Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 311:

  X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are highenergy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

  

  There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for "breaking radiation." X-rays are emitted by the electrons as they are brought to rest by

  interactions with the positive nuclei of the anode.

  The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the

  energy difference is emitted.

  X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV =

  1.6 x 1019 J; Planck's constant h = 4.1 x 10–15 eV•s; speed of light c = 3 x 108 m/s.)

  What is the minimum potential difference required to produce a 0.06 nm X-ray from an electron transition in a metal?

  A. 15,000 V

  B. 20,000 V

  C. 20,500 V

  D. 21,500 V

  Correct Answer: C

  In the passage, we are told that an X-ray is emitted when an electron makes a transition from a higher to a lower energy level in an atom. This question asks us to calculate the minimum potential difference required to produce a 0.06-nm X-ray from an electron transition. First, we have to determine the energy of the X-ray produced. We use the equation E = hc/, where E is the energy, h is Planck's constant, c is the speed of light in a vacuum, and is the wavelength. Substituting in, we get that E = (4.1 x 10-15 x 3 x 108 )/(0.06 x 10-9), which equals 20,500 electron-volts. This is the energy of an X-ray whose wavelength is 0.06 nm, and it is therefore the minimum energy required to create an X-ray with a wavelength of

  0.06 nanometers. This energy comes from accelerating electrons through a potential difference. Now, recall that 1 electron- volt is the energy of an electron accelerated through a potential difference of 1 volt. Therefore 20,500 electron-volts requires that an electron be accelerated through a potential difference of 20,500 volts. So this represents the minimum potential difference required to produce an X-ray with a wavelength of 0.06 nanometers, and answer choice C is correct.

• Question 312:

  X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are high-energy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

  

  There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for "breaking radiation." X-rays are emitted by the electrons as they are brought to rest by

  interactions with the positive nuclei of the anode.

  The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the

  energy difference is emitted.

  X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV =

  1.6 x 1019 J; Planck's constant h = 4.1 x 10–15 eV•s; speed of light c = 3 x 108 m/s.)

  An X-ray source produces X-rays with a maximum frequency of 6 x 108 Hz. If the cathode current is doubled so that twice as many electrons are emitted per unit time, what is the new maximum frequency of the X-rays produced?

  A. 3 x 1018 Hz

  B. 6 x 1018 Hz

  C. 12 x 1018 Hz

  D. 24 x 1018 Hz

  Correct Answer: B

  In the question we're told that a cathode produces electrons. We are asked to determine what the new maximum frequency of the X-rays produced will be if the current is doubled, so that twice as many electrons are emitted per unit time. By doubling the current, we double the rate at which electrons are produced. However, the energy of each electron does not change. The energy is controlled by the potential difference through which the electrons are accelerated, and this remains constant. In addition, since the maximum frequency of an X-ray depends only on the electron's energy, increasing the number of electrons produced will have no effect on the maximum frequency of the X-rays produced. The maximum frequency will remain 6 x 1018 Hz, and this is answer choice B.

• Question 313:

  X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are high-energy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

  

  There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for "breaking radiation." X-rays are emitted by the electrons as they are brought to rest by

  interactions with the positive nuclei of the anode.

  The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the

  energy difference is emitted.

  X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV =

  1.6 x 1019 J; Planck's constant h = 4.1 x 10–15 eV•s; speed of light c = 3 x 108 m/s.)

  What is the direction of the electric field that accelerates the electrons?

  A. From the anode toward the cathode

  B. From the cathode toward the anode

  C. Into the page

  D. Out of the page

  Correct Answer: A

  This question involves the use of information in the passage along with some basic knowledge of electric fields and forces. In the second paragraph of the passage, we are told that electrons are accelerated from the cathode to the anode by an electric field. Now we have to figure out the direction of the electric field that would cause the electrons to move in that way. When the electric field lines are drawn, the arrows always point from the positive charge to the negative charge. So the direction of the electric field is the direction in which a positive charge would accelerate in that field. Since electrons are negative charges, they will be accelerated in the direction opposite to that of the electric field. We are pointing from the anode toward the cathode. Therefore, answer choice A is correct.

• Question 314:

  X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are high energy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

  

  There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for "breaking radiation." X-rays are emitted by the electrons as they are brought to rest by

  interactions with the positive nuclei of the anode.

  The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the

  energy difference is emitted.

  X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV =

  1.6 x 1019 J; Planck's constant h = 4.1 x 10–15 eV•s; speed of light c = 3 x 108 m/s.)

  How does the wavelength of an X-ray produced from a K-alpha transition in molybdenum compare to that produced from a lower energy K-alpha transition in copper?

  A. It is shorter.

  B. It is the same.

  C. It is longer.

  D. It depends on the energy of the incoming electron.

  Correct Answer: A

  From the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency, we know that energy and frequency are proportional. However, the question asks about wavelength, so we have to convert frequency into wavelength. The speed of any wave is equal to its frequency times its wavelength. So solving for frequency, we get that frequency equals the speed of the wave over the wavelength. So frequency is inversely proportional to wavelength. Since we know that frequency and energy are proportional, we can predict that as the energy difference increases, the wavelength of the emitted photon must decrease. This implies that the wavelengths of the photons emitted from molybdenum will be shorter than that of the photons emitted from copper, since wavelength and energy are inversely proportional -- as one gets bigger, the other gets smaller. Therefore, the correct answer is choice A.

• Question 315:

  Several techniques have been developed to determine the order of a reaction. The rate of a reaction cannot be predicted on the basis of the overall equation, but can be predicted on the basis of the rate-determining step. For instance, the following reaction can be broken down into three steps.

  

  Step 1

  

  (Slow) Step 2

  

  (fast) Step 3

  

  (fast)

  Reaction 1 In this case, the first step in the reaction pathway is the rate-determining step. Therefore, the overall rate of the reaction must equal the rate of the first step, k1 [A] where k1 is the rate constant for the first step. (Rate constants of the different steps are denoted by kx , where x is the step number.)

  In some cases, it is desirable to measure the rate of a reaction in relation to only one species. In a second-order reaction, for instance, a large excess of one species is included in the reaction vessel. Since a relatively small amount of this large concentration is reacted, we assume that the concentration essentially remains unchanged. Such a reaction is called a pseudo first-order reaction. A new rate constant, k', is established, equal to the product of the rate constant of the original reaction, k, and the concentration of the species in excess. This approach is often used to analyze enzyme activity.

  In some cases, the reaction rate may be dependent on the concentration of a short-lived intermediate. This can happen if the rate-determining step is not the first step. In this case, the concentration of the intermediate must be derived from the equilibrium constant of the preceding step. For redox reactions, the equilibrium can be correlated with the voltage produced by two half-cells by means of the Nernst equation. This equation states that at any given moment:

  

  Equation 1 When

  

  Reaction 2

  Note: R = 8.314 J/K•mol; F = 9.6485 x 104 C/mol.)

  Catalysts are effective in increasing the rate of a reaction because they:

  A. increase the energy of the activated complex.

  B. increase the value of the equilibrium constant.

  C. decrease the number of collisions between reactant molecules.

  D. lower the activation energy

  Correct Answer: D

  Activation energy is the minimum energy needed for a reaction to occur. Catalysts are effective because they lower the activation energy of the reaction; as a result, there are more reactant molecules with sufficient kinetic energy to collide and react with each other. An increase in the number of colliding molecules will result in a faster reaction, so choice D is correct. Choice C is incorrect because a catalyst ensures that the number of collisions between reactant molecules increases. In a chemical equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. If a catalyst is added to increase the rate of the forward reaction, the system adjusts so that the reverse rate can rise to match the forward rate. A catalyst, therefore, does not affect the value of the equilibrium constant, but does affect the speed at which the equilibrium is reached. So, choice B is incorrect. Finally, choice A is incorrect because increasing the energy of the activated complex is the same as increasing the activation energy of the reaction. An activated complex forms when reactant molecules meet. These molecules can combine to form products, or fall apart into their original reactants. Obviously, if the energy barrier to form this complex is higher, the reactant molecules would decrease the rate of reaction, and choice A is incorrect.

• Question 316:

  X-rays are produced by a device which beams electrons with an energy between 103 and 106 eV at a metal plate. The electrons interact with the metal plate and are stopped by it. Much of the energy of the incoming electrons is released in the form of X-rays, which are high-energy photons of electromagnetic radiation. An example of such a device is shown below. Electrons are accelerated from the cathode towards the anode by an electric field.

  

  There are two mechanisms by which the X-rays are produced within the metal. The first mechanism is called bremsstrahlung, which is German for "breaking radiation." X-rays are emitted by the electrons as they are brought to rest by

  interactions with the positive nuclei of the anode.

  The second mechanism occurs when an incoming electron knocks an inner electron out of one of the metal atoms of the anode. This electron is replaced by an electron from a higher energy level of the atom, and a photon making up the

  energy difference is emitted.

  X-rays are absorbed by a material when they pass through it. The amount of X-rays absorbed increases with the density of the material. In addition, lower energy X-rays are more likely to be absorbed than higher energy X-rays. (Note: 1 eV =

  1.6 x 1019 J; Planck's constant h = 4.1 x 10-15 eV•s; speed of light c = 3 x 108 m/s.)

  An electron is accelerated through a distance of 0.1 m by a potential difference of 10,000 volts. What is the electron's energy as it strikes the anode?

  A. 100 eV

  B. 1,000 eV

  C. 10,000 eV

  D. 1 J

  Correct Answer: C

  The key to answering this question is knowing the definition of the electron-volt. One electron- volt is defined as the change in potential energy of a single electron as it moves through a potential difference of 1 volt. In this question we have a single electron that is accelerated through a distance of 0.1 meters by a potential difference of 10,000 volts, so the electron's energy as it strikes the anode is simply 10,000 electron-volts, and this is choice C. The important thing to note is that it doesn't matter what distance the electron is accelerated through. All that matters are the magnitude of the potential difference that accelerates the charge, and the magnitude of the charge itself.

• Question 317:

  Several techniques have been developed to determine the order of a reaction. The rate of a reaction cannot be predicted on the basis of the overall equation, but can be predicted on the basis of the rate-determining step. For instance, the following reaction can be broken down into three steps.

  

  Step 1

  

  (Slow) Step 2

  

  (fast) Step 3

  

  (fast)

  Reaction 1 In this case, the first step in the reaction pathway is the rate-determining step. Therefore, the overall rate of the reaction must equal the rate of the first step, k1 [A] where k1 is the rate constant for the first step. (Rate constants of the different steps are denoted by kx , where x is the step number.)

  In some cases, it is desirable to measure the rate of a reaction in relation to only one species. In a second-order reaction, for instance, a large excess of one species is included in the reaction vessel. Since a relatively small amount of this large concentration is reacted, we assume that the concentration essentially remains unchanged. Such a reaction is called a pseudo first-order reaction. A new rate constant, k', is established, equal to the product of the rate constant of the original reaction, k, and the concentration of the species in excess. This approach is often used to analyze enzyme activity.

  In some cases, the reaction rate may be dependent on the concentration of a short-lived intermediate. This can happen if the rate-determining step is not the first step. In this case, the concentration of the intermediate must be derived from the equilibrium constant of the preceding step. For redox reactions, the equilibrium can be correlated with the voltage produced by two half-cells by means of the Nernst equation. This equation states that at any given moment:

  

  Equation 1 When

  

  Reaction 2

  Note: R = 8.314 J/K•mol; F = 9.6485 104 C/mol.)

  What would be the cell emf of the following system at 298K?

  

  A. 1.07 V

  B. 1.10 V

  C. 1.13 V

  D. 1.20 V

  Correct Answer: A

  In order to answer this question, you have to manipulate Equation 1, which states that the cell potential, E, is equal to the standard cell potential, Eo, minus the product of RT/nF times the natural log of the concentration of products divided by reactants. The term ([C]c [D]d /[A]a [B]b) is often defined as Q, the mass action expression, where the reactants and products are raised to a power equal to their stoichiometric coefficients in the balanced reaction equation. The balanced reaction equation for the cell is as follows:

  

  How can we tell that copper is reduced and zinc is oxidized? The first clue is the standard cell potential: +1.10 V. A positive cell potential means that the reaction is spontaneous and the cell must be galvanic. In galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. The species on the left-hand side of the cell diagram, i.e., zinc, is the anode, and the species on the right-hand side, i.e., copper, is the cathode. In other words, zinc is oxidized and copper is reduced. So, plugging our values into Equation 1, we get E = +1.10 ?[(8.314 x 298)/(2 x 96485)] 1n ([Zn2+]/[Cu2+]). Notice that Zn(s) and Cu(s) are not inserted into the equation because they are pure solids with a molar concentration of 1. Also notice that n = 2, since two electrons are exchanged in the reaction. The natural log of 0.2/0.02 is equal to 2.3, so the equation can be further simplified to E = +1.10 ?[(8.314 x 298)/(2 x 96485)] x 2.3. The value of E works out to be +1.07 V, choice A. Choice B is the cell potential under standard conditions of 1 atmosphere pressure and 1 molar concentration for each reacting species. Since the reaction takes place under non-standard conditions of 1 atmosphere pressure and 1 molar concentration for each reacting species, choice B is incorrect. If you incorrectly identified copper as the species being oxidized and zinc as the species being reduced, you would have ended up with choice C. There is a short cut to getting the right answer that doesn't involve too much calculation: the standard cell potential, +1.10V, corresponds to concentrations of 1M for Zn2+ and Cu2 +. Therefore, the cell potential must be equal to the standard cell potential, since the term RT/nF 1n [Zn2+]/[ Cu2+ ] is equal to zero. Now, with concentrations of 0.2 M and 0.02 M for Zn2+ and Cu2 +, you should be able to see that the term RT/nF 1n [Zn2+]/[ Cu2+ ] is greater than zero. Since this term is subtracted from the standard cell potential, the actual cell potential must be lower than +1.10V. The only answer choice that has a value lower than +1.10V is choice A.

• Question 318:

  Several techniques have been developed to determine the order of a reaction. The rate of a reaction cannot be predicted on the basis of the overall equation, but can be predicted on the basis of the rate-determining step. For instance, the following reaction can be broken down into three steps.

  

  Step 1

  

  (Slow) Step 2

  

  (fast) Step 3

  

  (fast)

  Reaction 1 In this case, the first step in the reaction pathway is the rate-determining step. Therefore, the overall rate of the reaction must equal the rate of the first step, k1 [A] where k1 is the rate constant for the first step. (Rate constants of the different steps are denoted by kx, where x is the step number.)

  In some cases, it is desirable to measure the rate of a reaction in relation to only one species. In a second-order reaction, for instance, a large excess of one species is included in the reaction vessel. Since a relatively small amount of this large concentration is reacted, we assume that the concentration essentially remains unchanged. Such a reaction is called a pseudo first-order reaction. A new rate constant, k', is established, equal to the product of the rate constant of the original reaction, k, and the concentration of the species in excess. This approach is often used to analyze enzyme activity.

  In some cases, the reaction rate may be dependent on the concentration of a short-lived intermediate. This can happen if the rate-determining step is not the first step. In this case, the concentration of the intermediate must be derived from the equilibrium constant of the preceding step. For redox reactions, the equilibrium can be correlated with the voltage produced by two half-cells by means of the Nernst equation. This equation states that at any given moment:

  

  Equation 1 When

  

  Reaction 2

  Note: R = 8.314 J/K•mol; F = 9.6485 x 104 /mol.)

  What is the effect of increasing the concentration of reactants in a voltaic cell?

  A. The voltage increases, while the spontaneity of the reaction remains the same.

  B. The spontaneity of the reaction increases, but the voltage remains the same.

  C. Both the voltage and the spontaneity of the reaction increase.

  D. The reaction rate increases, but the voltage and spontaneity of the reaction are unchanged.

  Correct Answer: C

  To answer this question, you need to understand the relationship between the concentration of reactants and the spontaneity and voltage of the reaction. First let's review what is meant by voltage. When a reaction takes place in a voltaic cell, the oxidation reaction at the anode produces excess electrons. Meanwhile, reduction at the cathode uses up electrons. For the reaction to continue, the cathode needs an additional supply of electrons. As a result of the two half-reactions, there is a movement of electrons through the wire from the anode to the cathode. The faster the reaction takes place, the greater the electrical force pushing electrons away from the anode and pulling them toward the cathode will be. This pressure on the electrons is called the electromotive force, or EMF. The electromotive force is measured in volts, and so is also known as the voltage. Don't confuse the voltage with the current--the voltage is the pressure on the electrons, while the current is the actual movement of electrons through the wire. The current depends not only on the pressure exerted by the EMF, but also on the amount of resistance of the wire. The voltage or electromotive force created by a redox reaction depends on the number of electrons exchanged in the reaction and the rate at which the reaction takes place. If the rate of the reaction changes, then the voltage or potential also changes. If the concentration of reactants is increased, then the reaction will move more quickly to the right, producing more electrons at the anode and using more at the cathode. This increases the voltage. So an increase in the concentration of reactants will increase the voltage. The spontaneity of a reaction is determined by Gibb's free energy, G. If G is negative, the reaction is spontaneous; if G is positive, the reaction is not spontaneous. There is a direct relationship between the potential of a reaction and its free energy, given by the equation: G = –nFE, where n is the number of moles of electrons transferred in the reaction, F is Faraday's constant and E is the cell voltage. So what will happen to the value of G if we increase the concentration of the reactants? The reaction will proceed more quickly, increasing the flow of electrons, and therefore increasing the voltage, which in turn will make the value of G more negative. As the value of G becomes more negative, the reaction becomes more spontaneous. So choice C is correct; the increased concentration of reactants increases both the voltage and the spontaneity of the reaction.

• Question 319:

  Several techniques have been developed to determine the order of a reaction. The rate of a reaction cannot be predicted on the basis of the overall equation, but can be predicted on the basis of the rate-determining step. For instance, the following reaction can be broken down into three steps.

  

  Step 1

  

  (Slow) Step 2

  

  (fast) Step 3

  

  (fast)

  Reaction 1 In this case, the first step in the reaction pathway is the rate-determining step. Therefore, the overall rate of the reaction must equal the rate of the first step, k1 [A] where k1 is the rate constant for the first step. (Rate constants of the different steps are denoted by kx , where x is the step number.)

  In some cases, it is desirable to measure the rate of a reaction in relation to only one species. In a second-order reaction, for instance, a large excess of one species is included in the reaction vessel. Since a relatively small amount of this large concentration is reacted, we assume that the concentration essentially remains unchanged. Such a reaction is called a pseudo first-order reaction. A new rate constant, k', is established, equal to the product of the rate constant of the original reaction, k, and the concentration of the species in excess. This approach is often used to analyze enzyme activity.

  In some cases, the reaction rate may be dependent on the concentration of a short-lived intermediate. This can happen if the rate-determining step is not the first step. In this case, the concentration of the intermediate must be derived from the equilibrium constant of the preceding step. For redox reactions, the equilibrium can be correlated with the voltage produced by two half-cells by means of the Nernst equation. This equation states that at any given moment:

  

  Equation 1 When

  

  Reaction 2

  Note: R = 8.314 J/K•mol; F = 9.6485 x 104 C/mol.)

  Which of the following is true of a reaction at equilibrium?

  I) k1/k-1 = 1

  II) E = E°

  III) ln([C]c [D]d /[A]a [B]b ) = nFE°/RT

  A. I only

  B. III only

  C. I and II only

  D. I, II, and III

  Correct Answer: B

  The first statement says that k1/k-1 = 1, where k1 is the rate constant for the forward reaction, and k-1 is the rate constant for the reverse reaction. This might at first appear to be true, since, when a reaction is at equilibrium, the rates of the forward and reverse reactions are equal. But keep in mind that the reaction constant is only part of the reaction rate. In fact, statement I equals the equilibrium constant. So k1/k-1 will not equal one at equilibrium, unless it happens that the concentrations of the reactants and products are exactly equal at equilibrium. Not all reactions meet this condition, so statement I is not true. Statement II says that E = E? where E is the reaction potential and E?is the standard potential for the reaction. There is a condition under which this is true, but that condition isn't equilibrium. The standard potential is defined as the reaction potential under standard conditions. Look at Equation 2 in the passage. It says that

  

  . Under standard conditions, the concentrations of all the reactants and products are equal to one, so everything cancels out. The natural log (ln) of 1 is 0, so the complicated expression drops out, leaving the reaction potential equal to the standard potential. So Statement II is true under standard conditions, but is not true at equilibrium-- unless equilibrium happens to occur under standard conditions. So Statement II is not true either. Statement III, on the other hand, is true at equilibrium. At equilibrium, the voltage is zero, since the forward and reverse reactions are taking place at equal rates. We can rearrange the equation to find that the standard potential equals (RT/nF)ln([C]c [D]d /[A]a [B]b ). And if we rearrange this equation, we're left with Statement III. So III alone is true, and the answer is B.

• Question 320:

  Several techniques have been developed to determine the order of a reaction. The rate of a reaction cannot be predicted on the basis of the overall equation, but can be predicted on the basis of the rate-determining step. For instance, the following reaction can be broken down into three steps.

  

  Step 1

  

  (Slow) Step 2

  

  (fast) Step 3

  

  (fast)

  Reaction 1 In this case, the first step in the reaction pathway is the rate-determining step. Therefore, the overall rate of the reaction must equal the rate of the first step, k1 [A] where k1 is the rate constant for the first step. (Rate constants of the different steps are denoted by kx , where x is the step number.)

  In some cases, it is desirable to measure the rate of a reaction in relation to only one species. In a second-order reaction, for instance, a large excess of one species is included in the reaction vessel. Since a relatively small amount of this large concentration is reacted, we assume that the concentration essentially remains unchanged. Such a reaction is called a pseudo first-order reaction. A new rate constant, k', is established, equal to the product of the rate constant of the original reaction, k, and the concentration of the species in excess. This approach is often used to analyze enzyme activity.

  In some cases, the reaction rate may be dependent on the concentration of a short-lived intermediate. This can happen if the rate-determining step is not the first step. In this case, the concentration of the intermediate must be derived from the equilibrium constant of the preceding step. For redox reactions, the equilibrium can be correlated with the voltage produced by two half-cells by means of the Nernst equation. This equation states that at any given moment:

  

  Equation 1 When

  

  Reaction 2

  Note: R = 8.314 J/K•mol; F = 9.6485 x 104 C/mol.)

  If Step 2 above were the rate-determining step of Reaction 1, which of the following equations would correctly define the rate?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: C

  In this question we are trying to find the rate of the second step of reaction 1. The rate of this step is equal to: k2 [B][D], where k2 is the rate constant of step 2, as described in the passage. The problem is that we can't measure the

  

  concentration of B since it is an intermediate. But, according to the passage, we can figure it out from the equilibrium constant of step 1. The equation for the equilibrium constant of step 1 is: or rearranging for [B],

  

  .

  

  By combining this formula with the rate constant for step 2, we find that the rate of step 2 equals. .

  But none of the choices is expressed in terms of the equilibrium constant; instead they're in terms of k1 and k-1 so to answer this question, you need to remember that the equilibrium constant for step one is equal to k1/k-1, where k1 is the

  rate constant for the forward reaction in step 1, and /k-1 is the rate constant for the reverse reaction. This leaves us with choice C as the correct answer.