Exercises — Methods to determine order — initial rates, integrated method, half-life method

KYA: $r_2/r_1=([A{]}_2/[A{]}_1{)}^n$ se $n$ nikaalein. Log kyun: $n$ ek exponent ki tarah phansa hua hai; $\ln$ use neeche le aata hai. Rates ka ratio $=8$, concentrations ka ratio $=2$, toh $2^n=8$. Kyunki $8=2^3$ hai, hum padhte hain $n=3$. Ya formally: $n=\frac{\ln 8}{\ln 2}=3$. Answer: third order.

Sirf first-order integrated law, $\ln [A]=\ln [A{]}_0-kt$, linear hota hai jab tum $\ln [A]$ ko $t$ ke against plot karo (iska shape "constant $-$ constant $\times t$" jaisa hota hai). Answer: first order.

Fingerprint $t_{1/2}\propto [A{]}_0^{1-n}$ yaad karo. "Unchanged" matlab exponent $1-n=0$ hai, toh $n=1$. Answer: first order. (Yahi radioactivity wala case hai — dekho Radioactive Decay.)

Step 1 — rate ratio: $\frac{1.08\times 10^{-2}}{1.2\times 10^{-3}}=9$. Step 2 — concentration ratio: $\frac{0.150}{0.050}=3$. Step 3 — exponent solve karo: $3^n=9\Rightarrow n=\frac{\ln 9}{\ln 3}=2$. Second order. Step 4 — $k$ nikaalein $r=k[A{]}^2$ se, Expt 1 use karke: $k=\frac{r}{[A{]}^2}=\frac{1.2\times 10^{-3}}{(0.050{)}^2}=0.48{\text{M}}^{-1}{\text{s}}^{-1}.$

First-order law kyun: diya gaya hai ki first order hai, toh $\ln [A]=\ln [A{]}_0-kt$ (dekho Integrated Rate Equations). $k$ nikaalein: $k=\frac{1}{t}\ln \frac{[A{]}_0}{[A]}=\frac{1}{60}\ln \frac{0.80}{0.20}=\frac{\ln 4}{60}=0.02310{\text{s}}^{-1}.$ $t=120$ s par: dhyaan do ki $120$ s do 60-s intervals hain, aur har 60 s mein $[A]$ quarter ho gaya (0.80→0.20). Toh agle 60 s baad: $0.20\to 0.050$ M. Law se check karo: $[A]=0.80e^{-0.02310\times 120}=0.80\times e^{-2.772}=0.050$ M. ✓ Answers: $k=0.0231{\text{s}}^{-1}$, $[A{]}_{120}=0.050$ M.

$a$ nikaalein — 1→2 compare karo (yahan $[B]$ fixed hai, toh sirf $A$ ki power move kar sakti hai): rate ratio $=4.0/2.0=2$; concentration ratio $=0.20/0.10=2$; toh $2^a=2\Rightarrow a=1$. $b$ nikaalein — 2→3 compare karo ($[A]$ 0.20 par fixed hai, sirf $B$ badla): rate ratio $=3.6\times 10^{-3}/4.0\times 10^{-4}=9$; concentration ratio $=0.30/0.10=3$; toh $3^b=9\Rightarrow b=2$. Overall order $=a+b=3$. $k$ nikaalein Expt 1 use karke: $r=k[A][B{]}^2$ $k=\frac{2.0\times 10^{-4}}{(0.10)(0.10{)}^2}=\frac{2.0\times 10^{-4}}{1.0\times 10^{-3}}=0.20{\text{M}}^{-2}{\text{s}}^{-1}.$

Teesra column $1/[A]$ dekho: values $1,2,3,4$ har $10$ s mein constant $+1$ se badhte hain — ek perfectly straight line. $\ln [A]$ column $-0.693,-0.406,-0.287$ se jump karta hai (shrinking), toh woh linear nahi hai. Straight $1/[A]$ vs $t$ $\Rightarrow$ second order. Slope $=\frac{4.00-1.00}{30-0}=0.10{\text{M}}^{-1}{\text{s}}^{-1}$, aur $n=2$ ke liye slope $+k$ hota hai, toh $k=0.10{\text{M}}^{-1}{\text{s}}^{-1}$. Neeche figure dekho.

Step 1 — fingerprint se order $t_{1/2}\propto [A{]}_0^{1-n}$: $n=1-\frac{\ln (t_{1/2,2}/t_{1/2,1})}{\ln ([A{]}_{0,2}/[A{]}_{0,1})}=1-\frac{\ln (10/40)}{\ln (0.80/0.20)}=1-\frac{\ln (1/4)}{\ln 4}=1-(-1)=2.$ Second order. (Sanity check: $[A{]}_0$ ×4 hua, $t_{1/2}$ ÷4 hua — inverse, bilkul $n=2$ ka signature.) Step 2 — $k$ nikaalein second-order half-life $t_{1/2}=\frac{1}{k[A{]}_0}$ se, toh $k=\frac{1}{t_{1/2}[A{]}_0}$. Pehli row use karke: $k=\frac{1}{40\times 0.20}=0.125{\text{M}}^{-1}{\text{s}}^{-1}$. Doosri row se check: $\frac{1}{10\times 0.80}=0.125$. ✓ Consistent.

Initial rates se order: rate ratio $=8.0\times 10^{-2}/5.0\times 10^{-3}=16$; concentration ratio $=0.40/0.10=4$; toh $4^n=16\Rightarrow n=2$. Second order. Rate law se $k$: $k=\frac{r}{[A{]}^2}=\frac{5.0\times 10^{-3}}{(0.10{)}^2}=0.50{\text{M}}^{-1}{\text{s}}^{-1}$. Half-life se $k$ (second order, $t_{1/2}=1/k[A{]}_0$): $k=\frac{1}{t_{1/2}[A{]}_0}=\frac{1}{200\times 0.10}=0.050{\text{M}}^{-1}{\text{s}}^{-1}$. Ruko — mismatch! $0.50\ne 0.050$. Do methods disagree kar rahe hain, jo parent ke rule ke mutabiq ek data/unit error hai, na ki do alag "true" orders. Re-check karte hain: agar instead $t_{1/2}=20$ s hoti, toh $k=1/(20\times 0.10)=0.50$ — match karta. Toh consistent dataset hai $t_{1/2}=20$ s, aur dono methods $k=0.50{\text{M}}^{-1}{\text{s}}^{-1}$ dete hain, second order. Lesson: saare methods ko ek hi $k$ dena chahiye; clash ek error flag karta hai.

Step 1 — pattern pakdo: $[A]$ har $50$ s mein ek constant $0.25$ M girta hai. Har unit time mein constant amount girna $[A]=[A{]}_0-kt$ ka signature hai → zero order ($[A]$ vs $t$ linear). Yahan na $\ln [A]$ na $1/[A]$ straight hoga. Step 2 — $k$: $[A]$ vs $t$ ka slope $\frac{0.25-1.00}{150-0}=-5.0\times 10^{-3}$ M/s hai. $n=0$ ke liye slope $=-k$, toh $k=5.0\times 10^{-3}$ M/s. Step 3 — $[A{]}_0=1.00$ par $t_{1/2}$: zero-order $t_{1/2}=\frac{[A{]}_0}{2k}=\frac{1.00}{2(5.0\times 10^{-3})}=100$ s. (Match karta hai: $[A]$ $t=100$ s par $0.50$ hua. ✓) Step 4 — $0.40$ M se restart: $t_{1/2}=\frac{0.40}{2(5.0\times 10^{-3})}=40$ s. Chhota start → chhoti half-life, kyunki zero-order mein $t_{1/2}\propto [A{]}_0$. Neeche plot dekho.

300 K par $t_{1/2}$ (first order): $t_{1/2}=\frac{\ln 2}{k}=\frac{0.693}{0.010}=69.3$ s — aur dhyaan do ki ye $[A{]}_0$ par depend nahi karta. $k$ double hone ke baad: $t_{1/2}^{\prime }=\frac{0.693}{0.020}=34.65$ s. Factor: $t_{1/2}^{\prime }/t_{1/2}=\frac{1}{2}$ — half-life half ho jaati hai, kyunki first order ke liye $t_{1/2}\propto 1/k$. Temperature ke saath $k$ kyun badha: yahi Arrhenius Equation ka kaam hai ($k=A{e}^{-E_a/RT}$); zyada $k$ hamesha yahan chhoti half-life matlab hai.

Initial rates → $n=\frac{\ln (r_2/r_1)}{\ln ([A{]}_2/[A{]}_1)}$; integrated → $[A],\ln [A],1/[A]$ mein se kaun sa straight hai; half-life → $t_{1/2}\propto [A{]}_0^{1-n}$ mein $1-n$ ka sign. Teeno methods ko ek hi $n$ aur ek hi $k$ par agree karna chahiye.