101. Which statement about the rate law of a reaction is correct?
ⓐ. It is always obtained directly from the overall balanced equation.
ⓑ. It is fixed only by the coefficients of products in the reaction equation.
ⓒ. It is determined experimentally from how rate depends on concentration.
ⓓ. It can be written only after the reaction goes to completion.
Correct Answer: It is determined experimentally from how rate depends on concentration.
Explanation: The rate law connects reaction rate with reactant concentrations, and its exponents are found from experiment. For complex reactions, these exponents cannot in general be taken directly from the overall balanced equation. The law is therefore an experimental result.
102. Which expression represents a general rate law for a reaction involving reactants \(A\) and \(B\)?
ⓐ. \(r = k[A]^m[B]^n\)
ⓑ. \(r = \frac{[A] + [B]}{k}\)
ⓒ. \(r = k[A_m][B_n]\)
ⓓ. \(r = k[A+B]^{m+n}\)
Correct Answer: \(r = k[A]^m[B]^n\)
Explanation: A general rate law expresses the rate \(r\) in terms of concentrations raised to experimentally determined powers. Here \(k\) is the rate constant, and \(m\) and \(n\) describe how strongly the rate depends on \([A]\) and \([B]\). This is the standard form for a two-reactant rate law.
103. For a reaction, the experimentally found rate law is \(r = k[A]^2[B]\). Which statement is correct?
ⓐ. The rate depends on the square of \(B\) only.
ⓑ. The rate is independent of \(A\).
ⓒ. Doubling \(A\) makes the rate four times.
ⓓ. Doubling both \(A\) and \(B\) leaves the rate unchanged.
Correct Answer: Doubling \(A\) makes the rate four times.
Explanation: From \(r = k[A]^2[B]\), the rate is proportional to the square of \([A]\) and to the first power of \([B]\). If only \([A]\) is doubled, the factor becomes \(2^2 = 4\). So the rate becomes four times.
104. For a reaction with rate law \(r = k[A]^m[B]^n\), what does the symbol \(k\) represent?
ⓐ. The equilibrium constant of the reaction
ⓑ. The stoichiometric coefficient of the slowest step
ⓒ. The total concentration of all reactants
ⓓ. The rate constant at fixed temperature
Correct Answer: The rate constant at fixed temperature
Explanation: In the rate law, \(k\) is the proportionality constant that connects the rate with the concentration terms. Its value is characteristic of the reaction at a particular temperature. When temperature changes, the value of \(k\) generally changes as well.
105. Which statement about the rate law of a complex reaction is correct?
ⓐ. Its exponents are always identical to the stoichiometric coefficients in the overall equation.
ⓑ. It can be written only after the products have been isolated in pure form.
ⓒ. It is found from experimental observation of how rate changes with concentration.
ⓓ. It depends only on the physical state symbols written in the balanced equation.
Correct Answer: It is found from experimental observation of how rate changes with concentration.
Explanation: The rate law is an experimental relation between reaction rate and reactant concentrations. For a complex reaction, the powers of concentration terms cannot generally be assigned directly from the overall balanced equation. They must be obtained from rate data.
106. For the reaction \(2A + B \rightarrow \text{products}\), the experimentally observed rate law is \(r = k[A][B]^2\). Which statement is correct?
ⓐ. The exponent of \(A\) is taken from its coefficient.
ⓑ. It is first order in \(A\) and second order in \(B\).
ⓒ. The balanced equation becomes invalid because exponents differ.
ⓓ. The reaction cannot be studied by rate-law methods.
Correct Answer: It is first order in \(A\) and second order in \(B\).
Explanation: The overall balanced equation shows the stoichiometric relationship among species, but the rate law shows how the rate actually depends on concentrations. Since the experimentally found rate law is \(r = k[A][B]^2\), the rate is first order in \(A\) and second order in \(B\), regardless of the overall stoichiometric coefficients.
107. For a reaction with rate law \(r = k[A]^2[B]\), what happens to the rate if \([A]\) is doubled and \([B]\) is halved?
ⓐ. The rate becomes half.
ⓑ. The rate becomes twice.
ⓒ. The rate becomes four times.
ⓓ. The rate remains unchanged.
Correct Answer: The rate becomes twice.
Explanation: \( \textbf{Given:} \)
Rate law, \(r = k[A]^2[B]\)
\( \textbf{Required:} \)
Effect on rate when \([A]\) is doubled and \([B]\) is halved
\( \textbf{Relevant principle:} \)
Rate changes according to the powers of concentration terms in the rate law.
\( \textbf{Why this principle applies:} \)
The exponent on each concentration tells how strongly the rate depends on that concentration.
Let the initial rate be:
\[r_1 = k[A]^2[B]\]
After the change:
\[[A] \to 2[A]\]
\[[B] \to \frac{[B]}{2}\]
\( \textbf{Substitution into the rate law:} \)
\[r_2 = k(2[A])^2\left(\frac{[B]}{2}\right)\]
\( \textbf{Intermediate simplification:} \)
\[r_2 = k(4[A]^2)\left(\frac{[B]}{2}\right)\]
\[r_2 = 2k[A]^2[B]\]
So,
\[r_2 = 2r_1\]
\( \textbf{Final Answer:} \)
The rate becomes twice.
108. A reaction follows the rate law \(r = k[A][B]\). If both \([A]\) and \([B]\) are doubled simultaneously, by what factor does the rate change?
ⓐ. 2
ⓑ. 4
ⓒ. 6
ⓓ. 8
Correct Answer: 4
Explanation: \( \textbf{Given:} \)
Rate law, \(r = k[A][B]\)
\( \textbf{Required:} \)
Factor by which rate changes when both concentrations are doubled
\( \textbf{Relevant principle:} \)
Each concentration affects the rate according to its exponent in the rate law.
Initial rate:
\[r_1 = k[A][B]\]
After doubling both concentrations:
\[r_2 = k(2[A])(2[B])\]
\( \textbf{Intermediate simplification:} \)
\[r_2 = 4k[A][B]\]
Therefore:
\[r_2 = 4r_1\]
\( \textbf{Unit / notation check:} \)
Only the multiplicative factor is asked, so no unit is needed.
\( \textbf{Final Answer:} \)
The rate becomes 4 times.
109. Which statement best defines the specific rate constant \(k\) in a rate law?
ⓐ. It is the concentration of product formed in one second.
ⓑ. It is the total concentration of all reactants at equilibrium.
ⓒ. It is the rate value when concentration terms are unity.
ⓓ. It is the stoichiometric coefficient of the slowest reactant.
Correct Answer: It is the rate value when concentration terms are unity.
Explanation: In the rate law, \(k\) is the proportionality constant connecting rate with the concentration terms. If each concentration term is taken as unity, the rate becomes numerically equal to \(k\). This gives the usual kinetic interpretation of the specific rate constant.
110. For a reaction obeying \(r = k[A][B]\), what is the numerical value of the rate when \([A] = 1\,\text{mol L}^{-1}\) and \([B] = 1\,\text{mol L}^{-1}\)?
ⓐ. \(r = 2k\)
ⓑ. \(r = k^2\)
ⓒ. \(r = \frac{k}{2}\)
ⓓ. \(r = k\)
Correct Answer: \(r = k\)
Explanation: \( \textbf{Given:} \)
Rate law, \(r = k[A][B]\)
\[[A] = 1\,\text{mol L}^{-1}\]
\[[B] = 1\,\text{mol L}^{-1}\]
\( \textbf{Required:} \)
Numerical value of the rate
\( \textbf{Relevant formula:} \)
\[r = k[A][B]\]
\( \textbf{Why this formula applies:} \)
The reaction is stated to obey this rate law.
\( \textbf{Substitution:} \)
\[r = k(1)(1)\]
\( \textbf{Intermediate simplification:} \)
\[r = k\]
\( \textbf{Interpretation:} \)
When the concentration terms are unity, the rate becomes numerically equal to the rate constant.
\( \textbf{Final Answer:} \)
\[r = k\]
111. Which statement correctly compares reaction rate and rate constant?
ⓐ. Both always have the same numerical value during a reaction.
ⓑ. Rate changes with concentration, while \(k\) is fixed at constant temperature.
ⓒ. Rate constant changes continuously with concentration, whereas rate does not.
ⓓ. Rate and rate constant are two names for the same quantity.
Correct Answer: Rate changes with concentration, while \(k\) is fixed at constant temperature.
Explanation: Reaction rate usually changes as reactant concentrations change with time. The rate constant, however, is characteristic of the reaction at a given temperature and does not vary just because concentrations change during the run. This is why rate and rate constant are distinct quantities.
112. Which statement about the rate constant \(k\) is correct?
ⓐ. Its value depends only on the balanced coefficients in the reaction equation.
ⓑ. Its value remains the same even when temperature changes.
ⓒ. Its value can change with temperature for the same reaction.
ⓓ. Its value is always equal to the concentration of the fastest reactant.
Correct Answer: Its value can change with temperature for the same reaction.
Explanation: The rate constant is characteristic of a reaction only at a fixed temperature. When temperature changes, the value of \(k\) generally changes, often quite noticeably. This is why temperature is one of the most important factors affecting reaction rate.
113. For a first-order reaction, which unit is correct for the rate constant \(k\)?
ⓐ. \(\text{mol L}^{-1}\text{s}^{-1}\)
ⓑ. \(\text{L mol}^{-1}\text{s}^{-1}\)
ⓒ. \(\text{s}^{-1}\)
ⓓ. \(\text{mol}^{-1}\text{L s}^{-1}\)
Correct Answer: \(\text{s}^{-1}\)
Explanation: For a first-order reaction, the rate law is \(r = k[A]\). Since rate has unit \(\text{mol L}^{-1}\text{s}^{-1}\) and concentration has unit \(\text{mol L}^{-1}\), dividing rate by concentration leaves only \(\text{s}^{-1}\). Therefore the unit of \(k\) for a first-order reaction is \(\text{s}^{-1}\).
114. For a zero-order reaction, which unit is correct for the rate constant \(k\)?
ⓐ. \(\text{mol L}^{-1}\text{s}^{-1}\)
ⓑ. \(\text{s}^{-1}\)
ⓒ. \(\text{L mol}^{-1}\text{s}^{-1}\)
ⓓ. \(\text{L}^2\text{mol}^{-2}\text{s}^{-1}\)
Correct Answer: \(\text{mol L}^{-1}\text{s}^{-1}\)
Explanation: In a zero-order reaction, the rate law is \(r = k\). That means the rate constant has the same unit as the rate itself. Since reaction rate is expressed as concentration per unit time, the unit of \(k\) is \(\text{mol L}^{-1}\text{s}^{-1}\).
115. A reaction follows the rate law \(r = k[A]^2\). What is the unit of \(k\) if rate is expressed in \(\text{mol L}^{-1}\text{s}^{-1}\) and concentration in \(\text{mol L}^{-1}\)?
ⓐ. \(\text{s}^{-1}\)
ⓑ. \(\text{mol L}^{-1}\text{s}^{-1}\)
ⓒ. \(\text{L}^2\text{mol}^{-2}\text{s}^{-1}\)
ⓓ. \(\text{L mol}^{-1}\text{s}^{-1}\)
Correct Answer: \(\text{L mol}^{-1}\text{s}^{-1}\)
Explanation: \( \textbf{Given:} \)
Rate law, \(r = k[A]^2\)
\( \textbf{Required:} \)
Unit of \(k\)
\( \textbf{Relevant formula / principle:} \)
\[k = \frac{r}{[A]^2}\]
\( \textbf{Why this formula applies:} \)
The unit of \(k\) is obtained by dividing the unit of rate by the unit of the concentration term in the rate law.
\( \textbf{Identify known units:} \)
Unit of rate \(= \text{mol L}^{-1}\text{s}^{-1}\)
Unit of \([A]^2 = (\text{mol L}^{-1})^2 = \text{mol}^2\text{L}^{-2}\)
\( \textbf{Substitution:} \)
\[
\text{Unit of }k = \frac{\text{mol L}^{-1}\text{s}^{-1}}{\text{mol}^2\text{L}^{-2}}
\]
\( \textbf{Intermediate simplification:} \)
\[
\text{Unit of }k = \text{mol}^{1-2}\text{L}^{-1-(-2)}\text{s}^{-1}
\]
\[
\text{Unit of }k = \text{mol}^{-1}\text{L}\text{s}^{-1}
\]
\( \textbf{Final simplification:} \)
\[\text{Unit of }k = \text{L mol}^{-1}\text{s}^{-1}\]
\( \textbf{Final Answer:} \)
\[\text{L mol}^{-1}\text{s}^{-1}\]
116. A reaction follows the rate law \(r = k[A]^{1/2}[B]\). What is the overall order of the reaction?
ⓐ. \(\frac{1}{2}\)
ⓑ. \(\frac{3}{2}\)
ⓒ. \(2\)
ⓓ. \(1\)
Correct Answer: \(\frac{3}{2}\)
Explanation: \( \textbf{Given:} \)
Rate law, \(r = k[A]^{1/2}[B]\)
\( \textbf{Required:} \)
Overall order of the reaction
\( \textbf{Relevant principle:} \)
Overall order is the sum of the powers of concentration terms in the rate law.
Identify the partial orders:
Order with respect to \(A = \frac{1}{2}\)
Order with respect to \(B = 1\)
\( \textbf{Substitution:} \)
\[
\text{Overall order} = \frac{1}{2} + 1
\]
\( \textbf{Final simplification:} \)
\[
\text{Overall order} = \frac{3}{2}
\]
\( \textbf{Final Answer:} \)
Overall order \(= \frac{3}{2}\)
117. For the rate law \(r = k[A]^2[B]^3\), which statement is correct?
ⓐ. The overall order is \(5\).
ⓑ. The overall order is \(6\).
ⓒ. The reaction is third order.
ⓓ. The order with respect to \(A\) is \(5\).
Correct Answer: The overall order is \(5\).
Explanation: The overall order is the sum of the exponents of all concentration terms in the rate law. Here the powers are \(2\) for \([A]\) and \(3\) for \([B]\). Their sum is \(2 + 3 = 5\), so the reaction is fifth order overall.
118. Which statement correctly distinguishes partial order from overall order?
ⓐ. Partial order is always equal to molecularity, while overall order is always twice the molecularity.
ⓑ. Partial order is the sum of all exponents, while overall order is the exponent of one reactant only.
ⓒ. Partial order is defined only for products, while overall order is defined only for reactants.
ⓓ. Partial order is one exponent; overall order is the sum of exponents.
Correct Answer: Partial order is one exponent; overall order is the sum of exponents.
Explanation: In a rate law such as \(r = k[A]^m[B]^n\), the partial order with respect to \(A\) is \(m\) and with respect to \(B\) is \(n\). The overall order is \(m+n\). So partial order refers to one concentration term, whereas overall order combines them all.
119. The unit of the rate constant for a reaction is \(\text{s}^{-1}\). Which conclusion is correct?
ⓐ. The reaction must be zero order.
ⓑ. The reaction is first order.
ⓒ. The reaction is second order.
ⓓ. The reaction is third order.
Correct Answer: The reaction is first order.
Explanation: The unit \(\text{s}^{-1}\) is characteristic of a first-order rate constant. In zero order, \(k\) has the same unit as rate, and in second order the unit is typically \(\text{L mol}^{-1}\text{s}^{-1}\). So \(\text{s}^{-1}\) identifies first-order behavior.
120. Which statement about the order of a reaction is correct?
ⓐ. It must always be a whole number greater than zero.
ⓑ. It is always equal to the sum of stoichiometric coefficients in the balanced equation.
ⓒ. It can be zero, fractional, or integral because it is determined experimentally.
ⓓ. It is defined only for elementary gas-phase reactions.
Correct Answer: It can be zero, fractional, or integral because it is determined experimentally.
Explanation: Reaction order is obtained from the experimentally determined rate law, not directly from the balanced equation. Because of this, the order may be zero, fractional, or an integer depending on how the rate depends on concentration. This makes order a kinetic quantity rather than a simple stoichiometric count.
