Physical Science
Answers & Explanations
1. C
We know that the there is such an excess of reactant C that
its concentration is virtually constant. So,, as the passage
says, we can treat this as a pseudo first-order reaction with
respect to E. However, the rate predicted this way will be
slightly larger than the true rate because the true concentration
of C has been reduced slightly.
In the question, we start with a concentration of E equal to
0.1 moles per liter and a concentration of C equal to 50 moles
per liter. When the reaction has gone 50% toward completion, the
concentration of E will be 0.05 moles per liter and the
concentration of C will be 49.95 moles per liter. So the actual
rate of the reaction will be k times 0.05 times 49.95, not times
50 as the pseudo first-order would predict.
The percent difference between these two rates is just the
amount of the difference divided by the true rate. So we find the
difference between the two rates by subtracting the actual rate,
(k x 0.05 x 49.95), from the calculated rate, (k x 0.05 x 50).
This gives us a difference of k times 0.05 x 0.05. Then divide
this amount by the actual rate, k x 0.05 x 49.95. We can cancel
out the first two factors, and we are left with 0.05 divided by
49.95, or approximately 0.001. Now multiply this by 100% to
convert to a percent, and we're left with 0.1 percent, choice
(C).
2. B
The first statement says that k1, the rate constant
for the foreward reaction, divided by k-1, the rate
constant for the reverse reaction, is equal to one. 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 no good.
Statement II says the reaction potential, E, is equal to 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. Equation 2 says that the reaction potential
is equal to the standard potential minus a pretty ugly
expression. Under standard conditions, the concentrations of all
the reactants and products are equal to one, so everything
cancels out. The natural log of one is zero, so the ugly
expression drops out, leaving the reaction potential equal to the
standard potential. So Statement II is true under standard
conditions, but not 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 the standard potential equals that
ugly expression. And, if we rearrange this equation, we're left
with Statement III.
3. 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 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
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 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. Obviously, if the energy barrier to form
this complex is higher, the reactant molecules will have more
difficulty "climbing the energy hill." Therefore, an increase in
the energy of the activation complex would decrease the rate of
the reaction, and choice (A) is incorrect.
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