Q.In a reaction, $2A \rightarrow$ Products, the concentration of A decreases from $0.5\ \text{mol L}^{-1}$ to $0.4\ \text{mol L}^{-1}$ in 10 minutes. Calculate the rate during this interval?

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 $N_2O_5 = -\dfrac{\Delta[N_2O_5]}{\Delta t} = -\dfrac{-0.5}{100} = 5\times10^{-3}\ mol\,L^{-1}min^{-1}$. …

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: $\dfrac{|x-y|}{100}$. …

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 $=\dfrac{-\Delta[\mathrm{R}]}{\Delta t}=\dfrac{0.01}{2400}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. …

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 $= \dfrac{0.01}{1500} = 6.667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. …

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). …

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 $M\,s^{-1}$. …