Exercises — Pseudo-first-order kinetics

Answer: (b) aur (c). Pseudo-first-order behaviour ke liye humein chahiye ki flooded reactant watched wale se kam se kam $\approx 50$–$100\times$ zyaada ho.

• (a) ratio $=0.5/0.5=1$. Dono equally change karte hain → truly second-order, not pseudo.
• (b) ratio $=0.90/0.005=180$. Bahut bada excess → $[B]\approx [B{]}_0$ hold karta hai → pseudo-first-order. ✔
• (c) water $\approx 55.5\text{M}$ par hai jabki ester dilute hai → excess automatic hai → pseudo-first-order. ✔

Ek first-order constant ek concentration ko multiply karta hai, isliye uske units ${\text{s}}^{-1}$ hone chahiye. Ek second-order constant do concentrations ($\text{M}\times \text{M}$) ko multiply karta hai, isliye rate ${\text{M s}}^{-1}$ mein rakhne ke liye use ${\text{M}}^{-1}{\text{s}}^{-1}$ carry karna padta hai.

• $3.0\times 10^{-4}{\text{s}}^{-1}$ → yeh $k^{\prime }$ hai (pseudo, first-order units).
• $3.0\times 10^{-4}{\text{M}}^{-1}{\text{s}}^{-1}$ → yeh $k$ hai (true, second-order units).

Water flooded reactant hai, isliye $[B{]}_0=55.5\text{M}$. $k^{\prime }=k[B{]}_0$ ko rearrange karke: $k=\frac{k^{\prime }}{[B{]}_0}=\frac{2.5\times 10^{-4}}{55.5}=4.5\times 10^{-6}{\text{M}}^{-1}{\text{s}}^{-1}.$ Units par sanity check: $\frac{{\text{s}}^{-1}}{\text{M}}={\text{M}}^{-1}{\text{s}}^{-1}$ ✔ — true bimolecular units.

Time convert karo: $t=2\times 3600=7200\text{s}$. Exponential form use karo (yeh directly concentration deta hai, koi logs ki zaroorat nahi): $[A]=[A{]}_0{e}^{-k^{\prime }t}=0.50e^{-(6.0\times 10^{-5})(7200)}=0.50e^{-0.432}.$ Ab $e^{-0.432}=0.649$, isliye $[A]=0.50\times 0.649=0.325\text{M}.$ Bacha fraction $=0.325/0.50=0.65$, yaani $\approx 35\%$ react ho chuka. 2 h ke liye is rate par reasonable hai.

$t_{1/2}=\frac{0.693}{k^{\prime }}=\frac{0.693}{6.0\times 10^{-5}}=1.155\times 10^4\text{s}\approx 192.5\text{min}.$ Kyunki $t_{1/2}$ mein sirf $k^{\prime }$ hai, yeh $[A{]}_0$ se independent hai — first-order (aur isliye pseudo-first-order) kinetics ki pehchaan. Dekho Half-life.

Neeche figure mein dono plots dekho. Kya observe karna hai: deep-teal curve (Run 2, huge excess) bilkul seedhi descending line hai, jabki burnt-orange curve (Run 1, equal concentrations) neeche ki taraf bend karti aur steep hoti hai — woh bend genuine second-order behaviour ki visual tell hai jo $\ln [A]$ axis par forced hai.

• Run 1 curve karta hai kyunki dono concentrations drop karte hain, isliye reaction genuine second-order integrated law follow karta hai, jo $\ln [A]$ axis par linear nahi hai.
• Run 2 seedha jaata hai kyunki $[B{]}_0/[A{]}_0=100$ $[B]$ ko freeze kar deta hai; reaction $A$ mein first-order par isolated ho jaata hai (dekho Method of Isolation). Straight slope $-k^{\prime }$ hai, isliye $k^{\prime }=0.050{\text{s}}^{-1}$. $k=\frac{k^{\prime }}{[B{]}_0}=\frac{0.050}{1.0}=0.050{\text{M}}^{-1}{\text{s}}^{-1}.$ Overall order $=1$ ($A$ mein) $+1$ ($B$ mein) $=$ truly second order; Run 2 mein observed order first hai.

Maan lo $[A{]}_0=1$ unit, isliye $[B{]}_0=10$ units. $A$ ka aadha react karna $B$ ke $0.5$ units consume karta hai (ek $B$ per ek $A$). $\text{drop in }[B]=\frac{0.5}{10}=0.05=5\%.$ "Constant" mein $5\%$ ka wandering ek careful $\ln [A]$ plot par already visible hota hai — yeh thodi si curve produce karta hai. Isliye $10\times$ marginal hai: rough estimate ke liye theek hai, precise $k$ ke liye kharaab. $\ge 50\times$ ka aim karo.

$\ln [A]=\ln [A{]}_0-k^{\prime }t$ se compare karo. Term by term match karke:

• Intercept $\ln [A{]}_0=-4.605\Rightarrow [A{]}_0=e^{-4.605}=0.0100\text{M}.$
• Slope $=-k^{\prime }=-0.0300\Rightarrow k^{\prime }=0.0300{\text{s}}^{-1}.$

$0.020$ se $0.010\text{M}$ tak jaana exactly ek half-life hai, isliye $t_{1/2}=231\text{s}$. $k^{\prime }=\frac{0.693}{t_{1/2}}=\frac{0.693}{231}=3.0\times 10^{-3}{\text{s}}^{-1}.$ True constant: $k=\frac{k^{\prime }}{[{\text{H}}_2\text{O}]}=\frac{3.0\times 10^{-3}}{55.5}=5.4\times 10^{-5}{\text{M}}^{-1}{\text{s}}^{-1}.$ $0.020\text{M}$ se $0.0025\text{M}$ tak pahunchna $0.0025/0.020=1/8=(1/2{)}^3$ hai → teen half-lives: $t=3\times 231=693\text{s}.$ Log law se cross-check: $t=\frac{1}{k^{\prime }}\ln \frac{0.020}{0.0025}=\frac{\ln 8}{3.0\times 10^{-3}}=\frac{2.079}{3.0\times 10^{-3}}=693\text{s}$ ✔.

Saara $A$ consume karne par $B$ se $[A{]}_0$ worth nikalta hai. $B$ mein fractional drift: $\frac{[A{]}_0}{[B{]}_0}\le 0.01⟹\frac{[B{]}_0}{[A{]}_0}\ge 100.$ Isliye 100× excess error ko $1\%$ level par rakhta hai — precisely textbook threshold. Zyaada loose $2\%$ tolerance ke liye, $50\times$ kaafi hai.

Is low-substrate regime mein poora prefactor $[S]$ mein ek first-order constant ki tarah act karta hai: $k^{\prime }=\frac{V_{\max }}{K_M}=\frac{2.0\times 10^{-5}}{4.0\times 10^{-3}}=5.0\times 10^{-3}{\text{s}}^{-1}.$ $t_{1/2}=\frac{0.693}{k^{\prime }}=\frac{0.693}{5.0\times 10^{-3}}=138.6\text{s}.$ Yahan first-order kyun? $[S]\ll K_M$ ke saath enzyme mostly empty hai, isliye $[S]$ ko double karne par rate double hoti hai — first-order dependence ki pehchaan, exactly pseudo-first-order idea ek naye setting mein.

(a) $\ln [A]$ compute karo: $-4.605,-4.855,-5.104,-5.356$. Successive gaps: $-0.250,-0.249,-0.252$ — equal time step par constant, isliye $\ln [A]$ vs $t$ ek straight line hai → observed first-order. ✔

(b) Slope $=-k^{\prime }=\frac{-5.356-(-4.605)}{300-0}=\frac{-0.751}{300}=-2.50\times 10^{-3}{\text{s}}^{-1}$, isliye $k^{\prime }=2.50\times 10^{-3}{\text{s}}^{-1}.$ (c) $k=\frac{k^{\prime }}{[B{]}_0}=\frac{2.50\times 10^{-3}}{0.80}=3.13\times 10^{-3}{\text{M}}^{-1}{\text{s}}^{-1}.$ (d) $[A]=[A{]}_0{e}^{-k^{\prime }t}=0.0100e^{-(2.50\times 10^{-3})(500)}=0.0100e^{-1.25}=0.0100\times 0.2865=2.87\times 10^{-3}\text{M}.$

$[B{]}_0$ temperature ke saath essentially unchanged rehta hai, isliye $k^{\prime }\propto k$, aur unknown factor $A$ ratio mein cancel ho jaata hai (isliye hum ratio lete hain — yeh humein $A$ jaane bina predict karne deta hai): $\frac{k_2^{\prime }}{k_1^{\prime }}=\frac{k_2}{k_1}=\frac{A{e}^{-E_a/R{T}_2}}{A{e}^{-E_a/R{T}_1}}=\exp \left[-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right].$ $T_1=298\text{K},T_2=308\text{K}$ ke saath: $\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{298}-\frac{1}{308}=1.0896\times 10^{-4}{\text{K}}^{-1}.$ $\frac{k_2^{\prime }}{k_1^{\prime }}=\exp \left[\frac{50000}{8.314}\times 1.0896\times 10^{-4}\right]=\exp (0.6553)=\mathrm{1.93.}$ Isliye $k^{\prime }$ $10^{\circ }$ badhne par roughly double ho jaata hai — classic rule-of-thumb, ab derive kiya gaya. Naya $k_2^{\prime }\approx 1.0\times 10^{-3}\times 1.93=1.93\times 10^{-3}{\text{s}}^{-1}.$

Dono "constant" species ko apparent constant mein fold karo: $k^{\prime }=k[{\text{H}}_2\text{O}][{\text{H}}^{+}]$. Water genuinely $55.5\text{M}$ par fixed hai, isliye woh culprit nahi hai. Catalyst ${\text{H}}^{+}$ har cycle mein regenerate hota hai (catalyst consume nahi hota), lekin reaction acetic acid produce karta hai; agar weak buffer itna dilute ho ki us extra acid ko absorb na kar sake, toh solution ka $[{\text{H}}^{+}]$ dheere dheere badhta hai. Kyunki $k^{\prime }\propto [{\text{H}}^{+}]$, badhta $[{\text{H}}^{+}]$ $k^{\prime }$ ko time ke saath grow karata hai, aur badhta $k^{\prime }$ $\ln [\text{ester}]$ slope ko steep banata hai — exactly wahi observed drift.

• Diagnosis: pseudo-first-order approximation instant-by-instant valid hai, lekin $k^{\prime }$ ke andar chupi ek quantity (catalyst, solvent nahi) truly clamped nahi hai.
• Fix: ek stronger / more concentrated buffer use karo taaki produced acid $[{\text{H}}^{+}]$ ko barely shift kare; phir $k^{\prime }$ constant rahega aur poora run ek single straight line par collapse ho jaayega.