Chemical Kinetics Set-3

Q1. A first-order reaction has half-life $t_{1/2}=30\ \mathrm{min}$. How long will it take for the concentration of reactant to fall to $12.5\%$ of its initial value?

Q2. The bimolecular reaction $A + B \rightarrow \text{products}$ follows rate law $r = k[A][B]$ with $k = 0.05\ \mathrm{M^{-1}s^{-1}}$. If $[B]_0 = 0.20\ \mathrm{M}$ (kept in large excess) and $[A]_0 = 0.010\ \mathrm{M}$, use the pseudo-first-order approximation to estimate the time required for $[A]$ to decrease to half its initial value.

Q3. For a reaction measured rate constants are $k_1 = 2.0\times 10^{-4}\ \mathrm{s^{-1}}$ at $T_1 = 300\ \mathrm{K}$ and $k_2 = 8.0\times 10^{-4}\ \mathrm{s^{-1}}$ at $T_2 = 310\ \mathrm{K}$. Using the Arrhenius relation $k = A e^{-E_a/RT}$, estimate the activation energy $E_a$ (in $\mathrm{kJ\ mol^{-1}}$).

Q4. Consider the mechanism:
Step 1 (fast equilibrium): $A + A \rightleftharpoons D$ (equilibrium constant $K$)
Step 2 (slow): $D + B \xrightarrow{k_2} \text{products}$
Assertion (A): The rate law derived from this mechanism is $r = k_2 K [A]^2[B]$, so the overall reaction is third order (second order in $A$, first order in $B$).
Reason (R): Since Step 1 is a fast equilibrium, $[D] = K[A]^2$ and the slow step gives $r = k_2[D][B] = k_2K[A]^2[B]$.

Q5. The reaction $A + B \rightarrow \text{products}$ follows $r = k[A][B]$ with $k = 0.50\ \mathrm{M^{-1}s^{-1}}$. Initially $[A]_0 = 0.100\ \mathrm{M}$ and $[B]_0 = 0.400\ \mathrm{M}$. Calculate the time required for $[A]$ to decrease to $0.025\ \mathrm{M}$.

Q6. For a first-order reaction $A \rightarrow$ products, 40% of $A$ decomposes in 30 min. Using $N_t=N_0e^{-kt}$ and $t_{1/2}=\dfrac{\ln 2}{k}$, the half-life $t_{1/2}$ (in minutes) is:

Q7. The initial rate data for the reaction $A + B \rightarrow$ products are:
Experiment 1: $[A]_0=0.10\ \mathrm{M},\ [B]_0=0.10\ \mathrm{M},\ \text{rate}=2.0\times10^{-3}\ \mathrm{M\,s^{-1}}$;
Experiment 2: $[A]_0=0.20\ \mathrm{M},\ [B]_0=0.10\ \mathrm{M},\ \text{rate}=8.0\times10^{-3}\ \mathrm{M\,s^{-1}}$;
Experiment 3: $[A]_0=0.20\ \mathrm{M},\ [B]_0=0.20\ \mathrm{M},\ \text{rate}=1.6\times10^{-2}\ \mathrm{M\,s^{-1}}$.
Determine the empirical rate law and the value of the rate constant $k$ (including units).

Q8. Consider the second-order reaction $2A\rightarrow$ products with $r=k[A]^2$. At $T=300\ \mathrm{K}$ the uncatalyzed path has activation energy $E_a^{\rm unc}=60\ \mathrm{kJ\,mol^{-1}}$ and pre-exponential $A_{\rm unc}$. A catalyst provides a new path with $E_a^{\rm cat}=40\ \mathrm{kJ\,mol^{-1}}$ and $A_{\rm cat}=A_{\rm unc}/2$. Using $k=Ae^{-E_a/(RT)}$ and the second-order half-life $t_{1/2}=\dfrac{1}{k[A]_0}$, estimate how $t_{1/2}$ changes on adding the catalyst at 300 K (take $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$).

Q9. Assertion (A): For the unimolecular decomposition treated by the Lindemann–Hinshelwood mechanism

if $k_{-1}[M]\gg k_2$ the observed rate is first order in $[A]$ and effectively independent of $[M]$.
Reason (R): Using steady-state/pre-equilibrium one obtains

and when $k_{-1}[M]\gg k_2$ the denominator $\approx k_{-1}[M]$, giving $r\approx\dfrac{k_1k_2}{k_{-1}}[A]$, which is independent of $[M]$.

Q10. Consider the mechanism:
Step 1 (fast equilibrium): $A + B \rightleftharpoons I$ with equilibrium constant $K=\dfrac{[I]}{[A][B]}=10\ \mathrm{M^{-1}}$.
Step 2 (slow): $I + B \xrightarrow{k_2} P$ with $k_2=0.5\ \mathrm{M^{-2}\,s^{-1}}$.
(a) Derive the initial rate expression in terms of $K$, $k_2$, $[A]$ and $[B]$.
(b) Calculate the initial rate when $[A]=0.020\ \mathrm{M}$ and $[B]=0.030\ \mathrm{M}$.