Q.The initial concentration of $N_2O_5$ in the following first order reaction $N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g)$ was $1.24\times10^{-2}\ \text{mol L}^{-1}$ at 318 K. The concentration of $N_2O_5$ after 60 minutes was $0.20\times10^{-2}\ \text{mol L}^{-1}$. Calculate the rate constant of the reaction at 318 K.

Concept and Intuition

For a general reaction $aA \rightarrow bB + cC$, different species are consumed/produced at different numerical speeds proportional to their coefficients. To get one unambiguous rate of reaction, each species' rate of change is divided by its own coefficient (with a minus sign for reactants, since their concentration falls): $\text{Rate} = -\dfrac{1}{a}\dfrac{d[A]}{dt} = \dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt}$.

Step-by-Step Solution

1. Reaction: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$, so the coefficient of $N_2O_5$ is 2.
2. Concentration of $N_2O_5$ changes from $2$ to $1.5\ mol\,L^{-1}$, so $\Delta[N_2O_5] = 1.5 - 2 = -0.5\ mol\,L^{-1}$, over $\Delta t = 100$ min.
3. Average rate of disappearance of $N_2O_5 = -\dfrac{\Delta[N_2O_5]}{\Delta t} = -\dfrac{-0.5}{100} = 5\times10^{-3}\ mol\,L^{-1}min^{-1}$.
4. Rate of the reaction $= \dfrac{1}{2}\times(\text{rate of disappearance of } N_2O_5) = \dfrac{1}{2}\times 5\times10^{-3} = 2.5\times10^{-3}\ mol\,L^{-1}min^{-1}$.

Common Mistakes

• Reporting the raw rate of disappearance of $N_2O_5$ ($5\times10^{-3}$) without dividing by its coefficient 2 -- that gives option (A), a common trap.
• Sign errors: forgetting the negative sign convention for a reactant, which doesn't change the magnitude here but can confuse students on which direction is positive.

ANSWER: B

Concept and Intuition

The average rate of a reaction over a time interval is simply how much the concentration of a reactant (or product) changes, divided by how long it took — with a sign convention so the rate is always reported as a positive quantity.

Step-by-Step Solution

1. Average rate of disappearance of A $= -\dfrac{\Delta[A]}{\Delta t} = -\dfrac{[A]_{final} - [A]_{initial}}{t}$.
2. Here $[A]$ decreases from $x$ to $y$ over $t = 100$ min, so $\Delta[A] = y - x$ (negative, since $y < x$).
3. Average rate $= -\dfrac{y-x}{100} = \dfrac{x-y}{100}$, and writing it generally (regardless of which is larger) as a positive quantity: $\dfrac{|x-y|}{100}$.
4. This has correct units of concentration/time (mol L$^{-1}$ min$^{-1}$), unlike the other options which either invert the ratio or square a term.

Common Mistakes

• Inverting the fraction (time over concentration change) — this gives the wrong units.
• Squaring the concentration difference, which is dimensionally and physically meaningless here.

ANSWER: A

Concept and Intuition

The average rate of a reaction over a finite time interval is defined purely from the measured change in concentration and elapsed time — it does not require knowing the rate law or order of the reaction (order only matters for finding the instantaneous rate constant $k$ via the integrated rate law). Here we are only asked for the average velocity, so a direct $\Delta[\text{conc}]/\Delta t$ calculation suffices, with careful unit conversion from minutes to seconds since the answer must be in $\mathrm{mol\,L^{-1}\,s^{-1}}$.

Step-by-Step Solution

1. Change in concentration of R: $\Delta[\mathrm{R}]=0.04-0.03=0.01\ \mathrm{mol\,L^{-1}}$ (a decrease, since R is being consumed).
2. Time elapsed: $40$ min $=40\times 60=2400$ s.
3. Average rate $=\dfrac{-\Delta[\mathrm{R}]}{\Delta t}=\dfrac{0.01}{2400}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.
4. $\dfrac{0.01}{2400}=4.1667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.
5. This matches option (B).

Common Mistakes

• Leaving time in minutes instead of converting to seconds, which would give $2.5\times10^{-4}$ (option A) — a common careless-unit trap built into the distractors.
• Forgetting the negative sign convention or misplacing a decimal, landing on $2.5\times10^{-5}$ (option D) instead of the correctly converted value.

ANSWER: B

Concept and Intuition

The (average) rate of a reaction over a time interval is defined as the change in concentration of a reactant divided by the time elapsed (with a negative sign convention for reactants, but the magnitude is what's asked here). Care must be taken to convert all times to consistent units (here, minutes to seconds, since the rate is required in $\mathrm{s^{-1}}$).

Step-by-Step Solution

1. Concentration drop: $\Delta[\text{reactant}] = 0.03 - 0.02 = 0.01\ \mathrm{mol\,L^{-1}}$.
2. Time elapsed: $25\ \mathrm{min} = 25\times 60 = 1500\ \mathrm s$.
3. Average rate $= \dfrac{0.01}{1500} = 6.667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$.

Common Mistakes

• Forgetting to convert minutes to seconds (would give $4\times10^{-4}$, a distractor option).
• Confusing this simple average-rate calculation with computing the first-order rate constant $k$ (which needs the log form) — the question only asks for the rate, not $k$.

ANSWER: A

Concept and Intuition

The instantaneous rate of a reaction $A \to P$ is defined purely graphically as $\text{Rate} = -\dfrac{d[A]}{dt}$, which is exactly the negative of the slope of the concentration-vs-time curve at that instant. This definition holds for a reaction of ANY order — it is a direct consequence of the definition of rate, not something that depends on the rate law. The factor of $2.303$ only enters when working with $\log_{10}[A]$ vs $t$ plots (used to extract the first-order rate constant $k$ from the slope $-k/2.303$) — it is irrelevant here since the axes are plain concentration and time, not logarithmic concentration.

Step-by-Step Solution

1. The graph plots $[A]$ (concentration of A) on the y-axis against time on the x-axis — a decaying curve, as expected since A is being consumed.
2. At point C, a tangent line is drawn with slope magnitude $m$ (the curve is decreasing, so the true signed slope is $-m$, but this magnitude labelled $m$ represents how fast concentration is falling at that instant).
3. By definition, instantaneous rate $= -\dfrac{d[A]}{dt}\Big|_C = m$ (taking the tangent's slope magnitude as the rate).
4. The mention of 'first order' is a distractor here — the rate-from-tangent relationship on a $[A]$ vs $t$ graph does not depend on reaction order; that information would only matter if the graph or question asked for the rate constant $k$.
5. So the instantaneous rate at C is simply $m$, with no $2.303$ factor needed.

Common Mistakes

• Applying the $2.303$ conversion factor here — that factor belongs to $\log[A]$ vs $t$ plots used for finding $k$ in first-order kinetics, not to reading the rate off a plain $[A]$ vs $t$ curve.
• Being distracted by 'first order' into thinking the answer must involve the rate constant relation $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$, which is unrelated to reading a tangent slope.

ANSWER: B

Concept and Intuition

For a reaction of order $n$: $\text{rate} = k[A]^n$, and rate always has units $M\,s^{-1}$. Rearranging for $k$:

So the units of $k$ change with order — this is exactly why chemists use the units of an experimentally measured $k$ to identify the reaction order.

Step-by-Step Solution

1. Rate of reaction / rate of disappearance of reactant: both are $\dfrac{\Delta[\text{conc}]}{\Delta t}$, units $M\,s^{-1}$ — (A), (B) correct.
2. Units of $k$ depend on order: for order $n$, units are $M^{1-n}s^{-1}$ — (C) correct.
3. For $n=1$ (first order): units of $k = M^{1-1}s^{-1} = M^0 s^{-1} = s^{-1}$, not $M\,s^{-1}$.
4. So statement (D), which claims $M\,s^{-1}$ for a first-order $k$, is false — that unit actually belongs to a zero-order reaction ($n=0 \Rightarrow M^{1}s^{-1}$).

Common Mistakes

• Confusing the zero-order rate-constant unit ($M\,s^{-1}$) with the first-order one ($s^{-1}$).
• Thinking rate constant units are always the same as rate units.

ANSWER: D