Easy Practice — Set 1

Quick Recap — Rate, Order & Molecularity

• Rate $=\dfrac{-d[\text{reactant}]}{dt}=\dfrac{+d[\text{product}]}{dt}$ (adjusted for stoichiometry); units mol L$^{-1}$ s$^{-1}$.
• Rate law: rate $=k[A]^m[B]^n$; order $=m+n$ (found experimentally, can be zero/fraction/integer).
• Molecularity = number of species in an elementary step (a positive integer $\le3$).
• $k$ depends on temperature and catalyst, not concentration; a catalyst speeds a reaction by lowering $E_a$ (does not change $\Delta H$).

Beyond-NCERT JEE Formulae

Everything below assumes constant temperature unless a temperature is quoted. Match the shape of the data to the correct law before you compute.

1. Integrated Rate Laws — pick by the data you are given

• Zero order ($A\to$ products): $[A]=[A]_0-kt$, half-life $t_{1/2}=\dfrac{[A]_0}{2k}$, and the reaction is complete at $t_c=\dfrac{[A]_0}{k}$. Reach for this when the rate is stated to be independent of concentration, or when a plot of $[A]$ versus $t$ is a straight line (surface-catalysed or photochemical reactions).
• First order ($A\to$ products): $k=\dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ and $t_{1/2}=\dfrac{0.693}{k}$. Reach for this when the half-life is quoted as constant, when data come as percentages, or when $\log[A]$ versus $t$ is linear.
• Second order (one reactant): $\dfrac{1}{[A]}=\dfrac{1}{[A]_0}+kt$ with $t_{1/2}=\dfrac{1}{k[A]_0}$. Signalled by a linear plot of $\dfrac{1}{[A]}$ versus $t$.

[JEE Tip] For first order, translate percentages into half-lives on sight: 50% done $=1\,t_{1/2}$, 75% $=2\,t_{1/2}$, 87.5% $=3\,t_{1/2}$, 90% $\approx 3.32\,t_{1/2}$, and 99.9% $=10\,t_{1/2}$. Because concentration units cancel inside $\log\dfrac{[A]_0}{[A]}$, you may feed in pressures, volumes of titrant, or angles of rotation directly, with no conversion to mol/L.

2. Units of the Rate Constant — one master formula

For an $n$-th order reaction the units of $k$ are $\text{(mol/L)}^{1-n}\text{s}^{-1}$. This one expression reproduces the whole table: order 0 gives $\text{mol L}^{-1}\text{s}^{-1}$, order 1 gives $\text{s}^{-1}$, order 2 gives $\text{L mol}^{-1}\text{s}^{-1}$, and order 3 gives $\text{L}^{2}\text{mol}^{-2}\text{s}^{-1}$.

[JEE Tip] Run it backwards. If a question hands you $k$ in $\text{s}^{-1}$ the order is 1; in $\text{L mol}^{-1}\text{s}^{-1}$ the order is 2. Reading the concentration exponent off the units is often the fastest route to the order.

3. Arrhenius Equation and Temperature Dependence

The rate constant grows with temperature through $k=Ae^{-E_a/RT}$, or in log form $\log k=\log A-\dfrac{E_a}{2.303RT}$, so a plot of $\log k$ versus $\dfrac{1}{T}$ is a straight line of slope $-\dfrac{E_a}{2.303R}$. For two temperatures,

$$\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$$

The temperature coefficient is the factor by which $k$ rises over a 10 K interval; across a change $\Delta T$ the rate is multiplied by $n^{\Delta T/10}$ (near room temperature $n\approx 2$ to $3$).

[JEE Tip] Commit $2.303R=19.147$ (J/mol per K) to memory so the two-temperature formula collapses in one step to $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{19.147}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$. Keep the bracket in the order $\dfrac{1}{T_1}-\dfrac{1}{T_2}$: it is positive when $T_2>T_1$, which guarantees $k_2>k_1$.

4. Half-life vs Order, Fraction Reacted, Graphical Order

• Master relation: $t_{1/2}\propto[A]_0^{\,1-n}$. So the half-life is independent of $[A]_0$ only for $n=1$, rises with $[A]_0$ when $n<1$ (e.g. zero order), and falls as $[A]_0$ increases when $n>1$ (e.g. second order).
• Order from two half-lives: $n=1+\dfrac{\log(t_{1/2,1}/t_{1/2,2})}{\log([A]_{0,2}/[A]_{0,1})}$.
• Fraction reacted (first order): the fraction remaining after time $t$ is $e^{-kt}$, and after $m$ half-lives it is $\left(\dfrac{1}{2}\right)^{m}$.
• Graphical order test: a straight line of $[A]$ vs $t$ means zero order, of $\log[A]$ vs $t$ means first order, and of $\dfrac{1}{[A]}$ vs $t$ means second order.

[JEE Tip] When two $(t_{1/2},[A]_0)$ pairs are supplied, do not chase $k$. Take the ratio, match it to $[A]_0^{\,1-n}$, and read the order straight off the exponent.

5. Molecularity vs Order and the Rate-Determining Step

Molecularity is a whole number from 1 to 3 belonging to a single elementary step (theoretical, taken from the mechanism); order is experimental and may be zero, fractional, or even negative. For an elementary step order equals molecularity, but for an overall reaction the two need not agree. The observed rate law is fixed by the slowest, rate-determining step, with any intermediate eliminated through a fast pre-equilibrium.

[JEE Tip] If the rate-law exponents differ from the balanced coefficients, the equation is not elementary and the reaction is multi-step. Classic case: $2\text{N}_2\text{O}_5\to 4\text{NO}_2+\text{O}_2$ is first order, not second.

Solved Examples — Beyond-NCERT Formulae

Take $2.303R=19.147$ (J/mol per K) and $R=8.314$ J/mol/K throughout.

Example 1 — First-order rate constant, then time for a target completion. A first-order reaction is 20% complete in 10 minutes. Find (a) the rate constant $k$ and (b) the time for 75% completion.

• (a) With 20% reacted, $[A]$ is 80% of $[A]_0$, so $k=\dfrac{2.303}{10}\log\dfrac{100}{80}=\dfrac{2.303}{10}(0.0969)=0.0223$ min$^{-1}$.
• (b) For 75% completion $[A]$ is 25% of $[A]_0$: $t=\dfrac{2.303}{k}\log\dfrac{100}{25}=\dfrac{2.303}{0.0223}(0.6021)=62.1$ minutes.
• Check: 75% is exactly two half-lives; $t_{1/2}=\dfrac{0.693}{0.0223}=31.1$ min, and $2\times 31.1=62.1$ min. Consistent.

Example 2 — Half-life across zero, first and second order. A reactant starts at $[A]_0=0.50$ mol/L and each version carries $k=0.05$ in its own units. Compare the half-lives.

• Zero order: $t_{1/2}=\dfrac{[A]_0}{2k}=\dfrac{0.50}{2(0.05)}=5.0$ min.
• First order: $t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{0.05}=13.9$ min.
• Second order: $t_{1/2}=\dfrac{1}{k[A]_0}=\dfrac{1}{(0.05)(0.50)}=40$ min.
• Takeaway: only the first-order half-life is independent of $[A]_0$; the zero-order value scales as $[A]_0$ and the second-order value as $\dfrac{1}{[A]_0}$, all consistent with $t_{1/2}\propto[A]_0^{\,1-n}$.

Example 3 — Activation energy from two rate constants. The rate constant of a reaction doubles when the temperature rises from 300 K to 310 K. Calculate $E_a$.

• Use $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ with $\dfrac{k_2}{k_1}=2$.
• $\dfrac{1}{300}-\dfrac{1}{310}=\dfrac{10}{93000}=1.075\times10^{-4}$ per K.
• $\log 2=0.3010$, so $0.3010=\dfrac{E_a}{19.147}(1.075\times10^{-4})$.
• $E_a=\dfrac{0.3010\times19.147}{1.075\times10^{-4}}=5.36\times10^{4}$ J/mol $=53.6$ kJ/mol.

Example 4 — Zero-order completion time. A zero-order reaction has $k=2.0\times10^{-2}$ mol L$^{-1}$ s$^{-1}$ and $[A]_0=0.50$ mol/L. Find (a) the time for the reactant to be used up and (b) the half-life.

• (a) All of it is gone at $t_c=\dfrac{[A]_0}{k}=\dfrac{0.50}{2.0\times10^{-2}}=25$ s.
• (b) $t_{1/2}=\dfrac{[A]_0}{2k}=\dfrac{0.50}{2(2.0\times10^{-2})}=12.5$ s, exactly half of $t_c$, which is a signature of zero order.

Example 5 — Order from half-life data. When $[A]_0=0.20$ mol/L the half-life is 40 min; when $[A]_0=0.40$ mol/L it falls to 20 min. Find the order.

• Apply $t_{1/2}\propto[A]_0^{\,1-n}$, so $\dfrac{t_{1/2,1}}{t_{1/2,2}}=\left(\dfrac{[A]_{0,1}}{[A]_{0,2}}\right)^{1-n}$.
• $\dfrac{40}{20}=\left(\dfrac{0.20}{0.40}\right)^{1-n}$, that is $2=(0.5)^{1-n}=2^{\,n-1}$.
• Matching exponents gives $n-1=1$, so $n=2$: the reaction is second order. (The half-life halving when $[A]_0$ doubles is the second-order fingerprint.)

Example 6 — Rate constant from the Arrhenius pre-exponential factor. For a first-order reaction $A=6.0\times10^{13}$ s$^{-1}$ and $E_a=100$ kJ/mol. Find $k$ at 300 K.

• Use the log form $\log k=\log A-\dfrac{E_a}{2.303RT}$.
• $\log A=\log(6.0\times10^{13})=13.78$.
• $\dfrac{E_a}{2.303RT}=\dfrac{100000}{19.147\times300}=17.41$.
• $\log k=13.78-17.41=-3.63$, hence $k=10^{-3.63}=2.3\times10^{-4}$ s$^{-1}$.

Example 7 — Predicting a rate constant at a new temperature. A reaction has $k_1=2.0\times10^{-3}$ s$^{-1}$ at 300 K and $E_a=50$ kJ/mol. Find $k_2$ at 320 K.

• $\dfrac{1}{T_1}-\dfrac{1}{T_2}=\dfrac{1}{300}-\dfrac{1}{320}=2.083\times10^{-4}$ per K.
• $\log\dfrac{k_2}{k_1}=\dfrac{50000}{19.147}(2.083\times10^{-4})=0.544$.
• $\dfrac{k_2}{k_1}=10^{0.544}=3.50$, so $k_2=3.50\times(2.0\times10^{-3})=7.0\times10^{-3}$ s$^{-1}$.

Example 8 — Temperature coefficient to activation energy. A reaction has temperature coefficient 3, meaning its rate constant triples per 10 K. (a) By what factor does the rate rise from 300 K to 340 K? (b) Estimate $E_a$ over this range.

• (a) The factor is $n^{\Delta T/10}=3^{40/10}=3^{4}=81$.
• (b) Treat $\dfrac{k_{340}}{k_{300}}=81$, so $\log 81=\dfrac{E_a}{19.147}\left(\dfrac{1}{300}-\dfrac{1}{340}\right)$.
• $\dfrac{1}{300}-\dfrac{1}{340}=3.922\times10^{-4}$ per K and $\log 81=1.908$, giving $E_a=\dfrac{1.908\times19.147}{3.922\times10^{-4}}=9.32\times10^{4}$ J/mol $=93.2$ kJ/mol.