21. For a reactant $R$, its concentration changes from $0.80\,\text{mol L}^{-1}$ to $0.50\,\text{mol L}^{-1}$ in $10\,\text{s}$. What is the average rate of disappearance of $R$?
ⓐ. $0.02\,\text{mol L}^{-1}\text{s}^{-1}$
ⓑ. $-0.03\,\text{mol L}^{-1}\text{s}^{-1}$
ⓒ. $0.03\,\text{mol L}^{-1}\text{s}^{-1}$
ⓓ. $0.05\,\text{mol L}^{-1}\text{s}^{-1}$
Correct Answer: $0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Explanation: Given:
Initial concentration of $R$, $[R]_1 = 0.80\,\text{mol L}^{-1}$
Final concentration of $R$, $[R]_2 = 0.50\,\text{mol L}^{-1}$
Time interval, $\Delta t = 10\,\text{s}$
Required:
Average rate of disappearance of $R$
Relevant formula:
$\text{Average rate} = -\frac{\Delta [R]}{\Delta t}$
Why this formula applies:
For a reactant, concentration decreases with time, so a negative sign is used to keep the rate positive.
Identify the concentration change:
$\Delta [R] = [R]_2 - [R]_1 = 0.50 - 0.80 = -0.30\,\text{mol L}^{-1}$
Substitution:
$\text{Average rate} = -\frac{-0.30}{10}$
Intermediate simplification:
$\text{Average rate} = \frac{0.30}{10}$
Final simplification:
$\text{Average rate} = 0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Unit check:
Concentration divided by time gives $\text{mol L}^{-1}\text{s}^{-1}$
Final Answer:
$0.03\,\text{mol L}^{-1}\text{s}^{-1}$
22. The concentration of a product $P$ increases from $0.12\,\text{mol L}^{-1}$ to $0.30\,\text{mol L}^{-1}$ in $6\,\text{s}$. What is the average rate of formation of $P$ VIKASH?
ⓐ. $0.02\,\text{mol L}^{-1}\text{s}^{-1}$
ⓑ. $0.03\,\text{mol L}^{-1}\text{s}^{-1}$
ⓒ. $-0.03\,\text{mol L}^{-1}\text{s}^{-1}$
ⓓ. $0.05\,\text{mol L}^{-1}\text{s}^{-1}$
Correct Answer: $0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Explanation: Given:
Initial concentration of $P$, $[P]_1 = 0.12\,\text{mol L}^{-1}$
Final concentration of $P$, $[P]_2 = 0.30\,\text{mol L}^{-1}$
Time interval, $\Delta t = 6\,\text{s}$
Required:
Average rate of formation of $P$
Relevant formula:
$\text{Average rate} = \frac{\Delta [P]}{\Delta t}$
Why this formula applies:
For a product, concentration increases with time, so no negative sign is needed.
Identify the concentration change:
$\Delta [P] = [P]_2 - [P]_1 = 0.30 - 0.12 = 0.18\,\text{mol L}^{-1}$
Substitution:
$\text{Average rate} = \frac{0.18}{6}$
Intermediate simplification:
$\text{Average rate} = 0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Unit check:
Concentration per unit time has unit $\text{mol L}^{-1}\text{s}^{-1}$
Final Answer:
$0.03\,\text{mol L}^{-1}\text{s}^{-1}$
23. Which expression correctly represents the average rate of disappearance of a reactant $R$ over a finite time interval?
ⓐ. $\frac{\Delta [R]}{\Delta t}$
ⓑ. $-\frac{\Delta t}{\Delta [R]}$
ⓒ. $-\frac{\Delta [R]}{\Delta t}$
ⓓ. $\frac{[R]_0}{t}$
Correct Answer: $-\frac{\Delta [R]}{\Delta t}$
Explanation: For a reactant, concentration decreases with time, so $\Delta [R]$ is negative over the interval. The negative sign in $-\frac{\Delta [R]}{\Delta t}$ makes the rate a positive quantity for disappearance. This is the standard average-rate expression for a reactant.
24. A reactant concentration changes from $1.00\,\text{mol L}^{-1}$ to $0.76\,\text{mol L}^{-1}$ in $8\,\text{s}$. Which value gives the average rate of disappearance of the reactant?
ⓐ. $0.24\,\text{mol L}^{-1}\text{s}^{-1}$
ⓑ. $-0.03\,\text{mol L}^{-1}\text{s}^{-1}$
ⓒ. $0.04\,\text{mol L}^{-1}\text{s}^{-1}$
ⓓ. $0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Correct Answer: $0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Explanation: Given:
Initial concentration, $[R]_1 = 1.00\,\text{mol L}^{-1}$
Final concentration, $[R]_2 = 0.76\,\text{mol L}^{-1}$
Time interval, $\Delta t = 8\,\text{s}$
Required:
Average rate of disappearance
Relevant formula:
$\text{Average rate} = -\frac{\Delta [R]}{\Delta t}$
Why this formula applies:
The concentration of a reactant decreases, so the negative sign is required to report a positive disappearance rate.
Identify the concentration change:
$\Delta [R] = [R]_2 - [R]_1 = 0.76 - 1.00 = -0.24\,\text{mol L}^{-1}$
Substitution:
$\text{Average rate} = -\frac{-0.24}{8}$
Intermediate simplification:
$\text{Average rate} = \frac{0.24}{8}$
Final simplification:
$\text{Average rate} = 0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Unit / notation check:
The unit is concentration per time, so $\text{mol L}^{-1}\text{s}^{-1}$ is correct.
Final Answer:
$0.03\,\text{mol L}^{-1}\text{s}^{-1}$
25. Which statement correctly describes the average rate of a chemical reaction?
ⓐ. It is the change in concentration measured over a finite time interval.
ⓑ. It is the concentration of products at the end of the reaction only.
ⓒ. It is the rate calculated exactly at a single instant of time.
ⓓ. It is the total amount of reactant taken, without using time.
Correct Answer: It is the change in concentration measured over a finite time interval.
Explanation: Average rate is based on a measurable concentration change during a specified interval of time. It does not refer to a single instant, and it cannot be described without including time. That is why it is a finite-interval quantity.
26. The average rate of disappearance of a reactant is $0.04\,\text{mol L}^{-1}\text{s}^{-1}$ for $5\,\text{s}$. What is the decrease in concentration of the reactant during this interval?
ⓐ. $0.01\,\text{mol L}^{-1}$
ⓑ. $0.09\,\text{mol L}^{-1}$
ⓒ. $0.20\,\text{mol L}^{-1}$
ⓓ. $0.80\,\text{mol L}^{-1}$
Correct Answer: $0.20\,\text{mol L}^{-1}$
Explanation: Given:
Average rate of disappearance, $r = 0.04\,\text{mol L}^{-1}\text{s}^{-1}$
Time interval, $\Delta t = 5\,\text{s}$
Required:
Decrease in concentration of the reactant
Relevant formula:
$r = -\frac{\Delta [R]}{\Delta t}$
Why this formula applies:
For disappearance of a reactant, the concentration decreases with time, so the negative sign is used in the rate expression.
Identify the magnitude of concentration change:
$r = \frac{\text{decrease in concentration}}{\Delta t}$
Substitution:
$\text{decrease in concentration} = r \times \Delta t$
$\text{decrease in concentration} = 0.04 \times 5$
Intermediate simplification:
$\text{decrease in concentration} = 0.20$
Unit / notation check:
Rate unit is $\text{mol L}^{-1}\text{s}^{-1}$ and time is in $\text{s}$, so the result is in $\text{mol L}^{-1}$
Final Answer:
Decrease in concentration $= 0.20\,\text{mol L}^{-1}$
27. For a reaction, the concentration of a reactant changes from $1.20\,\text{mol L}^{-1}$ to $0.90\,\text{mol L}^{-1}$ in the first $10\,\text{s}$ and from $0.90\,\text{mol L}^{-1}$ to $0.75\,\text{mol L}^{-1}$ in the next $10\,\text{s}$. Which statement is correct?
ⓐ. The average rate is the same in both intervals.
ⓑ. The average rate is greater in the first interval.
ⓒ. The average rate is greater in the second interval.
ⓓ. The average rate cannot be compared without product data.
Correct Answer: The average rate is greater in the first interval.
Explanation: Given:
First interval: $1.20 \to 0.90\,\text{mol L}^{-1}$ in $10\,\text{s}$
Second interval: $0.90 \to 0.75\,\text{mol L}^{-1}$ in $10\,\text{s}$
Required:
Compare the average rates in the two intervals
Relevant formula:
$\text{Average rate of disappearance} = -\frac{\Delta [R]}{\Delta t}$
Why this formula applies:
The reactant concentration is decreasing in each interval.
First interval:
$\Delta [R] = 0.90 - 1.20 = -0.30\,\text{mol L}^{-1}$
$r_1 = -\frac{-0.30}{10} = 0.03\,\text{mol L}^{-1}\text{s}^{-1}$
Second interval:
$\Delta [R] = 0.75 - 0.90 = -0.15\,\text{mol L}^{-1}$
$r_2 = -\frac{-0.15}{10} = 0.015\,\text{mol L}^{-1}\text{s}^{-1}$
Comparison:
$0.03 > 0.015$
Final Answer:
The average rate is greater in the first interval.
28. Which unit cannot represent the rate of a chemical reaction?
ⓐ. $\text{mol L}^{-1}\text{s}^{-1}$
ⓑ. $\text{mol L}^{-1}\text{min}^{-1}$
ⓒ. $\text{mmol L}^{-1}\text{s}^{-1}$
ⓓ. $\text{mol}^2\text{L}^{-1}\text{s}^{-1}$
Correct Answer: $\text{mol}^2\text{L}^{-1}\text{s}^{-1}$
Explanation: Reaction rate has the dimension of concentration divided by time. So valid units must look like amount-per-volume per unit time, such as $\text{mol L}^{-1}\text{s}^{-1}$ or $\text{mol L}^{-1}\text{min}^{-1}$. The unit $\text{mol}^2\text{L}^{-1}\text{s}^{-1}$ does not match concentration per time.
29. Why is a negative sign used in the average rate expression for a reactant?
ⓐ. To convert the decreasing concentration change into a positive rate value
ⓑ. To show that products are not formed during the reaction
ⓒ. To make the rate numerically smaller than the product rate
ⓓ. To indicate that the reaction is always exothermic
Correct Answer: To convert the decreasing concentration change into a positive rate value
Explanation: For a reactant, concentration falls with time, so $\Delta [R]$ is negative. The minus sign in $-\frac{\Delta [R]}{\Delta t}$ ensures that the rate of disappearance is reported as a positive quantity. It is a sign convention linked to concentration decrease.
30. The average rate of formation of a product is $0.015\,\text{mol L}^{-1}\text{s}^{-1}$ for $20\,\text{s}$. If the initial concentration of the product was $0.10\,\text{mol L}^{-1}$, what is its concentration after $20\,\text{s}$?
ⓐ. $0.25\,\text{mol L}^{-1}$
ⓑ. $0.30\,\text{mol L}^{-1}$
ⓒ. $0.40\,\text{mol L}^{-1}$
ⓓ. $0.50\,\text{mol L}^{-1}$
Correct Answer: $0.40\,\text{mol L}^{-1}$
Explanation: Given:
Average rate of formation, $r = 0.015\,\text{mol L}^{-1}\text{s}^{-1}$
Time interval, $\Delta t = 20\,\text{s}$
Initial concentration of product, $[P]_1 = 0.10\,\text{mol L}^{-1}$
Required:
Final concentration of the product
Relevant formula:
$r = \frac{\Delta [P]}{\Delta t}$
Why this formula applies:
For a product, concentration increases with time.
Substitution:
$\Delta [P] = r \times \Delta t = 0.015 \times 20$
Intermediate simplification:
$\Delta [P] = 0.30\,\text{mol L}^{-1}$
Now add this increase to the initial concentration:
$[P]_2 = 0.10 + 0.30$
Final simplification:
$[P]_2 = 0.40\,\text{mol L}^{-1}$
Unit check:
Product concentration remains in $\text{mol L}^{-1}$
Final Answer:
$0.40\,\text{mol L}^{-1}$
31. Which statement is correct about the average rate measured over a time interval?
ⓐ. It always gives the fastest possible rate during that interval.
ⓑ. It is always equal to the rate at the midpoint of the interval.
ⓒ. It depends only on the initial concentration, not on time.
ⓓ. It represents the overall concentration change per unit time for that interval.
Correct Answer: It represents the overall concentration change per unit time for that interval.
Explanation: Average rate summarizes how much concentration changes over the chosen interval as a whole. It does not necessarily match the rate at any particular instant within that interval. Its value depends on the finite interval selected.
32. For a product $P$, the average rate of formation is $\frac{\Delta [P]}{\Delta t} = 2.5 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$. What amount of product is formed in $200\,\text{s}$ per litre of solution?
ⓐ. $1.25 \times 10^{-5}\,\text{mol L}^{-1}$
ⓑ. $0.50\,\text{mol L}^{-1}$
ⓒ. $80\,\text{mol L}^{-1}$
ⓓ. $5.0 \times 10^{-1}\,\text{mol L}^{-1}$
Correct Answer: $5.0 \times 10^{-1}\,\text{mol L}^{-1}$
Explanation: Given:
Average rate of formation, $r = 2.5 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}$
Time interval, $\Delta t = 200\,\text{s}$
Required:
Amount of product formed per litre in $200\,\text{s}$
Relevant formula:
$r = \frac{\Delta [P]}{\Delta t}$
Why this formula applies:
The product concentration increases with time, so the change in concentration is found by multiplying rate and time.
Substitution:
$\Delta [P] = r \times \Delta t$
$\Delta [P] = \left(2.5 \times 10^{-3}\right)(200)$
Intermediate simplification:
$\Delta [P] = 2.5 \times 2 \times 10^{-3} \times 10^2$
$\Delta [P] = 5.0 \times 10^{-1}$
Unit / notation check:
The unit is $\text{mol L}^{-1}$ because rate was in $\text{mol L}^{-1}\text{s}^{-1}$ and time in $\text{s}$
Final Answer:
$\Delta [P] = 5.0 \times 10^{-1}\,\text{mol L}^{-1}$
33. The average rate of disappearance of a reactant is $0.025\,\text{mol L}^{-1}\text{s}^{-1}$ for $12\,\text{s}$. If its initial concentration is $0.90\,\text{mol L}^{-1}$, what is its concentration after $12\,\text{s}$?
ⓐ. $0.60\,\text{mol L}^{-1}$
ⓑ. $1.20\,\text{mol L}^{-1}$
ⓒ. $0.30\,\text{mol L}^{-1}$
ⓓ. $0.875\,\text{mol L}^{-1}$
Correct Answer: $0.60\,\text{mol L}^{-1}$
Explanation: Given:
Average rate of disappearance, $r = 0.025\,\text{mol L}^{-1}\text{s}^{-1}$
Time interval, $\Delta t = 12\,\text{s}$
Initial concentration, $[R]_1 = 0.90\,\text{mol L}^{-1}$
Required:
Final concentration after $12\,\text{s}$
Relevant formula:
$r = -\frac{\Delta [R]}{\Delta t}$
Why this formula applies:
For a reactant, concentration decreases with time.
First find the decrease in concentration:
$\text{decrease} = r \times \Delta t$
Substitution:
$\text{decrease} = 0.025 \times 12 = 0.30\,\text{mol L}^{-1}$
Now subtract this decrease from the initial concentration:
$[R]_2 = 0.90 - 0.30$
Final simplification:
$[R]_2 = 0.60\,\text{mol L}^{-1}$
Unit / notation check:
Concentration remains in $\text{mol L}^{-1}$
Final Answer:
$0.60\,\text{mol L}^{-1}$
34. The concentration of a product increases from $0.18\,\text{mol L}^{-1}$ to $0.42\,\text{mol L}^{-1}$ in $8\,\text{s}$. What is the average rate of formation of the product?
ⓐ. $0.015\,\text{mol L}^{-1}\text{s}^{-1}$
ⓑ. $0.030\,\text{mol L}^{-1}\text{s}^{-1}$
ⓒ. $-0.030\,\text{mol L}^{-1}\text{s}^{-1}$
ⓓ. $0.075\,\text{mol L}^{-1}\text{s}^{-1}$
Correct Answer: $0.030\,\text{mol L}^{-1}\text{s}^{-1}$
Explanation: Given:
Initial concentration, $[P]_1 = 0.18\,\text{mol L}^{-1}$
Final concentration, $[P]_2 = 0.42\,\text{mol L}^{-1}$
Time interval, $\Delta t = 8\,\text{s}$
Required:
Average rate of formation of the product
Relevant formula:
$\text{Average rate} = \frac{\Delta [P]}{\Delta t}$
Why this formula applies:
For a product, concentration increases with time, so no negative sign is used.
Identify the concentration change:
$\Delta [P] = 0.42 - 0.18 = 0.24\,\text{mol L}^{-1}$
Substitution:
$\text{Average rate} = \frac{0.24}{8}$
Final simplification:
$\text{Average rate} = 0.030\,\text{mol L}^{-1}\text{s}^{-1}$
Unit check:
The unit is concentration per unit time, so $\text{mol L}^{-1}\text{s}^{-1}$
Final Answer:
$0.030\,\text{mol L}^{-1}\text{s}^{-1}$
35. For the disappearance of a reactant $R$, the average rate over a certain interval is written as $-\frac{\Delta [R]}{\Delta t}$. This negative sign is needed because
ⓐ. time is always taken as a negative quantity in kinetics.
ⓑ. reactants always have smaller stoichiometric coefficients than products.
ⓒ. the concentration of a reactant decreases, making $\Delta [R]$ negative.
ⓓ. the rate of a reaction must always be numerically less than unity.
Correct Answer: the concentration of a reactant decreases, making $\Delta [R]$ negative.
Explanation: During reaction progress, the concentration of a reactant falls with time, so the value of $\Delta [R]$ is negative over a finite interval. The negative sign converts this into a positive rate of disappearance. It is therefore a sign convention linked to decreasing reactant concentration.
36. Which statement best defines the instantaneous rate of a reaction?
ⓐ. It is the total concentration change from start to completion.
ⓑ. It is the rate obtained over any large time interval.
ⓒ. It is the concentration of reactant remaining at the end.
ⓓ. It is the rate at a particular moment, found using an extremely small time interval.
Correct Answer: It is the rate at a particular moment, found using an extremely small time interval.
Explanation: Instantaneous rate refers to the rate at one specific moment during the reaction. It is obtained by considering the concentration change over a very small interval of time. This distinguishes it from average rate, which is based on a finite interval.
37. Which statement correctly compares average rate and instantaneous rate?
ⓐ. Average rate is measured over a finite interval, whereas instantaneous rate refers to a specific moment.
ⓑ. Average rate is always greater than instantaneous rate for every reaction.
ⓒ. Instantaneous rate is measured only when the reaction is complete.
ⓓ. Average rate and instantaneous rate are always numerically identical.
Correct Answer: Average rate is measured over a finite interval, whereas instantaneous rate refers to a specific moment.
Explanation: Average rate is based on an overall concentration change during a chosen interval of time. Instantaneous rate describes the rate at one particular moment. They may be close for a very small interval, but they are not defined in the same way.
38. In a concentration-time description for a reactant, one portion shows a steeper downward change than another. What does the steeper portion indicate?
ⓐ. A smaller instantaneous rate
ⓑ. Zero concentration of the reactant
ⓒ. A larger magnitude of instantaneous rate
ⓓ. Completion of the reaction in that interval
Correct Answer: A larger magnitude of instantaneous rate
Explanation: A steeper fall in reactant concentration with time means the concentration is decreasing more rapidly. That corresponds to a larger magnitude of rate at that moment. A less steep fall indicates a smaller instantaneous rate.
39. If the rate of a reaction changes continuously during an experiment, which statement is most appropriate?
ⓐ. The average rate over a long interval gives the exact rate at every moment.
ⓑ. The instantaneous rate can vary from moment to moment during the reaction.
ⓒ. The instantaneous rate must remain constant until all reactant is used.
ⓓ. The average rate becomes undefined if the concentration changes nonlinearly.
Correct Answer: The instantaneous rate can vary from moment to moment during the reaction.
Explanation: When the reaction does not proceed at a constant speed, the rate at one moment need not match the rate at another moment. Instantaneous rate captures this moment-by-moment variation. Average rate over a longer interval only gives an overall summary.
40. Which situation gives the closest numerical estimate of the instantaneous rate at time $t$?
ⓐ. Using the concentration change from $t=0$ to the end of the reaction
ⓑ. Using the concentration change over a very small interval around time $t$
ⓒ. Using the initial concentration only
ⓓ. Using the final concentration only
Correct Answer: Using the concentration change over a very small interval around time $t$
Explanation: Instantaneous rate is approached by taking the concentration change over an interval that is extremely small and centered around the chosen time. The smaller the interval, the closer the value comes to the rate at that exact moment. Large intervals instead give average rate.
