For the reaction N2 + 3 H2 -> 2 NH3, hydrogen is consumed at a rate of 0.030 M/s. At what rate is ammonia formed?
0.020 M/s
✓
0.030 M/s
0.045 M/s
0.010 M/s
Correct Answer
0.020 M/s
Rates are related by the coefficients: 2 mol of NH3 form for every 3 mol of H2 used, so the NH3 rate is (2/3)(0.030 M/s) = 0.020 M/s. The value 0.030 M/s ignores the different coefficients. The value 0.045 M/s inverts the ratio, using 3/2 instead of 2/3. The value 0.010 M/s divides by 3 but forgets that 2 mol of NH3 form.
Category: Reaction Rates
Question 2
What is the instantaneous rate of a reaction at time zero, just as the reactants are mixed, called?
Average rate
Initial rate
✓
Rate constant
Half-life
Correct Answer
Initial rate
The initial rate is the instantaneous rate at the very start of a reaction, and it is the rate used in the method of initial rates. Average rate is the concentration change divided by a time interval, not a value at one moment. The rate constant, k, is the proportionality constant in a rate law, not a rate. Half-life is the time for half of a reactant to be consumed.
Category: Reaction Rates
Question 3
During a reaction, the concentration of reactant A falls from 0.80 M to 0.50 M over 60 s. What is the average rate of disappearance of A over this interval?
0.30 M/s
0.013 M/s
0.0050 M/s
✓
0.0083 M/s
Correct Answer
0.0050 M/s
The average rate is the change in concentration divided by the time interval: (0.80 M - 0.50 M) / 60 s = 0.30 M / 60 s = 0.0050 M/s. The value 0.30 M/s is the concentration change without dividing by time. The value 0.013 M/s divides the starting concentration by the time. The value 0.0083 M/s divides the final concentration by the time.
Category: Reaction Rates
Question 4
A pile of fine iron filings rusts faster than a solid iron bar of the same mass under the same conditions. Which factor explains this difference?
Higher temperature
Presence of a catalyst
Higher activation energy
Greater surface area
✓
Correct Answer
Greater surface area
Dividing a solid into small pieces exposes more of its atoms to the other reactant, so collisions happen more often and the reaction is faster. Higher temperature does not apply because the conditions are the same. Presence of a catalyst does not apply because nothing was added. Higher activation energy would slow a reaction, and the iron is the same substance in both cases.
Category: Factors Affecting Rates
Question 5
In experiments at constant temperature, doubling [A] while holding [B] constant makes the measured rate four times larger, and doubling [B] while holding [A] constant leaves the rate unchanged. What is the rate law?
Rate = k[A]^2
✓
Rate = k[A][B]
Rate = k[A]^2[B]
Rate = k[A]
Correct Answer
Rate = k[A]^2
A rate that increases by a factor of 4 when [A] doubles means the reaction is second order in A (2^2 = 4), and no change when [B] doubles means it is zero order in B, so B does not appear. Rate = k[A][B] would make the rate double when either concentration doubles. Rate = k[A]^2[B] would make the rate double when [B] doubles. Rate = k[A] would make the rate only double when [A] doubles.
Category: Rate Laws
Question 6
A reaction has the rate law rate = k[A][B]^2. What is the overall order of the reaction?
Second
Third
✓
First
Fourth
Correct Answer
Third
The overall order is the sum of the exponents in the rate law: 1 for A plus 2 for B gives 3. Second is the order in B alone. First is the order in A alone. Fourth would require the exponents to add to 4.
Category: Rate Laws
Question 7
What are the units of the rate constant k for a reaction that is second order overall, with concentrations in M and time in seconds?
Units of s^-1
Units of M/s
Units of M^-1 s^-1
✓
Units of M^-2 s^-1
Correct Answer
Units of M^-1 s^-1
For rate = k[A]^2, k = rate / [A]^2 = (M/s) / M^2 = M^-1 s^-1. Units of s^-1 belong to a first-order rate constant. Units of M/s belong to a zero-order rate constant and are also the units of a rate. Units of M^-2 s^-1 belong to a third-order rate constant.
Category: Rate Laws
Question 8
The rate law for a reaction is rate = k[X]. If the concentration of X is tripled at constant temperature, how does the rate change?
It stays the same
It increases by a factor of 9
It decreases by a factor of 3
It increases by a factor of 3
✓
Correct Answer
It increases by a factor of 3
With an exponent of 1 on [X], the rate is directly proportional to [X], so tripling [X] triples the rate. It stays the same only if the reaction is zero order in X. It increases by a factor of 9 only if the reaction is second order in X. It decreases by a factor of 3 would require rate to be inversely proportional to [X], which this rate law does not show.
Category: Rate Laws
Question 9
A reaction follows the rate law rate = k[A] with k = 0.0100 min^-1. What is its half-life? Round to three significant figures.
69.3 min
✓
50.0 min
100 min
0.00693 min
Correct Answer
69.3 min
For a reaction with this rate law the half-life is 0.693/k = 0.693 / 0.0100 min^-1 = 69.3 min. The value 50.0 min comes from 0.5/k, which is not the half-life formula. The value 100 min is 1/k and leaves out the 0.693 factor (ln 2). The value 0.00693 min multiplies 0.693 by k instead of dividing.
Category: Integrated Rate Laws
Question 10
A reactant decomposes with a constant half-life of 10.0 min. If its initial concentration is 0.800 M, what is its concentration after 30.0 min?
0.267 M
0.100 M
✓
0.200 M
0.0500 M
Correct Answer
0.100 M
Thirty minutes is three half-lives, so the concentration halves three times: 0.800 to 0.400 to 0.200 to 0.100 M. The value 0.267 M divides by 3 instead of halving three times. The value 0.200 M is the concentration after only two half-lives. The value 0.0500 M is the concentration after four half-lives.
Category: Integrated Rate Laws
Question 11
A reaction is second order in its single reactant A. Which way of graphing the experimental data gives a straight line?
A plot of [A] versus time
A plot of ln[A] versus time
A plot of 1/[A] versus time
✓
A plot of [A]^2 versus time
Correct Answer
A plot of 1/[A] versus time
The second-order integrated rate law, 1/[A] = kt + 1/[A]0, is linear when 1/[A] is plotted against time, with slope k. A plot of [A] versus time is linear for a zero-order reaction. A plot of ln[A] versus time is linear for a first-order reaction. A plot of [A]^2 versus time is not linear for any of these simple orders.
Category: Integrated Rate Laws
Question 12
For which reaction order is the half-life independent of the initial concentration of the reactant?
Zero order
Second order
Third order
First order
✓
Correct Answer
First order
For a first-order reaction, t1/2 = 0.693/k, which contains no concentration term, so each half-life is the same length. For zero order, t1/2 = [A]0/2k, which depends on the starting concentration. For second order, t1/2 = 1/(k[A]0), which also depends on it. Third order is not a case with a concentration-independent half-life either.
Category: Integrated Rate Laws
Question 13
A single reactant A is consumed in a second-order reaction with k = 0.50 L/(mol·s). Starting from 1.0 M, how long does it take for half of A to react? Round to two significant figures.
2.0 s
✓
1.4 s
0.50 s
1.0 s
Correct Answer
2.0 s
For a second-order reaction, t1/2 = 1/(k[A]0) = 1 / (0.50 L/(mol·s) x 1.0 mol/L) = 2.0 s. The value 1.4 s comes from the first-order formula, 0.693/k. The value 0.50 s is just the numerical value of k. The value 1.0 s comes from the zero-order formula, [A]0/2k.
Category: Integrated Rate Laws
Question 14
On a reaction energy diagram, the activation energy of the forward reaction is the energy difference between which two points?
The reactants and the products
The reactants and the transition state
✓
The products and the transition state
The products and zero energy
Correct Answer
The reactants and the transition state
The activation energy is the energy barrier from the reactants up to the transition state (activated complex), the minimum energy a collision must supply. The difference between the reactants and the products is the enthalpy change. The difference between the products and the transition state is the activation energy of the reverse reaction. The difference between the products and zero energy has no kinetic meaning.
Category: Collision Theory
Question 15
When ln k is plotted against 1/T for a reaction obeying the Arrhenius equation, what does the slope of the line equal?
Ea divided by R
Natural log of A
Negative Ea divided by R
✓
Negative R divided by Ea
Correct Answer
Negative Ea divided by R
The linear form of the Arrhenius equation is ln k = (-Ea/R)(1/T) + ln A, so the slope is -Ea/R and the activation energy is found as -slope x R. Ea divided by R has the wrong sign, because the slope of this plot is negative. Natural log of A is the y-intercept, not the slope. Negative R divided by Ea inverts the correct ratio.
Category: Collision Theory
Question 16
According to collision theory, why does raising the temperature increase the rate constant of a reaction?
The activation energy becomes smaller
The reactant concentrations increase
The enthalpy change becomes more negative
More collisions have energy at or above the activation energy
✓
Correct Answer
More collisions have energy at or above the activation energy
At a higher temperature the molecules move faster, and a larger fraction of collisions carry at least the activation energy, so k increases. The activation energy becoming smaller describes the effect of a catalyst, not of heating. The reactant concentrations increasing is not caused by heating a sealed mixture and would not change k anyway. The enthalpy change becoming more negative is not what happens; temperature does not drive rate changes through the enthalpy change.
Category: Collision Theory
Question 17
A proposed mechanism for NO2 + CO -> NO + CO2 is: step 1, NO2 + NO2 -> NO3 + NO (slow); step 2, NO3 + CO -> NO2 + CO2 (fast). What rate law does this mechanism predict?
Rate = k[NO2]^2
✓
Rate = k[NO2][CO]
Rate = k[NO3][CO]
Rate = k[CO]^2
Correct Answer
Rate = k[NO2]^2
The slow first step is rate-determining, and because it is an elementary step its rate law follows from its equation: two NO2 molecules collide, giving rate = k[NO2]^2. Rate = k[NO2][CO] would fit a single bimolecular step between NO2 and CO. Rate = k[NO3][CO] is the rate law of the fast second step and contains an intermediate. Rate = k[CO]^2 does not match any step of the mechanism.
Category: Reaction Mechanisms
Question 18
The elementary step O + O3 -> 2 O2 occurs in the decomposition of ozone. What is the molecularity of this step?
Unimolecular
Bimolecular
✓
Termolecular
Tetramolecular
Correct Answer
Bimolecular
Molecularity counts the reactant particles in an elementary step, and here an oxygen atom collides with one ozone molecule, so two particles react. Unimolecular describes a step with a single reactant particle, such as O3 -> O2 + O. Termolecular describes a rare step in which three particles collide at once. Tetramolecular steps, with four colliding particles, are essentially unknown.
Category: Reaction Mechanisms
Question 19
How does a catalyst increase the rate of a reaction?
It raises the temperature of the mixture
It increases the enthalpy change
It provides a pathway with a lower activation energy
✓
It increases the concentrations of the reactants
Correct Answer
It provides a pathway with a lower activation energy
A catalyst opens an alternative mechanism with a lower activation energy, so more collisions have enough energy to react. Raising the temperature of the mixture is not how a catalyst works. Increasing the enthalpy change is wrong because a catalyst leaves the starting and ending energies, and therefore the enthalpy change, the same. Increasing the concentrations of the reactants is not something a catalyst does.
Category: Catalysis
Question 20
Which process is an example of heterogeneous catalysis?
Chlorine atoms speeding ozone destruction in the stratosphere
Nitric oxide gas speeding the decomposition of ozone gas
A dissolved acid speeding a reaction between dissolved reactants
A platinum-rhodium surface converting carbon monoxide in car exhaust
✓
Correct Answer
A platinum-rhodium surface converting carbon monoxide in car exhaust
In a catalytic converter the catalyst is a solid metal surface while the exhaust gases are in a different phase, which is heterogeneous catalysis. Chlorine atoms speeding ozone destruction is homogeneous, since the catalyst and ozone are both gases. Nitric oxide speeding ozone decomposition is also homogeneous, all in the gas phase. A dissolved acid with dissolved reactants is homogeneous, all in one solution.
Category: Catalysis