CHEMISTRY MIDTERM REVIEW
Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: . Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst. What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more catalyst is added?
(f) no changes.
Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3. How could you determine whether the solution is unsaturated, saturated, or supersaturated?
Add a small crystal of Na2S2O3. It will dissolve in an unsaturated solution, remain apparently unchanged in a saturated solution, or initiate precipitation in a supersaturated solution.
Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
After finding k at several different temperatures, a plot of ln k versus gives a straight line with the slope , from which Ea may be determined.
Why are elementary reactions involving three or more reactants very uncommon?
Although some termolecular reactions are known, it is very rare for three or more molecules come together at exactly the same instant and with the proper orientation required for a reaction to occur.
Explain why the ions Na+ and Cl are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.
Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals.
Distinguish between dispersion methods and condensation methods for preparing colloidal systems.
Dispersion methods use a grinding device or some other means to bring about the subdivision of larger particles. Condensation methods bring smaller units together to form a larger unit. For example, water molecules in the vapor state come together to form very small droplets that we see as clouds.
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here: Temperature (K) k (M-1 s-1) 555 6.23 10-7 575 2.42 10-6 645 1.44 10-4 700 2.01 10-3 What is the value of the activation energy (in kJ/mol) for this reaction?
Ea may be determined from a plot of ln k against that gives a straight line whose slope is : T (K) k (M−1 s−1) ln k 555 1.802 6.23 10−7 −14.289 575 1.739 2.42 10−6 −12.932 645 1.550 1.44 10−4 −8.846 700 1.429 2.42 10−3 −6.210 A plot of this data shows a straight line. Two points marked by an X are picked for convenience of reading and are used to determine the slope of the line: Ea = -2.13 104 8.314 J/mol = 177 kJ/mol
How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.
It increases the rate of reaction by increasing the average kinetic energy of the molecules involved. This results in a larger fraction of collisions producing activated complexes.
Solutions of hydrogen in palladium may be formed by exposing Pd metal to H2 gas. The concentration of hydrogen in the palladium depends on the pressure of H2 gas applied, but in a more complex fashion than can be described by Henry's law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal. (a) Determine the molarity of this solution (solution density = 1.8 g/cm3). (b) Determine the molality of this solution (solution density = 1.8 g/cm3). (c) Determine the percent by mass of hydrogen atoms in this solution (solution density = 1.8 g/cm3).
(a) (b) (c)
In the PhET Reactions & Rates (https://openstaxcollege.org/l/16PHETrecation)interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select "Show Bonds" under Options. (a) Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow? (b) Click "Pause" and then "Reset All," and then enter 15 molecules of A and 10 molecules of BC once again. Select "Show Bonds" under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction.
(a) At the default temperature, the rate of reaction is very slow. Even though A molecules collide with BC molecules quite frequently, very few of them have enough energy to bond. (b) As the temperature is increased, the reaction proceeds at a faster rate. The amount of reactants decreases, and the amount of products increases. After a while, there is a roughly equal amount of BC, AB, and C in the mixture and a slight excess of A.
Give an example of each of the following types of solutions: (a) a gas in a liquid (b) a gas in a gas (c) a solid in a solid
(a) CO2 in water; (b) O2 in N2 (air); (c) bronze (solution of tin or other metals in copper)
Go to the Reactions & Rates (https://openstaxcollege.org/l/16PHETrecation) interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on "Reload Launcher" and change to "Angled shot" to see the difference. (a) What happens when the angle of the collision is changed? (b) Explain how this is relevant to rate of reaction.
(a) Depending on the angle selected, the atom may take a long time to collide with the molecule and, when a collision does occur, it may not result in the breaking of the bond and the forming of the other. (b) Particles of reactant must come into contact with each other before they can react.
(a) Determine the order for each of the reactants, NO and H2, from the data given and show your reasoning. (b) Write the overall rate law for the reaction. (c) Calculate the value of the rate constant, k, for the reaction. Include units. (d) For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H2 had been consumed. (e) The following sequence of elementary steps is a proposed mechanism for the reaction. Step 1: Step 2: Step 3: Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.
(a) Doubling [H2] doubles the rate. [H2] must enter the rate equation to the first power. Doubling [NO] increases the rate by a factor of 4. [NO] must enter the rate law to the second power. (b) The rate law is Rate = k[NO]2[H2]. (c) 1.8 10−4 mol/L/ min = k[0.0060 mol/L]2[0.0010 mol/L], k = 5.0 103 mol−2 L−2 min−1; (d) The reaction has consumed 0.0010 mol/L of H2. The amount of NO consumed is the same, 0.0010 mol/L of NO. Thus 0.0060 - 0.0010 = 0.0050 mol/L remains. (e) Step II is the rate-determining step. If step I gives N2O2 in adequate amount, steps 1 and 2 combine to give . This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3, which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.
Refer to Figure 11.11. (a) How did the concentration of dissolved CO2 in the beverage change when the bottle was opened? (b) What caused this change? (c) Is the beverage unsaturated, saturated, or supersaturated with CO2?
(a) It decreased as some of the CO2 gas left the solution (evidenced by effervescence). (b) Opening the bottle released the high-pressure CO2 gas above the beverage. The reduced CO2 gas pressure, per Henry's law, lowers the solubility for CO2. (c) The dissolved CO2 concentration will continue to slowly decrease until equilibrium is reestablished between the beverage and the very low CO2 gas pressure in the opened bottle. Immediately after opening, the beverage, therefore, contains dissolved CO2 at a concentration greater than its solubility, a nonequilibrium condition, and is said to be supersaturated.
Which of the following gases is expected to be most soluble in water? Explain your reasoning. (a) CH4 (b) CCl4 (c) CHCl3
(c) CHCl3 is expected to be most soluble in water. Of the three gases, only this one is polar and thus capable of experiencing relatively strong dipole-dipole attraction to water molecules.
In terms of collision theory, to which of the following is the rate of a chemical reaction proportional? (a) the change in free energy per second (b) the change in temperature per second (c) the number of collisions per second (d) the number of product molecules
(c) the number of collisions per second
When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally (a) Is the dissolution of KNO3 an endothermic or an exothermic process? (b) What conclusions can you draw about the intermolecular attractions involved in the process? (c) Is the resulting solution an ideal solution?
(a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K+ and ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption.
The rate of a certain reaction doubles for every 10 C rise in temperature. (a) How much faster does the reaction proceed at 45 C than at 25 C? (b) How much faster does the reaction proceed at 95 C than at 25 C?
(a) The rate doubles for each 10 C rise in temperature; 45 C is a 20 C increases over 25 C. Thus, the rate doubles two times, or 22 (rate at 25 °C) = 4-times faster. (b) 95 C is a 70 C increases over 25 C. Thus the rate doubles seven times, or 27 (rate at 25 C) = 128-times faster.
The hydrolysis of the sugar sucrose to the sugars glucose and fructose, follows a first-order rate equation for the disappearance of sucrose: Rate = k[C12H22O11]. (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, k = 2.1 1011 s1 at 27 C and 8.5 1011 s1 at 37 C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature). (b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 107 M. How long will it take the solution to reach equilibrium at 27 C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible. (c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
(a) The text demonstrates that the value of Ea may be determined from a plot of log k against that gives a straight line whose slope is . This relationship is based on the equation or where . Only two data points are given, and these must determine a straight line when log k is plotted against 1/T. The values needed are: k1 = 2.1 10−11 logk1= −10.6778 k2 = 8.5 10−11 logk2= −10.0706 T1 = 27 °C = 300 K = 3.3333 10−3 T2 = 37 °C = 310 K = 3.2258 10−3 The slope of the line determined by these points is given by: Ea = 2.303(8.314 J/mol)(-5648) = 108,100 J = 108 kJ Whenever differences of very small numbers are taken, such as the reciprocals of T provided, an inherent problem occurs. To have accurate differences, a larger number of significant figures than justified by the data must be used. Thus five figures were used to obtain the value Ea = 108 kJ. This difficulty may be alleviated by the following approach. For only two data points, the Arrhenius equation may be used in an equally accurate, analytical solution for Ea. This application is possible because the value of A will be the same throughout the course of the reaction. Once the value of Ea is determined, the value of A may be determined from either Equation (1) or (2). Then k at 47 °C may be determined using the value of Ea and A so determined. The procedure is as follows: Equating the values of A as solved from equations (1) and (2): or . Taking common logs of both sides gives: The value of A may be found from either equation (1) or (2). Using equation (1): A = 2.1 10-11 s-1 10+18.91 = 2.1 10-11(9.55 1018 s-1) = 2.0 108 s-1 The value of k at 47°C may be determined from the Arrhenius equation now that the values of EaandAhave been calculated: = 2.0 108 s-1 10-17.79 = 2.0 108 s-1(1.62 10-18) = 3.2 10-10 s-1 Using the earlier value of Ea = 108 kJ, the calculated value of A is 1.3 108 s-1, and k = 3.1 10-10 s-1. Either answer is acceptable. (b) Since this is a first-order reaction we can use the integrated form of the rate equation to calculate the time that it takes for a reactant to fall from an initial concentration [A]0 to some final concentration [A]: At 27 °C. k = 2.1 10−11 s−1. In this case, the initial concentration is 0.150 M and the final concentration is 1.65 10-7M. We can now solve for the time t: or 1.81 108 h or 7.6 106 day. (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state.
Define these terms: (a) unimolecular reaction (b) bimolecular reaction (c) elementary reaction (d) overall reaction
(a) Unimolecular reaction: A reaction in which a single molecule or ion produces one or more molecules or ions of product. (b) Bimolecular reaction: A collision and combination of two reactants to give an activated complex in an elementary reaction. (c) Elementary reaction: A reaction that occurs in a single step. One or more elementary reactions combine to form a reaction mechanism. (d) Overall reaction: An addition of all steps that excludes the intermediates. It indicates the stoichiometry of the reactants and the products, but not the mechanism.
Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C7H16, nonpolar solvent): (a) vegetable oil (nonpolar) (b) isopropyl alcohol (polar) (c) potassium bromide (ionic)
(a) heptane; (b) water; (c) water
What is the expected electrical conductivity of the following solutions? (a) NaOH(aq) (b) HCl(aq) (c) C6H12O6(aq) (glucose) (d) NH3(l)
(a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved)
Indicate the most important types of intermolecular attractions in each of the following solutions: (a) The solution in Figure 11.2 (b) NO(l) in CO(l) (c) Cl2(g) in Br2(l) (d) HCl(aq) in benzene C6H6(l) (e) Methanol CH3OH(l) in H2O(l)
(a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding
Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions: (a) the solutions in Figure 11.8 (b) methanol, CH3OH, dissolved in ethanol, C2H5OH (c) methane, CH4, dissolved in benzene, C6H6 (d) the polar halocarbon CF2Cl2 dissolved in the polar halocarbon CF2ClCFCl2 (e) O2(l) in N2(l)
(a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces
The reaction of CO with Cl2 gives phosgene (COCl2), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises: (fast, k1 represents the forward rate constant, k-1 the reverse rate constant) (slow, k2 the rate constant) (fast, k3 the rate constant) (a) Write the overall reaction. (b) Identify all intermediates. (c) Write the rate law for each elementary reaction. (d) Write the overall rate law expression.
(a) overall reaction: ; (b) identify all intermediates—Cl(g), COCl(g); (c) write the rate law for each elementary reaction: Rate = k2[CO][Cl], Rate = k3[COCl][Cl]; (d) Write the overall rate law expression: The overall rate law expression is derived from the slow step, which is the rate-determining step. In this case Rate = k2[CO][Cl]. Since Cl is an intermediate, algebraic manipulation is required to eliminate [Cl] from the rate law expression. Use the first equilibrium reaction to derive an expression that represents [Cl]: now divide each side by k-1: then take the square root of each side: . Now substitute into—
The Henry's law constant for O2 is 1.3 10-3 M/atm at 25 °C. What mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O2 is 0.21 atm?
. C(O2) = 1.3 10-3 M/atm 0.21 atm = 2.7 10-4 mol/L. The total amount is 2.7 10-4 mol/L 40 L = 1.08 10-2 mol. The mass of oxygen is 1.08 10-2 mol 32.0 g/mol = 0.346 g or, using two significant figures, 0.35 g.
Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain.
A medium must contain freely mobile, charged entities to be electrically conductive. The ions present in a typical ionic solid are immobilized in a crystalline lattice and so the solid is not able to support an electrical current. When the ions are mobilized, either by melting the solid or dissolving it in water to dissociate the ions, current may flow and these forms of the ionic compound are conductive.
How do solutions differ from compounds? From other mixtures?
A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous.
In the PhET Reactions & Rates (https://openstaxcollege.org/l/16PHETrecation) interactive, use the "Many Collisions" tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select "Show bonds" under Options. How is the rate of the reaction affected by concentration and temperature?
As temperature is raised, the molecules move more quickly and more reactions take place. As temperature is lowered, the molecules move more slowly and fewer reactions take place. Adding molecules to the chamber also increases the number of collisions and the rate of reaction.
Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C. See Figure 11.17 for useful data, and report the computed percentage to one significant digit.
At 10 °C, the solubility of KBr in water is approximately 60 g per 100 g of water.
Compare the functions of homogeneous and heterogeneous catalysts.
Both change the mechanism to one with a lower activation energy, thus producing a faster reaction. Homogeneous catalysts work in the same phase as the reactants; heterogeneous catalysts work in a different phase than the reactants, generally providing a surface upon which the reaction takes place.
Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby.
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How do colloids differ from solutions with regard to dispersed particle size and homogeneity?
Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale.
Consider the solutions presented: (a) Which of the following sketches best represents the ions in a solution of Fe(NO3)3(aq)? (b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO3)3.
Fe(NO3)3 is a strong electrolyte, thus it should completely dissociate into Fe3+ and ( ) ions. Therefore, (z) best represents the solution. (b)
How many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20-M solution of hydrochloric acid?
First, calculate the moles of HCl needed. Then use the ideal gas law to find the volume required. M = mol L-1 x = 4.00 mol HCl Before using the ideal gas law, change pressure to atmospheres and convert temperature from °C to kelvins. x = 0.9803 atm
The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
For only two data points, the Arrhenius equation: may be used in an analytical solution for Ea. This approach is possible because the value of A will be constant throughout the course of the reaction. Once the value of Ea is determined, the value of A may be determined from either Equation (1) or (2). At 293 K or 298 K, the value of k may be determined using the value of Ea and A so determined. The procedure is as follows: At 293 K: Equating the values of A as calculated from equations (1) and (2), we have: or . Taking natural logarithms of both sides gives:
Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.
HBr is an acid and so its molecules react with water molecules to form H3O+ and Br- ions that provide conductivity to the solution. Though HBr is soluble in benzene, it does not react chemically but remains dissolved as neutral HBr molecules. With no ions present in the benzene solution, it is electrically nonconductive.
Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes.
Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat.
Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.
Higher molarity increases the rate of the reaction. Higher temperature increases the rate of the reaction. Smaller pieces of magnesium metal will react more rapidly than larger pieces because more reactive surface exists.
How can it be demonstrated that colloidal particles are electrically charged?
If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate.
The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO, in the gas phase is 1.1 × 10-2 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.
In the text, a graphical method was used to determine activation energies, but we are only given two data points in this problem. With only two data points available, it is not necessary to plot the points to calculate the slope of the line that would be generated if more points were plotted: T (K) (K-1) k (L/mol/s) Lnk 703 1.422 10-3 1.1 10-2 -4.50986 865 1.156 10-3 4.95 1.59939 Also, the , so Ea = slope REa = slope R = -(-2.297 104 K)(8.314 J/mol/K) = 1.91 105 J/mol 1.91 102 kJ/mol
Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same? (a) (b) (c) (d) (e)
In (b), (d), and (e), the elementary and overall reactions are likely to be the same. (a) An elementary reaction is unlikely to have a collision involving more than two reactants. Thus, it would be improbable to find the concentration in the rate equation raised to a power other than 1 or 2. (b) The rate expression indicates that both reactants are involved in the reaction. A binary collision is likely, leading to the possibility of an elementary reaction. (c) The rate equation does not correspond to the stoichiometry of the overall equation and therefore the reaction cannot be elementary. (d) This equation could correspond to a termolecular collision process, one not highly likely, but possible as an elementary process. (e) This equation corresponds to a simple bimolecular collision and could be an elementary reaction.
What is the rate equation for the elementary termolecular reaction ? For ?
In an elementary reaction, the rate constant is multiplied by the concentration of the reactant raised to the power of its stoichiometric coefficient. Rate = k[A][B]2; Rate = k[A]3
Account for the increase in reaction rate brought about by a catalyst.
The general mode of action for a catalyst is to provide a mechanism by which the reactants can unite more readily by taking a path with a lower reaction energy. The rates of both the forward and the reverse reactions are increased, leading to a faster achievement of equilibrium.
Compare the processes that occur when methanol (CH3OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.
Methanol, CH3OH, dissolves in water in all proportions, interacting via hydrogen bonding. Methanol: Hydrogen chloride, HCl, dissolves in and reacts with water to yield hydronium cations and chloride anions that are solvated by strong ion-dipole interactions. Hydrogen chloride: Sodium hydroxide, NaOH, dissolves in water and dissociates to yield sodium cations and hydroxide anions that are strongly solvated by ion-dipole interactions and hydrogen bonding, respectively. Sodium hydroxide:
In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction ? Can we predict the effect if the reaction is known to be an elementary reaction?
No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate equation. Yes if the reaction is an elementary reaction, then doubling the concentration of A doubles the rate.
An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 C to 37 C. What is the activation energy for the ALP-catalyzed conversion of PNPP to PNP and phosphate?
Note that . Changes in rate brought about by temperature changes are governed by the Arrhenius equation: . In this particular reaction, k increases by 1.47 as T changes from 30 °C (303 K). The Arrhenius equation may be solved for A under both sets of conditions and then A can be eliminated between the two equations. Eliminating k from both sides, taking logs, and rearranging gives: Ea(1.72366 10-4 - 1.68474 10-4) = 0.1673 J/mol 3.892 10-6Ea = 0.1673 J/mol Ea= 42986 J/mol = 43.0 kJ/mol
A study of the rate of the reaction represented as gave the following data: Time (s) 0.0 5.0 10.0 15.0 20.0 25.0 35.0 [A] (M) 1.00 0.952 0.625 0.465 0.370 0.308 0.230 (a) Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s. (b) Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate? (c) Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.
Plot the concentration against time and determine the required slopes: (a) Average rates are computed directly from the reaction's rate expression and the specified concentration/time data: average rate, average rate, ; (b) The instantaneous rate is estimated as the slope of a line tangent to the curve at 15 s. Such a line is drawn in the plot, and two concentration/time data pairs are used to estimate the line's slope: instantaneous rate, ; (c) To derive rates for the formation of B from the previously calculated rates for the disappearance of A, we consider the stoichiometry of the reaction, namely, B will be produced at one-half the rate of the disappearance of A: average rate for B formation = instantaneous rate for B formation =
A study of the rate of dimerization of C4H6 gave the data shown in the table: Time (s) [C4H6] (M) 0 1.00 10-2 1600 5.04 10-3 3200 3.37 10-3 4800 2.53 10-3 6200 2.08 10-3 (a) Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s. (b) Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C4H6]. What are the units of this rate? (c) Determine the average rate of formation of C8H12 at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).
Plot the concentration of C4H6 against time and determine the various slopes required: (a) ; (b) from the approximate points on tangent line in the figure at 3200 s: ; (c) average rate = Instantaneous rate = somewhat different values may be obtained depending upon the slope of the drawn line
Explain the cleansing action of soap.
Soap molecules have both a hydrophobic and a hydrophilic end. The charged (hydrophilic) end, which is usually associated with an alkali metal ion, ensures water solubility. The hydrophobic end permits attraction to oil, grease, and other similar nonpolar substances that normally do not dissolve in water but are pulled into the solution by the soap molecules.
Use the PhET Reactions & Rates interactive simulation (http://openstaxcollege.org/l/16PHETreaction) to simulate a system. On the "Single collision" tab of the simulation applet, enable the "Energy view" by clicking the "+" icon. Select the first reaction (A is yellow, B is purple, and C is navy blue). Using the "straight shot" default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?
The A atom continues to bounce around without reacting with the BC molecule when the total energy is less than the transition state; the reaction does not occur because there is not enough energy available to supply the activation energy of the reaction. When the total energy is greater than the transition state, the reaction is able to occur because more energy is available than is required by the activation energy of the reaction.
Use the PhET Reactions & Rates interactive simulation(http://openstaxcollege.org/l/16PHETreaction) to simulate a system. On the "Single collision" tab of the simulation applet, enable the "Energy view" by clicking the "+" icon. Select the first reaction (A is yellow, B is purple, and C is navy blue). Using the "angled shot" option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?
The A atom has enough energy to react with BC; however, the different angles at which it bounces off of BC without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.
What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
The activation energy is the minimum amount of energy necessary to form the activated complex in a reaction. It is usually expressed as the energy necessary to form one mole of activated complex.
At 0 °C and 1.00 atm, as much as 0.70 g of O2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O2 dissolve in 1 L of water?
This problem requires the application of Henry's law. The governing equation is . Under the new conditions, = 0.70 g atm-1 4.00 atm = 2.80 g.
Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and that ethanethiol, C2H5SH, is soluble only to the extent of 1.5 g per 100 mL of water.
The hydrogen bonds between water and C2H5OH are much stronger than the intermolecular attractions between water and C2H5SH.
What is the difference between average rate, initial rate, and instantaneous rate?
The instantaneous rate is the rate of a reaction at any particular point in time, a period of time that is so short that the concentrations of reactants and products change by a negligible amount. The initial rate is the instantaneous rate of reaction as it starts (as product just begins to form). Average rate is the average of the instantaneous rates over a time period.
The rate constant at 325 C for the decomposition reaction is 6.1 108 s1, and the activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
The rate constant k is related to the activation energy Ea by a relationship known as the Arrhenius equation. Its form is: where A is the frequency factor. Using the data provided, and converting kilojoules to joules:
In an experiment, a sample of NaClO3 was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 C higher?
The rate doubles for each 10 C rise in temperature. Thus an increase of 20 C would increase the rate four times, thereby decreasing the time required to one-fourth its original value: .
Account for the relationship between the rate of a reaction and its activation energy.
The rate of reaction will increase as the activation energy decreases. This relationship is reasonable because a large activation energy that requires a large amount of energy is a hindrance to reaction.
Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
The reactants either may be moving too slowly to have enough kinetic energy to exceed the activation energy for the reaction, or the orientation of the molecules when they collide may prevent the reaction from occurring.
Nitrogen(II) oxide, NO, reacts with hydrogen, H2, according to the following equation: What would the rate law be if the mechanism for this reaction were:
The slow reaction is the rate-determining step: Therefore, the rate must be based on this equation.
Based on the diagrams in Exercise 17.85, which of the reactions has the fastest rate? Which has the slowest rate?
The smaller the activation energy, the faster the reaction. In this case, both have the same activation energy, so they would have the same rate.
Based on the diagrams in Exercise 17.84, which of the reactions has the fastest rate? Which has the slowest rate?
The smaller the activation energy, the faster the reaction: fastest: (b); slowest: (a)
Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures.
The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating.
Which of the principal characteristics of solutions can we see in the solutions of K2Cr2O7 shown in Figure 11.2?
The solutions are the same throughout (the color is constant throughout), and the composition of a solution of K2Cr2O7 in water can vary.
Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)
Water boils at a lower temperature at higher elevations. The higher cooking temperature at sea level makes the rate of cooking faster and the time to completion shorter.
In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl2 and F2 and the formation of ClF3.
Write the rate of change with a negative sign for substances decreasing in concentration (reactants) and a positive sign for those substances being formed (products). Multiply each term by the reciprocal of its coefficient:
Ozone decomposes to oxygen according to the equation Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O3 and the formation of oxygen.
Write the rate of change with a negative sign for substances decreasing in concentration (reactants) and a positive sign for those substances being formed (products). Multiply each term by the reciprocal of its coefficient:
Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl2F2, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is: (a) Explain why chlorine atoms are catalysts in the gas-phase transformation: (b) Nitric oxide is also involved in the decomposition of ozone by the mechanism: Is NO a catalyst for the decomposition? Explain your answer.
a) Chlorine atoms are a catalyst because they react in the second step but are regenerated in the third step. Thus, they are not used up, which is a characteristic of catalysts. (b) NO is a catalyst for the same reason as in part (a).
When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?
diffusion; in this example, since every collision between reactants leads to a reaction, the activation energy has been exceeded
Nitrogen and oxygen react at high temperatures. What will happen to the concentrations of N2, O2, and NO at equilibrium if a catalyst is added?
no changes occur
