H2O(l) +O2(g) delta H degrees = -196 kJ/mol rxn The decomposition of H2O2(aq) is represented by the equation above. A Student monitored the decomposition of a 1.0 L sample of H2O2(aq) at a constant temperature of 300. K and recorded the concentration of H2O2 as a function of time. The results are given on slide 5. Which of the following statements is a correct interpretation of the data regarding how the order of the reaction can be determined? a. The reaction must be first order because there is only one reactant species. b. The reaction is first order if the plot of ln[H2O2] versus time is a straight line. c. The reaction is first order if the plot of 1/[H2O2] versus time is a straight line. d. The reaction is second order because 2 is the coefficient of H2O2 in the chemical equation.
b. The reaction is first order if the plot of ln[H2O2] versus time is a straight line.
NO(g) + NO3(g) ---> 2NO2(g) rate = k[NO][NO3] The reaction represented above occurs in a single step that involves the collision between a particle of No and a particle of NO3. A scientist correctly calculates the rate of collisions between NO and NO3 that have sufficient energy to overcome the activation energy. The observed reaction rate is only a small fraction of the calculated collision rate. Which of the following best explains the discrepancy? a. The energy of collisions between two reactant particles is frequently absorbed by collision with a third particle. b. The two reactant particles must collide with a particular orientation in order to react. c. The activation energy for a reaction is dependent on the concentrations of the reactant particles. d. The activation energy for a reaction is dependent on the temperature.
b. The two reactant particles must collide with a particular orientation in order to react.
The radioactivity of a sample of I-131 was measured. The data collected are plotted on the graph on slide 7. c. Determine the half-life, t1/2, of I-131 using the graph. d. The data can be used to show that the decay of I-131 is a first-order reaction, as indicated on the graph below. i. Label the vertical axis of the graph above. ii. What are the units of the rate constant, k, for the decay reaction? iii. Explain how the half-life of I-131 can be calculated using the slope of the line plotted on the graph.
c. 8 days d. i. ln[rate of decay] ii. days-1 iii. The slope of the line could be calculated, used to determine the rate constant, and plugged into the half-life equation to solve for t1/2.
A kinetics experiment is set up to collect the gas is generated when a sample of chalk, consisting primarily of solid CaCO3, is added to a solution of ethanoic acid, CH3COOH. The rate of reaction between CaCO3 and CH3COOH is determined by measuring the volume of gas generated at 25 degrees Celsius and 1 atm as a function of time. Which of the following experimental conditions is most likely to increase the rate of gas production? a. Decreasing the volume of ethanoic acid solution used in the experiment. b. Decreasing the concentration of the ethanoic acid solution used in the experiment. c. Decreasing the temperature at which the experiment is performed. d. Decreasing the particle size of CaCO3 by grinding it into a fine powder.
c. Decreasing the temperature at which the experiment is performed.
What factors affect reaction rates?
concentration, temperature, surface area, catalysts
Which of the following best helps explain why an increase in temperature increases the rate of a chemical reaction? a. At higher temperatures, reactions have a lower activation energy. b. At higher temperatures, reactions have a higher activation energy. c. At higher temperatures, every collision results in the formation of product. d. At higher temperatures, high-energy collisions happen more frequently.
d. At higher temperatures, high-energy collisions happen more frequently.
As activation energy increases, the rate of the reaction rate....
decreases
Rate Law
expressed the rate of a reaction as proportional to the concentration of each reactant raised to a power
2N2O5(g)→4NO2(g)+O2(g) For the reaction represented by the equation above, the concentration of N2O5 was measured over time. The graphs on slide 18 were created using the data. Based on the graphs above, what is the order of the reaction with respect to N2O5?
first order
How does a catalyst increase the rate constant?
formation of a more stable activated complex; increased collision frequency; improved orientation effects
2NO(g) + O2(g) ---> 2NO2(g) A rate study of the reaction yielded the data recorded in the table on slide 3. Determine the order of the reaction with respect to each of the following reactants. i. NO ii. O2 Write the expression for the rate law for the reaction as determined from the experimental data. Determine the value of the rate constant for the reaction, clearly indicating the units.
i. NO - 2nd order ii. O2 - 1st order rate = k[NO]2[O2] k = 3.55x10^3 L^2 mol^-2 s^-1
The ammonium salt of isocyanic acid is a product of the decomposition of urea, CO(NH2)2, represented below. CO(NH2)2(aq)<---> NH4+(aq) + OCN-(aq) A student studying the decomposition reaction runs the reaction at 90 degrees Celsius. The student collects data in on the concentration of urea as a function of time, as shown on slide 9. a. The student proposes that the rate law is rate = k[CO(NH2)2] i. Explain how the data support the student's proposed rate law. ii. Using the proposed rate law and the student's results, determine the value of the rate constant, k. Include units with your answer.
i. The data in both the table and graph show the [CO(NH2)2] decreased by half from 0 to 10 hours and again decreased by half from 10 to 20 hours, indicating a constant half-life of 10 hours, which indicates that the reaction is a first-order reaction.
Catalyst
increases the reaction rate by providing an alternate pathway with lower potential energy for the activated complex; often leads to the proposal of a new mechanism
The initial-rate data in the table on slide 3 were obtained for the reaction represented below. What is the experimental rate law for the reaction? 2NO(g) + O2(g) ---> 2NO2(g)
k[NO][O2]2
What graph will be linear for a first order reaction?
ln([X])
Bismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle. What quantity plotted versus time will produce a straight line? (n = the number of moles in the sample)
ln(nBi)
Rate Constant Units in zero order
mol L^-1 s^-1 (Ms^-1)
Step 1:2NO⇄(NO)2(fast) Step 2:(NO)2+O2⇄2NO2(slow) The elementary steps in a proposed mechanism for the reaction 2 NO(g)+O2(g)→2NO2(g) are represented by the equations above. What is the rate law for the overall reaction that is consistent with the proposed mechanism?
rate = k[NO]2[O2]
Step 1:2X(g)⇄X2(g)(fast) Step 2:X2(g)+Y(g)→X2Y(g)(slow) The rate law for the hypothetical reaction 2X(g)+Y(g)→X2Y(g) is consistent with the mechanism shown above. What is the rate law that is consistent with this mechanism?
rate = k[X]2[Y]
Rate Constant Units in first order
s^-1
Intermediates
substances that are neither reactants or products but are formed in one elementary reaction and consumed in the next; not included in the overall chemical equation
Reaction Mechanism
the process by which a reaction occurs
Rate Constant
the proportionality constant in the rate law is called the rate constant; value of this constant is temperature dependent and the units reflect the overall reaction order
What is the rate determining step?
the slowest step
Sketch a reaction profile for the mechanism on slide 13.
three activation energy "hills"; first step should have the highest activation energy; the potential energy values for intermediates and products should be lower than the previous for all three steps; the potential energy for the product should be lower than the reactant, showing that the reaction is overall exothermic
Endothermic reaction
when the potential energy of the products is greater than the reactants
Exothermic reaction
when the potential energy of the products is less than the reactants
