Predict how the reaction rate will be affected by doubling the concentration of thefirst species in each equation.

C2H5I → C2H4 + HI: rate = k[C2H5I]

SO + O2 → SO2 + O: rate = k[SO][O2]

2CH3 → C2H6: rate = k[CH3]2

ClOO → Cl + O2: rate = k

Answer:

C2H5I → C2H4 + HI: rate = k[C2H5I]. If the concentration is doubled, the rate will double (first order reaction).SO + O2 → SO2 + O: rate = k[SO][O2]. If the concentration of both is doubled, the rate will quadruple (overall second order reaction).2CH3 → C2H6: rate = k[CH3]

2. If the concentration is doubled, the rate will quadruple (second order reaction).ClOO → Cl + O2: rate = k. If the concentration is doubled, the rate will stay the same (zeroth order reaction).

Cleavage of C2H6 to produce two CH3· radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with k = 5.46 ×10−4 s−1. How long will it take for the reaction to go to 15% completion? to 50% completion?

Answer:

2 CH3· C2H6 (700 oC); first order: rate = k [CH3·].We know k = 5.46 x10-4 s-1. If 15% of the reaction has been completed, that means [A]t = 100-15 =85; and [A]0 = 100.

Then, knowing the integrated rate law for a first order reaction:

We can get t(15%); t = 4.96 minAnd t(50%) = 21.16 min.

Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere.

Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to berate = k[N2

+][O2], which one of the steps is rate limiting?

Answer:

We know that the rate-determining step should predict the same rate law that is determined experimentally, so if the rate law for the overall reaction is rate = k1[N2

+][O2], then the rate limiting step is the first one (k1).

What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order?

Answer: The rate of a zeroth order reaction does not depend on the concentration of reactant(s) (Rate = k).Experimentally, one would have to perform the reaction with different concentration of reactants to observe that the rate is independent on their concentration.

Iodide reduces Fe(III) according to the following reaction:

Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?

Answer: the order of the reaction is first order in Fe3+; and second order in I−; third order overall;

rate = k[Fe3+][I−]2.

Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:

What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?

Experiment [Benzoyl Peroxide]0 (M) Initial Rate (M/s)

1 1.00 2.22 x 10-4

2 0.70 1.64 x 10-4

3 0.50 1.12 x 10-4

4 0.25 0.59 x 10-4

Answer: the general rate law is: rate = k [benzoyl peroxide]m. To find “m”, we do the following:

Rate2/rate1 = (k2/k1) {([benzoyl peroxide]2/[benzoyl peroxide]1) }m

Substituting, we obtain m = 0.85; so the rate law is:

Rate = k [benzoyl peroxide]0.85

Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?

Answer: The activation energy is required for a collision between molecules to result in a chemical reaction, as well it is related to the rate due to the Arrhenius equation. The activation energy is the threshold of energy needed in order for a reaction to occur. Reactant particles must collide with enough energy to be able to break chemical bonds, that will then allow the creation of new bonds. If the particles do not have enough energy when they collide, they will simply bounce off one another.

The increase in T increases the rate of reaction despite the fact the average kinetic is less than the activation energy since it increases the rate of collision between molecules.

With an increase in T, there is a greater distribution of kinetic energy among reactant particles. This increase in T allows the rate in which particles collide with one another to increase. Although the average kinetic energy is still lower than theactivation energy, the increase of collisions among particles increases the chance of particles that contain enough energy to overcome the energy barrier to collide. Thus, the reaction rate increases due to the increased rate of collisions.

Additionally, there is a higher amount of molecules that have sufficient kinetic energy to overcome the energy barrier, despite the average energy of all these molecules still being lower than the activation energy.

An enzyme-catalyzed reaction has an activation energy of 15 kcal/mol. Howwould the value of the rate constant differ between 20°C and 30°C? If theenzyme reduced the Ea from 25 kcal/mol to 15 kcal/mol, by what factor has the

enzyme increased the reaction rate at each temperature?

Answer: We use this equation, to determine the k at the different T, knowing Ea (15 kcal/mol) which we need to convert into kJ/mol 62.76 kJ/mol, R = 8.314472 J/(K mol) and T1 = 20 oC and T2 = 30 oC, which we need to convert to K. Then we should be able to calculate k1 and k2 and answer to the first question.k1 = 1.206k2 = 1.205; difference between both = 0.001.

For the second question, we would have to determine k1 and k2 for both Ea. That would give us the factor for each temperature (independently of order of reaction).

The reaction rate constant at 25°C is 1.0 × 10−4 M/s. Increasing the temperature to 75°Ccauses the reaction rate to increase to 7.0 × 10−2 M/s. Estimate Ea for this process.If Ea were 25 kJ/mol and the reaction rate at 25°C is 1.0 × 10−4 M/s, what would be the

reaction rate at 75°C?

Answer: First question: we are asked to provide Ea for the process, knowing Rate (298 K) = 10-4 M/S, and Rate (348.15 K) = 7 x10-2 M/s.

Here we know that rate = k[A]x, then we can write:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2), If we do this, we can determine Ea, substituting k1, k2, T1, T2 and R.

Above approximately 500 K, the reaction between NO2 and CO to produce CO2 and NO follows the second-order rate law Δ[CO2]/Δt = k[NO2][CO]. At lower temperatures, however, the rate law is Δ[CO2]/Δt = k′[NO2]

2, for which it is known that NO3 is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest?

Answer: Given that NO3 is an intermediate, we automatically know two things: 1) NO3 will not show up in the final overall reaction, and 2) NO3 will be in the products of the first reaction of the mechanism, and in the reactants of the second reaction. We can also assume that since we are given an intermediate from the problem, this is the only intermediate. Then the lower T reaction mechanism will look like this:

And the overall reaction will look like this:

We have now the reaction mechanism, so we can go about determining which step is the slowest. It would be difficult to do this without any further information, however, we know two more things: 1) the high T mechanism rate =k[NO2][CO] meaning that it took place in one step (given that the overall reaction is also equal to this rate) and 2) the low T mechanism’s rate = k’[NO2]2. These two things being said, we have both confirmed that our proposed low T reaction mechanism is in fact two steps, and that we have a means to find which step is slower.

By using the “guess and check” method we can label each step reaction one at a time as the “slow reaction” and see if the rate matches up with the rate given to us. Let’s first try the second reaction: by using rate laws we can determine that the rate of the reaction must be in the terms of its reactants, which follows: k = [CO][NO3].But we can’t have the overall reaction in terms of an intermediate! By looking at the first reaction, we can determine that we can substitute “[NO2]/[NO]” for “[NO3]” since by writing the full reaction rate of the first step and solving for [NO3] this is equivalent. So, rate = k{[CO][NO2]2}/[NO].

So clearly the second reaction is not the slow reaction since the rate is not equivalent to what we are given! Checking the first reaction: rate = k[NO2]2, which matches. We have now confirmed that the first reaction is the slow reaction equation, since its rate is equivalent to the overall reaction rate.

Assume that the rates of the forward and reverse reactions in the first equation are equal.

(fast)

(slow)

(fast)

Nitramide (O2NNH2) decomposes in aqueous solution to N2O and H2O. What is the experimental rate law (Δ[N2O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows?

Answer: We know that the slowest step of the reaction is the rate determining step, since it usually has the highest activation energy requirement. As a result, the slowest step of the reaction is the experimental rate law we are looking for. Since the slowest step is the rate determining step, that usually means there is some intermediate in between. Intermediates should never be a part of the rate law mechanism.

rate = rate2 = k2 [O2NNH-]Since nitramide is an intermediate, we must find some way to substitute it. To solve that problem, we look for where nitramide is produced and consumed. We see that:

rate1 = k1[O2NNH2]rate-1 = k-1[O2NNH-][H+]

Since these two rates produce and consume the same amount of O2NNH- over the same time period, we can set them equal to each other and solve for the intermediate.

rate1 = rate-1; k1[O2NNH2] = k-1[O2NNH-][H+]; [O2NNH-]= k1[O2NNH2]/k-1[H+]

Substituting this equation back into our original equation gives us: rate = rate2 = k2(k1[O2NNH2])/k-1[H+]With all the rate constants, we can clean up our equation by saying: k = k2k1/k-1

Leaving us with:rate = k[O2NNH2]/[H+]

The activation energy of an elementary chemical reaction is 30 kcal /mol. Determine:

a) The increase in the reaction rate when the temperature increases from 300 oC to 360 oCb) The increase in temperature necessary to double the rate of the reaction at 300 oC.

Answer: a) rate constant will increase by a factor of 12.

b) T must be increased 15.5 oC.

A reaction A + B + C Products yielded the following data:

a) Write the rate law for this reactionb) If the initial reactant concentrations were [A]0 = 2 mmol/L, [B]0= 1.5 mmol/L and [C]0 = 1.15 mmol/L,

indicate what would be the initial rate of disappearance of A.

Answer: a) –d[A]/dt = k [A][B]2[C]2

b) -d[A]/dt = 16.97 mmol L-1 s-1

The following reaction A B + C has a rate constant of 0.02 L /mol . min. A daily production of 9,000 mol/day is aimed for, using a batch reactor. To avoid secondary reactions, the maximum conversion of 75% is the limit. Knowing that the initial concentration of A is 1 mol/L and the conditioning time for the reactor is 1.5 h, calculate the minimum volume of the reactor.

Answer:

Volume = 2000 L.

Half-life time for substance A in a second order reaction A Products at 20 oC is 25 seconds when [A]0 = 0.8 mol/L.

a) Calculate the time required so that the concentration of A is reduced to one fifth of its initial value.b) If the rate of the reaction is doubled when temperature increases from 20 oC to 30 oC, calculate the activation

energy of the process.

Answer:

a) Knowing the half-life equation for a second order reaction, we can determine k, since we are given [A]0 and t1/2. Once calculated k, and knowing the integrated rate law, we can determine t= 100 s

b) Knowing that rate (20 oC) = k[A]2, and rate (30 oC) = 2 x rate (20 oC); that means that k(30 oC) = 2 x k(20 oC), we should be able to get to:

Ea = 51.2 kJ/mol

The initial rate of the reaction A+B C+D is determined for different initial conditions, with the results listed in the table:

a) What is the order of the reaction with respect of A and B? Write out respective rate laws for A and B.

b) What is the overall reaction order?c) What is the value of k?

Answer:

The first order reaction has t1/2 of 250 s:

a) What % of a sample of A remains unreacted 1,500 s after a reaction starts.b) What if the rate of the reaction is ½ the rate of disappearance of A.

Answer:

The reaction A to product is first order in A.

a) If 2.4 g of A is allowed to decompose for 30 minutes, the mass of A remaining undecomposed is found to be .6 g. What is the half-life of this reaction.

b) Starting with 2.4g of A, what is the mass of A remaining undecomposed after 1 h?

Answer: a) Mass has decreased to ¼ of the original mass. Since (1/2) (1/2) = ¼, two halflives have elapsed. As a result 2 x t1/2

= 30 min; t ½ = 15 min.

b) k= 0.693 / (15 min) = 0.0462 min-1.

If we use the equation ln[A]t/ln[A]0 = -kt; [A]t= 0.146 g A.

Acetoacetic acid, CH3COCH2COOH (aq), a reagent used in organic synthesis, decomposes in acidic solution, producing acetone and carbon dioxide gas:

CH3COCH2COOH (aq) CH3COCH3 (aq) +CO2 (g)

This is a first order decomposition with a half-life of 144 min. How long will it take for a sample to be 55% decomposed?

Answer: Start with the decomposed = 100 -55 = 45 % decomposed.

Ln [(45/100)[A]0]/[A]0 = ln 0.45 = -kt;

Find the value of k; k = ln2 /t1/2 = ln 2/144 min = 0.00481 min-1

Ln 0.45 = (-0.00481 min-1) (t); t – 166 seconds.

The rate constant for the reactions H2 (g) + F2 (g) 2HF has been determined at the following T: 650 oC: k = 4.8 x10-4 M-1 s-1; 700 oC: k= 3.0 x10-2 M-1 s-1. Calculate the activation energy for the reaction.

Answer: Use Arrhenius equation

ln (k1/k2) = Ea/R(1/T2-1/T1) ; Ea = 313 kJ/mol