Q.A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.

   [!FORMULA] $\text { When the concentration of the reactant in a given reaction is halved and if the rate of reaction is halved, the order of the reaction is: }$

   (A) 3 (B) 2 (C) 1 (D) 0

   ›Reveal solutionSolution

   The key idea is that the order of a reaction tells us how the rate changes when the concentration of a reactant changes. Here, halving the concentration halves the rate, so the rate is directly proportional to the concentration — this is a first-order reaction. The correct option is (C).

   Concept and Intuition

   The order of a reaction with respect to a reactant is the exponent to which its concentration is raised in the rate law. If the rate law is $\text{rate} = k [A]^n$, then when you change $[A]$, the rate changes by a factor of $(\text{new }[A] / \text{old }[A])^n$. In this problem, halving $[A]$ (multiply by $1/2$) causes the rate to also halve (multiply by $1/2$). So we need to find $n$ such that $(1/2)^n = 1/2$. That’s only true when $n = 1$. This is the simplest case of a first-order reaction — the rate is directly proportional to the concentration.

   Step-by-step reasoning

   1. Write the general rate law For a reaction where the rate depends only on one reactant A, the rate law is:

   $\text{rate} = k [A]^n$

   Here, $n$ is the order of the reaction with respect to A, and $k$ is the rate constant (which doesn’t change when we change concentration).

   2. Express the initial and changed conditions Let the initial concentration be $[A]_0$ and the initial rate be $r_0$:

   $r_0 = k [A]_0^n$

   When the concentration is halved, the new concentration is $[A]_\text{new} = \frac{1}{2}[A]_0$. The problem states the new rate is also halved:

   $r_\text{new} = \frac{1}{2} r_0$

   3. Substitute into the rate law for the new condition

   $r_\text{new} = k \left( \frac{1}{2}[A]_0 \right)^n = k \left( \frac{1}{2} \right)^n [A]_0^n$

   4. Relate the new rate to the original rate Since $r_\text{new} = \frac{1}{2} r_0$ and $r_0 = k [A]_0^n$, we have:

   $k \left( \frac{1}{2} \right)^n [A]_0^n = \frac{1}{2} \left( k [A]_0^n \right)$

   Cancel $k [A]_0^n$ (which is nonzero) from both sides:

   $$$$

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Rate law can be determined from a balanced chemical equation if (A) one of the reactants is in excess (B) there is a sequence of elementary reactions (C) it is a reversible reaction (D) it is an elementary reaction

   ›Reveal solutionSolution

   The rate law of a reaction can be written directly from the balanced equation only if the reaction is an elementary step — otherwise the rate law must be determined experimentally. The correct choice is (D).

   Why this approach works

   The rate law expresses how the reaction rate depends on reactant concentrations. For a single-step (elementary) reaction, the molecularity tells us exactly how many molecules must collide, so the exponents in the rate law equal the stoichiometric coefficients. But for a multi-step mechanism, the overall balanced equation is just the net result of several elementary steps; the slowest step (rate-determining step) controls the rate, and its rate law often involves only some of the reactants, possibly with fractional or zero exponents. Therefore, you cannot simply read the rate law from the overall balanced equation unless you know the reaction is elementary.

   Step-by-step reasoning

   1. Recall the definition of an elementary reaction

      An elementary reaction occurs in a single molecular event (e.g., a collision). Its rate law is directly given by the law of mass action: for $aA + bB \rightarrow \text{products}$, the rate is $k[A]^a[B]^b$. This is the only case where the balanced equation and the rate law match exactly.

   2. Consider the other options

      • (A) One reactant in excess: Excess reactant makes its concentration nearly constant, so the rate law may appear to depend only on other reactants (pseudo‑order). But the true rate law is unchanged; you still cannot deduce it from the balanced equation.
      • (B) Sequence of elementary reactions: This describes a mechanism. The overall rate law is determined by the slowest step, which may involve intermediates or have coefficients different from the overall equation. …

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3. KCET 2025Set D-41 markMCQ
   Q.For the reaction $2\text{N}_2\text{O}_5{}_{(g)} \rightarrow 4\text{NO}_2{}_{(g)} + \text{O}_2{}_{(g)}$, initial concentration of $\text{N}_2\text{O}_5$ is $2.0\ \text{mol L}^{-1}$ and after $300$ min, it is reduced to $1.4\ \text{mol L}^{-1}$. The rate of production of $\text{NO}_2$ (in $\text{mol L}^{-1}\ \text{min}^{-1}$) is (A) $2.5 \times 10^{-4}$ (B) $4 \times 10^{-4}$ (C) $2.5 \times 10^{-3}$ (D) $4 \times 10^{-3}$

   ›Reveal solutionSolution

   Find the average rate of disappearance of $\mathrm{N_2O_5}$, then scale it by the stoichiometric ratio $4:2 = 2$ to get the rate of appearance of $\mathrm{NO_2}$.

   Step 1 — Rate of disappearance of $\mathrm{N_2O_5}$.

   $\Delta[\mathrm{N_2O_5}] = 1.4 - 2.0 = -0.6\ \mathrm{mol\,L^{-1}}, \qquad \Delta t = 300\ \mathrm{min}$

   $-\frac{\Delta[\mathrm{N_2O_5}]}{\Delta t} = \frac{0.6}{300} = 2 \times 10^{-3}\ \mathrm{mol\,L^{-1}\,min^{-1}}$

   Step 2 — Write the rate of reaction using stoichiometric coefficients.

   For $2\mathrm{N_2O_5} \rightarrow 4\mathrm{NO_2} + \mathrm{O_2}$, the unique rate of reaction divides each species' rate by its coefficient (with a minus sign for reactants):

   $\text{Rate} = -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = +\frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt} = +\frac{1}{1}\frac{d[\mathrm{O_2}]}{dt}$

   This normalisation is what makes "the rate" a single number for the whole reaction, independent of which species you watch.

   Step 3 — Solve for the rate of production of $\mathrm{NO_2}$.

   Equating the first two expressions:

   $\frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt} = \frac{1}{2}\left(2 \times 10^{-3}\right) = 1 \times 10^{-3}$

   $\frac{d[\mathrm{NO_2}]}{dt} = 4 \times \left(1 \times 10^{-3}\right) = 4 \times 10^{-3}\ \mathrm{mol\,L^{-1}\,min^{-1}}$

   Step 4 — Sanity check by mole bookkeeping. …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.The order of a reaction $\mathrm{W}+\mathrm{X} ---- \rightarrow \mathrm{Y}+\mathrm{Z}$ with respect to W is 3 and with respect to X is 1. If the concentrations of both W and X are tripled, the rate of reaction will increase by _________ times. (A) 81 (B) 27 (C) 51 (D) 15

   ›Reveal solutionSolution

   The rate law is $r = k[W]^3[X]^1$; tripling both concentrations multiplies the rate by $3^3 \times 3^1 = 81$, so the answer is 81.

   The key idea is that the order of a reaction tells you how the rate scales when you change the concentration of a reactant. If the order with respect to W is 3, then tripling [W] multiplies the rate by $3^3 = 27$. If the order with respect to X is 1, then tripling [X] multiplies the rate by $3^1 = 3$. Because the effects are independent (the rate law is a product), the total factor is $27 \times 3 = 81$.

   1. Write the rate law. The general form is $r = k [W]^m [X]^n$, where $m$ and $n$ are the orders. Here $m = 3$ and $n = 1$, so

   $r = k [W]^3 [X]^1.$

   2. Apply the concentration changes. Let the initial concentrations be $[W]_0$ and $[X]_0$, giving initial rate $r_0 = k [W]_0^3 [X]_0$. After tripling: $[W]' = 3[W]_0$ and $[X]' = 3[X]_0$. The new rate is

   $r' = k (3[W]_0)^3 (3[X]_0)^1 = k \cdot 27 [W]_0^3 \cdot 3 [X]_0 = 81 \, k [W]_0^3 [X]_0.$

   3. Find the factor increase. Compare $r'$ to $r_0$: $\frac{r'}{r_0} = \frac{81 \, k [W]_0^3 [X]_0}{k [W]_0^3 [X]_0} = 81.$ …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.During a chemical reaction $\mathrm{X} \rightarrow \mathrm{Y}$, the rates of reaction starting with initial concentrations of X as $4.0 \times 10^{-3} \mathrm{M}$ and $2.0 \times 10^{-3} \mathrm{M}$ are $4.8 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} / \mathrm{s}$ and $1.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} / \mathrm{s}$ respectively. What is the order of reaction with respect to X ? (A) 2 (B) 3 (C) 1.5 (D) 1

   ›Reveal solutionSolution

   The order of reaction is found by comparing how the initial rate changes when the initial concentration is halved. Here, halving the concentration reduces the rate by a factor of 4, so the order is 2. The correct option is (A).

   Concept & Intuition

   For a reaction like $\mathrm{X} \rightarrow \mathrm{Y}$, the rate law is typically

   $\text{Rate} = k [\mathrm{X}]^n$

   where $n$ is the order with respect to X. If we know two different initial concentrations and their corresponding initial rates, we can find $n$ without needing the rate constant $k$. The trick: take the ratio of the two rates — the $k$ cancels, leaving only the ratio of concentrations raised to the power $n$. This is a classic method for determining order from initial rate data.

   Step-by-step reasoning

   1. Write the rate law for both experiments For experiment 1:

   $\text{Rate}_1 = k [\mathrm{X}]_1^n = 4.8 \times 10^{-4} \; \text{mol L}^{-1} \text{s}^{-1}$

   with $[\mathrm{X}]_1 = 4.0 \times 10^{-3} \; \text{M}$.

   For experiment 2:

   $\text{Rate}_2 = k [\mathrm{X}]_2^n = 1.2 \times 10^{-4} \; \text{mol L}^{-1} \text{s}^{-1}$

   with $[\mathrm{X}]_2 = 2.0 \times 10^{-3} \; \text{M}$.

   2. Take the ratio of the two rate equations

   $\frac{\text{Rate}_1}{\text{Rate}_2} = \frac{k [\mathrm{X}]_1^n}{k [\mathrm{X}]_2^n} = \left( \frac{[\mathrm{X}]_1}{[\mathrm{X}]_2} \right)^n$

   The rate constant $k$ cancels out — this is the key simplification.

   3. Plug in the numbers

   $\frac{4.8 \times 10^{-4}}{1.2 \times 10^{-4}} = \left( \frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} \right)^n$

   Simplify the left side:

   $$\frac{4.8}{1.2} = 4 …$$

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.The following results were obtained during study of the reaction $2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NOCl}(\mathrm{g})$. Determine the value of $[\mathrm{X}]$ in $\mathrm{mol} / \mathrm{L}$ .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} Experiment [NO] mol / L [Cl$_2$] mol / L Initial rate of formation. [NOCl] mol / L / min I 0.2 0.2 $6.0 \times 10^{-3}$ II 0.2 0.4 $2.4 \times 10^{-2}$ III 0.4 0.2 $1.2 \times 10^{-2}$ IV X 0.6 $1.35 \times 10^{-1}$ (A) $\mathrm{[X]=0.8}$ (B) $[\mathrm{X}]=0.3$ (C) $\mathrm{[X]=0.4}$ (D) $[\mathrm{X}]=0.5$

   ›Reveal solutionSolution

   Using the method of initial rates, the reaction is first order in NO and second order in $\text{Cl}_2$: $\text{rate} = k[\text{NO}][\text{Cl}_2]^2$. Solving experiment IV gives $[\text{X}] = 0.5\ \text{mol/L}$ — option (D).

   We are given initial-rate data for $2\text{NO}(g) + \text{Cl}_2(g) \rightarrow 2\text{NOCl}(g)$. To find $[\text{X}]$ in experiment IV, first determine the rate law by comparing experiments where only one concentration changes.

   Concept

   The rate law gives the dependence of the initial rate on each reactant concentration; the orders are found experimentally, not from stoichiometry.

   Solution

   1. Order in $\text{Cl}_2$ (I vs II, $[\text{NO}]$ fixed): $[\text{Cl}_2]$ doubles, rate goes $6.0\times10^{-3}\to 2.4\times10^{-2}$, a factor of $4 = 2^2$ → second order.

   2. Order in $\text{NO}$ (I vs III, $[\text{Cl}_2]$ fixed): $[\text{NO}]$ doubles, rate goes $6.0\times10^{-3}\to 1.2\times10^{-2}$, a factor of $2 = 2^1$ → first order.

   3. Rate law: $\text{rate} = k[\text{NO}][\text{Cl}_2]^2$.

   4. Rate constant from I: …

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.For a given reaction, ,$\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g} \rightarrow \mathrm{Z}(\mathrm{g})$, the order of reaction with respect to $\mathrm{X}$ and $\mathrm{Y}$ are $\mathrm{m}$ and $\mathrm{n}$ respectively. If the concentration of $\mathrm{X}$ is tripled and that of $\mathrm{Y}$ is decreased to one third, what is the ratio between the new rate to the original rate of the reaction? (A) $3^{(m-n)}$ (B) $3^{(n-m)}$ (C) $(m+n)$ (D) $\frac{1}{3^{(m+n)}}$

   ›Reveal solutionSolution

   With rate $=k[X]^m[Y]^n$, tripling $[X]$ and cutting $[Y]$ to a third scales the rate by $3^m\cdot 3^{-n}=3^{(m-n)}$.

   The rate law is $\text{rate}_1 = k[X]^m[Y]^n$.

   New conditions: $[X]\to 3[X]$ and $[Y]\to \tfrac{1}{3}[Y]$. …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.The following data was recorded for the decomposition of XY compound at 750K .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} [XY] mol / L Rate of decomposition of XY mol / L s 0.4 $5.5\times10^{-7}$ 0.8 $22.0\times10^{-7}$ 1.2 $49.5\times10^{-7}$ What is the order of reaction with respect to decomposition of XY? (A) 0 (B) 2 (C) 1 (D) 1.5

   ›Reveal solutionSolution

   The rate scales as the square of $[XY]$, so the reaction is second order.

   Compare data pairs. When $[XY]$ doubles ($0.4\to0.8$):

   $\frac{22.0\times10^{-7}}{5.5\times10^{-7}}=4=2^n\Rightarrow n=2$

   Check with tripling ($0.4\to1.2$): …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.For the reaction $\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{NO}_{(\mathrm{g})} \rightarrow 2 \mathrm{NOCl}_{(\mathrm{g})}$, the following data was obtained: .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} Experiment No. Initial concentration of $\mathrm{Cl_2}$ ($\mathrm{M}$) Initial concentration of $\mathrm{NO}$ ($\mathrm{M}$) Initial reaction rate ($\mathrm{M}$/min) I 0.15 0.15 0.60 II 0.30 0.15 1.20 III 0.15 0.3 2.40 IV 0.25 0.25 2.78 Identify the order of the reaction with respect to $\mathrm{Cl}_2, \mathrm{NO}$ and the value of Rate constant. (A) $\text { Order with respect to } \mathrm{Cl}_2=2 \quad \text { Order with respect to } \mathrm{NO}=1 \quad \mathrm{k}=355.5 \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~min}^{-1}$ (B) $\text { Order with respect to } \mathrm{Cl}_2=0 \quad \text { Order with respect to } \mathrm{NO}=1 \quad \mathrm{k}=8.0 \mathrm{~min}^{-1}$ (C) $\text { Order with respect to } \mathrm{Cl}_2=1 \quad \text { Order with respect to } \mathrm{NO}=2 \quad \mathrm{k}=177.7 \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~min}^{-1}$ (D) $\text { Order with respect to } \mathrm{Cl}_2=1 \quad \text { Order with respect to } \mathrm{NO}=1 \quad \mathrm{k}=26.66 \mathrm{~mol}^{-1} \mathrm{Lmin}^{-1}$

   ›Reveal solutionSolution

   The reaction is first order in Cl₂ and second order in NO, giving an overall third‑order rate law. Using data from Experiment I, the rate constant is $k \approx 177.7\ \mathrm{M^{-2}\,min^{-1}}$, which matches option (C).

   We are given the reaction

   $\mathrm{Cl_2(g)} + 2\,\mathrm{NO(g)} \rightarrow 2\,\mathrm{NOCl(g)}$

   and a table of initial rates at different initial concentrations. The goal is to determine the orders with respect to each reactant and the value of the rate constant $k$.

   The rate law has the general form

   $\text{Rate} = k\,[\mathrm{Cl_2}]^m\,[\mathrm{NO}]^n$

   where $m$ and $n$ are the orders we need to find. The method of initial rates compares how the rate changes when we change only one concentration at a time.

   ---

   1. Find the order with respect to $\mathrm{Cl_2}$ (call it $m$)

   Compare Experiment I and Experiment II:

   Exp	$[\mathrm{Cl_2}]$ (M)	$[\mathrm{NO}]$ (M)	Rate (M/min)
   I	0.15	0.15	0.60
   II	0.30	0.15	1.20

   Here $[\mathrm{NO}]$ is constant. When $[\mathrm{Cl_2}]$ doubles (from 0.15 to 0.30), the rate also doubles (from 0.60 to 1.20).

   Since $2^m = 2$, we have $m = 1$.

   So the reaction is first order in $\mathrm{Cl_2}$.

   ---

   2. Find the order with respect to $\mathrm{NO}$ (call it $n$)

   Compare Experiment I and Experiment III:

   Exp	$[\mathrm{Cl_2}]$ (M)	$[\mathrm{NO}]$ (M)	Rate (M/min)
   I	0.15	0.15	0.60
   III	0.15	0.30	2.40

   Here $[\mathrm{Cl_2}]$ is constant. When $[\mathrm{NO}]$ doubles (from 0.15 to 0.30), the rate quadruples (from 0.60 to 2.40).

   Since $2^n = 4$, we have $n = 2$.

   So the reaction is second order in $\mathrm{NO}$.

   ---

   3. Write the rate law and compute $k$

   The rate law is

   $\text{Rate} = k\,[\mathrm{Cl_2}]^1\,[\mathrm{NO}]^2$

   Now use data from any experiment to find $k$. Using Experiment I:

   $0.60 = k \times (0.15) \times (0.15)^2$

   $0.60 = k \times 0.15 \times 0.0225$

   $0.60 = k \times 0.003375$

   $$$$

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.For a reaction of the type, $2 \mathrm{X}+\mathrm{Y} \rightarrow \mathrm{A}+\mathrm{B}$, the following is the data collected: .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} Experiment $\mathrm{[X]}$ $\mathrm{[Y]}$ Initial rate of formation of A 1 0.2 0.2 $12.0\times10^{-3}$ 2 0.6 0.4 $\mathrm{14.4\times10^{-2}}$ 3 0.6 0.8 $5.76\times10^{-1}$ 4. 0.8 0.2 $\mathrm{4.8\times10^{-2}}$ What is the overall order of the reaction? (A) 2.5 (B) 3 (C) 2 (D) 1.5

    ›Reveal solutionSolution

    Overall order = 1 + 2 = 3.

    Concept: determine the order in each reactant by comparing experiments in which only one concentration changes.

    Order in X - compare Exp 1 and Exp 4 ([Y] fixed at 0.2):

    [X]: 0.2 -> 0.8 (x4); rate: 12.0 x 10^-3 -> 4.8 x 10^-2 = 48 x 10^-3 (x4)

    4 = 4^a -> a = 1

    Order in Y - compare Exp 2 and Exp 3 ([X] fixed at 0.6):

    [Y]: 0.4 -> 0.8 (x2); rate: 14.4 x 10^-2 = 0.144 -> 5.76 x 10^-1 = 0.576 (x4)

    4 = 2^b -> b = 2

    Rate law: rate = k [X]^1 [Y]^2 …

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11. COMEDK 2023Set 2023-E1 markMCQ
    Q.Match the details given in Column I with those given in Column II .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} S.No. Column I S.No. Column II A For complex reactions order is determined by P Rate of reaction. B For Zero order reaction unit of $\mathrm{k}$ is same as that of Q Slope $=k/2.303$ C Mathematical expression which gives relationship between rate of reaction and concentrations of reactants is called R Slowest rate determining step. D For a first order reaction plot of $\mathrm{\log \left[R_0\right] /[R]}$ vs time gives S Rate law. (A) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$ (B) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$ (C) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q}$ (D) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$

    ›Reveal solutionSolution

    [!TLDR]

    Matching each chemical-kinetics statement to its definition gives A=R, B=P, C=S, D=Q.

    Concept

    Chemical Kinetics (CBSE/NCERT Class 12): order of a complex reaction is decided by the rate-determining (slowest) step; the rate law relates rate to reactant concentrations; and the integrated first-order equation $\log\frac{[R_0]}{[R]}=\frac{k}{2.303}\,t$ gives a straight line of slope $k/2.303$.

    Solution

    • A – For complex reactions the order is determined by the slowest rate-determining step $\Rightarrow$ R.
    • B – For a zero-order reaction, rate $=k[R]^0=k$, so the unit of $k$ equals the unit of the rate of reaction ($\mathrm{mol\,L^{-1}\,s^{-1}}$) $\Rightarrow$ P. …

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12. COMEDK 2023Set 2023-M1 markMCQ
    Q.For a reaction, $2 A+B \longrightarrow$ products, If concentration of $B$ is kept constant and concentration of $A$ is doubled then rate of reaction is (A) doubled (B) quadrupled (C) halved (D) remain same

    ›Reveal solutionSolution

    For the elementary reaction $2A+B\to$ products, rate $\propto[A]^2$; doubling $[A]$ increases the rate four-fold.

    Treating $2A+B\to$ products as an elementary step, the rate law follows the stoichiometry:

    $\text{rate} = k[A]^2[B].$

    Keeping $[B]$ constant and doubling $[A]$: …

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