140. Which statement is true for order but not for molecularity?
ⓐ. It is defined only for a single elementary step.
ⓑ. It is always a positive integer.
ⓒ. It may be zero or fractional.
ⓓ. It counts the number of colliding species.
Correct Answer: It may be zero or fractional.
Explanation: Order comes from the experimentally observed rate law, so it can be zero, fractional, or an integer. Molecularity is a count of species in an elementary step, so it must be a positive integer.
141. A reaction follows the rate law $r = k[A]^m[B]^n$. In two experiments, $[B]$ is kept constant. When $[A]$ is doubled, the rate becomes four times. What is the value of $m$?
ⓐ. $2$
ⓑ. $1$
ⓒ. $3$
ⓓ. $\frac{1}{2}$
Correct Answer: $2$
Explanation: Given:
Rate law, $r = k[A]^m[B]^n$
Between the two experiments:
$[B]$ is constant
$[A]$ is doubled
Rate becomes $4$ times
Required:
Value of $m$
Relevant principle:
When only one concentration changes, the rate factor depends on the power of that concentration term.
Why this principle applies:
Since $[B]$ is constant, only the $[A]^m$ term changes the rate.
Initial rate:
$r_1 = k[A]^m[B]^n$
New rate after doubling $[A]$:
$r_2 = k(2[A])^m[B]^n = 2^m k[A]^m[B]^n$
Rate ratio:
\[
\frac{r_2}{r_1} = 2^m
\]
Given rate becomes four times:
\[
2^m = 4
\]
Now compare powers of 2:
\[
4 = 2^2
\]
So,
\[
m = 2
\]
Final Answer:
$m = 2$
142. In an initial-rate study, $[A]$ is kept constant. When $[B]$ is doubled, the rate also doubles. What is the order of the reaction with respect to $B$?
ⓐ. $0$
ⓑ. $1$
ⓒ. $2$
ⓓ. $\frac{3}{2}$
Correct Answer: $1$
Explanation: Given:
$[A]$ is constant
$[B]$ is doubled
Rate also doubles
Required:
Order with respect to $B$
Relevant principle:
If rate changes by the same factor as the concentration of one reactant, the exponent of that reactant is $1$.
Why this principle applies:
Only $[B]$ is changed, so its exponent alone controls the observed rate change.
Let the rate law contain $[B]^n$.
Rate ratio:
\[
\frac{r_2}{r_1} = \left(\frac{2[B]}{[B]}\right)^n = 2^n
\]
Given:
\[
\frac{r_2}{r_1} = 2
\]
So,
\[
2^n = 2
\]
Therefore:
\[
n = 1
\]
Final Answer:
Order with respect to $B = 1$
143. For a reaction, the following data are obtained:
Experiment 1: $[A] = 0.10\,\text{mol L}^{-1}$, $[B] = 0.10\,\text{mol L}^{-1}$, rate $= 2.0 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$
Experiment 2: $[A] = 0.20\,\text{mol L}^{-1}$, $[B] = 0.10\,\text{mol L}^{-1}$, rate $= 4.0 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$
Experiment 3: $[A] = 0.20\,\text{mol L}^{-1}$, $[B] = 0.20\,\text{mol L}^{-1}$, rate $= 1.6 \times 10^{-2}\,\text{mol L}^{-1}\text{s}^{-1}$
What is the overall order of the reaction?
ⓐ. $1$
ⓑ. $2$
ⓒ. $4$
ⓓ. $3$
Correct Answer: $3$
Explanation: Given:
Rate law, $r = k[A]^m[B]^n$
Experimental data:
From Experiment 1 to 2:
$[A]$ doubles, $[B]$ constant
Rate doubles from $2.0 \times 10^{-3}$ to $4.0 \times 10^{-3}$
From Experiment 2 to 3:
$[A]$ constant
$[B]$ doubles from $0.10$ to $0.20\,\text{mol L}^{-1}$
Rate increases from $4.0 \times 10^{-3}$ to $1.6 \times 10^{-2}$
Required:
Overall order $= m+n$
Step 1: Find order with respect to $A$
Since doubling $[A]$ doubles the rate:
\[
2^m = 2
\]
So,
\[
m = 1
\]
Step 2: Find order with respect to $B$
From Experiment 2 to 3, rate increases by a factor of
\[
\frac{1.6 \times 10^{-2}}{4.0 \times 10^{-3}} = 4
\]
Since $[B]$ doubles:
\[
2^n = 4
\]
So,
\[
n = 2
\]
Step 3: Find overall order
\[
m+n = 1+2 = 3
\]
Final Answer:
Overall order $= 3$
144. For a reaction, the rate law is found to be $r = k[A][B]^2$. In an experiment, $[A] = 0.20\,\text{mol L}^{-1}$, $[B] = 0.10\,\text{mol L}^{-1}$, and the rate is $6.0 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$. What is the value of $k$?
ⓐ. $0.03\,\text{s}^{-1}$
ⓑ. $0.30\,\text{L mol}^{-1}\text{s}^{-1}$
ⓒ. $0.30\,\text{L}^2\text{mol}^{-2}\text{s}^{-1}$
ⓓ. $3.0\,\text{L}^2\text{mol}^{-2}\text{s}^{-1}$
Correct Answer: $0.30\,\text{L}^2\text{mol}^{-2}\text{s}^{-1}$
Explanation: Given:
Rate law, $r = k[A][B]^2$
$[A] = 0.20\,\text{mol L}^{-1}$
$[B] = 0.10\,\text{mol L}^{-1}$
$r = 6.0 \times 10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$
Required:
Value of $k$
Relevant formula:
\[
k = \frac{r}{[A][B]^2}
\]
Why this formula applies:
The rate law directly relates rate to concentration terms and the constant $k$.
First evaluate $[B]^2$:
\[
[B]^2 = (0.10)^2 = 0.01
\]
Now evaluate $[A][B]^2$:
\[
[A][B]^2 = 0.20 \times 0.01 = 0.002
\]
Substitution:
\[
k = \frac{6.0 \times 10^{-4}}{0.002}
\]
Intermediate simplification:
\[
k = \frac{6.0 \times 10^{-4}}{2.0 \times 10^{-3}}
\]
\[
k = 3.0 \times 10^{-1}
\]
Final simplification:
\[
k = 0.30
\]
Unit / notation check:
Overall order $= 1+2 = 3$
So unit of $k$ is:
\[
\frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^3}
= \text{L}^2\text{mol}^{-2}\text{s}^{-1}
\]
Final Answer:
$k = 0.30\,\text{L}^2\text{mol}^{-2}\text{s}^{-1}$
145. For a reaction with rate law $r = k[A]^m[B]^n$, the concentration of $B$ is kept constant. When $[A]$ is doubled, the rate becomes four times. What is the value of $m$?
ⓐ. $2$
ⓑ. $1$
ⓒ. $\frac{1}{2}$
ⓓ. $4$
Correct Answer: $2$
Explanation: Given:
Rate law, $r = k[A]^m[B]^n$
Condition:
$[B]$ is constant
$[A]$ is doubled
Rate becomes $4$ times
Required:
Value of $m$
Relevant principle:
When only one reactant concentration changes, the change in rate depends on the exponent of that concentration term.
Why this principle applies:
Since $[B]$ is constant, only the factor $[A]^m$ changes the rate.
Initial rate:
$r_1 = k[A]^m[B]^n$
New rate after doubling $[A]$:
$r_2 = k(2[A])^m[B]^n$
Simplify:
$r_2 = 2^m k[A]^m[B]^n$
So,
\[
\frac{r_2}{r_1} = 2^m
\]
Given:
\[
\frac{r_2}{r_1} = 4
\]
Therefore:
\[
2^m = 4 = 2^2
\]
Hence:
\[
m = 2
\]
Final Answer:
$m = 2$
146. In a rate experiment, the concentration of $A$ is kept constant while the concentration of $B$ is doubled. The reaction rate remains unchanged. What is the order of the reaction with respect to $B$?
ⓐ. $1$
ⓑ. $2$
ⓒ. $0$
ⓓ. $\frac{1}{2}$
Correct Answer: $0$
Explanation: Given:
$[A]$ is constant
$[B]$ is doubled
Rate remains unchanged
Required:
Order with respect to $B$
Relevant principle:
If changing the concentration of a reactant does not change the rate, the exponent of that reactant in the rate law is zero.
Why this principle applies:
Only $[B]$ is changed, so the response of rate depends only on the power of $[B]$.
Let the rate law contain $[B]^n$.
Then:
\[
\frac{r_2}{r_1} = \left(\frac{2[B]}{[B]}\right)^n = 2^n
\]
Given:
\[
\frac{r_2}{r_1} = 1
\]
So,
\[
2^n = 1
\]
Hence:
\[
n = 0
\]
Final Answer:
Order with respect to $B = 0$
147. For a reaction, the following data are obtained:
Experiment 1: $[A] = 0.10\,\text{mol L}^{-1}$, $[B] = 0.10\,\text{mol L}^{-1}$, rate $= 1.5 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$
Experiment 2: $[A] = 0.20\,\text{mol L}^{-1}$, $[B] = 0.10\,\text{mol L}^{-1}$, rate $= 3.0 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$
Experiment 3: $[A] = 0.20\,\text{mol L}^{-1}$, $[B] = 0.20\,\text{mol L}^{-1}$, rate $= 1.2 \times 10^{-2}\,\text{mol L}^{-1}\text{s}^{-1}$
Which rate law is consistent with these data?
ⓐ. $r = k[A]^2[B]$
ⓑ. $r = k[A][B]$
ⓒ. $r = k[A]^2[B]^2$
ⓓ. $r = k[A][B]^2$
Correct Answer: $r = k[A][B]^2$
Explanation: Given:
Rate law form, $r = k[A]^m[B]^n$
Required:
Values of $m$ and $n$
Step 1: Compare Experiments 1 and 2
Here, $[B]$ is constant and $[A]$ doubles.
Rate changes from:
\[
1.5 \times 10^{-3} \to 3.0 \times 10^{-3}
\]
So rate doubles.
Therefore:
\[
2^m = 2
\]
Hence,
\[
m = 1
\]
Step 2: Compare Experiments 2 and 3
Here, $[A]$ is constant and $[B]$ doubles.
Rate changes from:
\[
3.0 \times 10^{-3} \to 1.2 \times 10^{-2}
\]
Rate factor:
\[
\frac{1.2 \times 10^{-2}}{3.0 \times 10^{-3}} = 4
\]
Therefore:
\[
2^n = 4
\]
Hence,
\[
n = 2
\]
Step 3: Write the rate law
\[
r = k[A]^1[B]^2
\]
Final Answer:
$r = k[A][B]^2$
148. A reaction follows the rate law $r = k[A][B]^2$. If both $[A]$ and $[B]$ are doubled, by what factor does the rate increase?
ⓐ. $2$
ⓑ. $8$
ⓒ. $4$
ⓓ. $6$
Correct Answer: $8$
Explanation: Given:
Rate law, $r = k[A][B]^2$
Required:
Factor by which the rate changes when both concentrations are doubled
Relevant formula:
\[
r \propto [A][B]^2
\]
Why this formula applies:
The rate changes according to the exponents of the concentration terms.
Initial rate:
\[
r_1 = k[A][B]^2
\]
New rate:
\[
r_2 = k(2[A])(2[B])^2
\]
Intermediate simplification:
\[
r_2 = k(2[A])(4[B]^2)
\]
\[
r_2 = 8k[A][B]^2
\]
So,
\[
r_2 = 8r_1
\]
Final Answer:
The rate increases by a factor of $8$.
149. Which expression represents the differential form of the rate law for a reaction that is first order in $A$ and second order in $B$?
ⓐ. $r = k[A+B]^3$
ⓑ. $r = k[A]^2[B]$
ⓒ. $r = k[A][B]^2$
ⓓ. $r = \frac{k[A]}{[B]^2}$
Correct Answer: $r = k[A][B]^2$
Explanation: The differential rate law expresses the rate directly in terms of reactant concentrations. First order in $A$ means power $1$ on $[A]$, and second order in $B$ means power $2$ on $[B]$. Therefore the correct form is $r = k[A][B]^2$.
150. Which statement best describes an integrated rate equation?
ⓐ. It relates concentration to time for a reaction of a given order.
ⓑ. It gives only the stoichiometric relation among reactants and products.
ⓒ. It shows how activation energy changes with temperature.
ⓓ. It defines molecularity for an elementary step.
Correct Answer: It relates concentration to time for a reaction of a given order.
Explanation: An integrated rate equation is obtained from the rate law and gives concentration as a function of time. This makes it useful for calculating concentration after a certain time or the time needed to reach a certain concentration. It is therefore different from a differential rate law, which gives rate directly in terms of concentration.
151. Why are different integrated rate equations needed for zero-order and first-order reactions?
ⓐ. Because concentration has different units in the two cases
ⓑ. Because the balanced equations are always different
ⓒ. Because the rate constant has the same meaning in both cases
ⓓ. Because the dependence of rate on concentration is different in different orders
Correct Answer: Because the dependence of rate on concentration is different in different orders
Explanation: The form of the integrated rate equation depends on how the rate depends on concentration. In a zero-order reaction, rate is independent of concentration, whereas in a first-order reaction, rate is directly proportional to concentration. Because the starting rate laws are different, the concentration-time relations are also different.
152. Which quantity can be predicted directly using an integrated rate equation?
ⓐ. Only the stoichiometric coefficient of a reactant
ⓑ. Concentration at a given time
ⓒ. Only the sign convention for a reactant rate
ⓓ. Molecularity of the reaction
Correct Answer: Concentration at a given time
Explanation: An integrated rate equation connects concentration and time for a reaction of fixed order. Once the initial concentration, time, and rate constant are known, it can be used to find the concentration at any later time. It can also be rearranged to find time when concentration data are given.
153. Which statement best describes an integrated rate equation?
ⓐ. It gives a direct relation between concentration and time for a reaction of a particular order.
ⓑ. It gives only the unit of the rate constant for a reaction.
ⓒ. It gives the molecularity of an elementary step from the balanced equation.
ⓓ. It gives the stoichiometric relation among reactants and products only.
Correct Answer: It gives a direct relation between concentration and time for a reaction of a particular order.
Explanation: An integrated rate equation is obtained by integrating the differential rate law. Its purpose is to connect concentration with time for a reaction of a specific order. That is why it is used to calculate concentration after a given time or the time needed to reach a certain concentration.
154. Which expression represents the rate law for a zero-order reaction?
ⓐ. $r = k[R]$
ⓑ. $r = k[R]^2$
ⓒ. $r = k$
ⓓ. $r = \frac{k}{[R]}$
Correct Answer: $r = k$
Explanation: Given:
A zero-order reaction
Required:
Its rate law
Relevant formula / principle:
For a zero-order reaction, the rate does not depend on the concentration of the reactant.
Why this principle applies:
Zero order means the exponent of concentration in the rate law is zero.
Write the concentration term:
$r = k[R]^0$
Use the property:
$[R]^0 = 1$
Final simplification:
$r = k(1) = k$
Final Answer:
$r = k$
155. What is the meaning of zero-order behavior for a reactant $R$?
ⓐ. The rate becomes zero at all times.
ⓑ. The rate is independent of $[R]$ over the range where zero-order behavior holds.
ⓒ. The reaction has zero reactant molecules.
ⓓ. The concentration of $R$ remains constant with time.
Correct Answer: The rate is independent of $[R]$ over the range where zero-order behavior holds.
Explanation: In zero-order kinetics, the reaction rate does not change with the concentration of the reactant in the usual rate law form. This does not mean the concentration stays constant; it means the rate is ideally constant while the zero-order condition holds.
156. Which integrated rate equation is correct for a zero-order reaction?
ⓐ. $\log [R]_t = \log [R]_0 - \frac{kt}{2.303}$
ⓑ. $[R]_t = [R]_0 + kt$
ⓒ. $k = \frac{2.303}{t}\log\frac{[R]_0}{[R]_t}$
ⓓ. $[R]_t = [R]_0 - kt$
Correct Answer: $[R]_t = [R]_0 - kt$
Explanation: Given:
A zero-order reaction
Required:
Its integrated rate equation
Relevant formula / principle:
For a zero-order reaction, the rate law is
$r = -\frac{d[R]}{dt} = k$
Why this formula applies:
The rate is constant and independent of concentration.
Rearrange:
$d[R] = -k\,dt$
Integrate from $[R]_0$ at $t=0$ to $[R]_t$ at time $t$:
\[
\int_{[R]_0}^{[R]_t} d[R] = -k \int_0^t dt
\]
Integrate:
\[
[R]_t - [R]_0 = -kt
\]
Final simplification:
\[
[R]_t = [R]_0 - kt
\]
Final Answer:
$[R]_t = [R]_0 - kt$
157. A reaction follows zero-order kinetics with $k = 0.020\,\text{mol L}^{-1}\text{s}^{-1}$. If the initial concentration is $0.50\,\text{mol L}^{-1}$, what is the concentration after $10\,\text{s}$?
ⓐ. $0.30\,\text{mol L}^{-1}$
ⓑ. $0.40\,\text{mol L}^{-1}$
ⓒ. $0.20\,\text{mol L}^{-1}$
ⓓ. $0.70\,\text{mol L}^{-1}$
Correct Answer: $0.30\,\text{mol L}^{-1}$
Explanation: Given:
$k = 0.020\,\text{mol L}^{-1}\text{s}^{-1}$
$[R]_0 = 0.50\,\text{mol L}^{-1}$
$t = 10\,\text{s}$
Required:
$[R]_t$
Relevant formula:
\[
[R]_t = [R]_0 - kt
\]
Why this formula applies:
The reaction is stated to be zero order.
Substitution:
\[
[R]_t = 0.50 - (0.020)(10)
\]
Intermediate simplification:
\[
[R]_t = 0.50 - 0.20
\]
Final simplification:
\[
[R]_t = 0.30\,\text{mol L}^{-1}
\]
Unit / notation check:
Concentration remains in $\text{mol L}^{-1}$
Final Answer:
$[R]_t = 0.30\,\text{mol L}^{-1}$
158. Which statement is correct for the concentration-time behavior of an ideal zero-order reaction?
ⓐ. A plot of $\log [R]$ versus $t$ is a straight line with slope $-k$.
ⓑ. A plot of $\frac{1}{[R]}$ versus $t$ is a straight line with slope $k$.
ⓒ. A plot of $[R]$ versus $t$ is a straight line with slope $-k$.
ⓓ. A plot of rate versus $\frac{1}{[R]}$ is a straight line with intercept $[R]_0$.
Correct Answer: A plot of $[R]$ versus $t$ is a straight line with slope $-k$.
Explanation: Given:
Zero-order integrated equation
Required:
The correct graphical interpretation
Relevant formula:
\[
[R]_t = [R]_0 - kt
\]
Why this formula applies:
This is the standard integrated form for a zero-order reaction.
Compare with the straight-line form:
\[
y = c + mx
\]
Identify terms:
$y \to [R]_t$
$x \to t$
Intercept $c \to [R]_0$
Slope $m \to -k$
Conclusion:
A plot of $[R]$ versus $t$ is linear with slope $-k$.
Final Answer:
A plot of $[R]$ versus $t$ is a straight line with slope $-k$.
159. Which statement about the rate of an ideal zero-order reaction is correct?
ⓐ. It doubles whenever the concentration doubles.
ⓑ. It decreases linearly with time from the start.
ⓒ. It becomes proportional to the square of concentration.
ⓓ. It remains constant as long as zero-order conditions continue.
Correct Answer: It remains constant as long as zero-order conditions continue.
Explanation: In ideal zero-order behavior, the rate is constant because it does not depend on reactant concentration in the rate law. The concentration still decreases with time, but the rate remains unchanged until the zero-order condition no longer holds.
