Exercises — Activation energy from Arrhenius plot; effect of catalyst

KYA: $\ln k$ vs $1/T$ ka slope $-E_a/R$ ke barabar hota hai. KYUN: kyunki $\ln k=-\frac{E_a}{R}\cdot \frac{1}{T}+\ln A$ mein, $\frac{1}{T}$ ko multiply karne wala number hi slope hai. $E_a=-m\times R=-(-7500)\times 8.314=62,355\text{J/mol}\approx 62.4kJ/mol$ Slope par minus sign aur $-m$ mein minus cancel ho jaate hain — $E_a$ positive aata hai, jaise hona chahiye (negative energy hill ho hi nahi sakti).

Slope $=-E_a/R$. Zyada badi $E_a$ slope ko aur negative banati hai, matlab dahine taraf (bade $1/T$ ki taraf) jaate waqt zyada tezi se neeche jaati hai. Toh steeper line wali $E_a=90\text{kJ/mol}$ wali reaction hai (figure mein red line); shallower line $40\text{kJ/mol}$ wali hai.

Galat. Catalyst activation energy (barrier ki unchaai) kam karta hai, lekin reactant aur product ke energy levels untouched rehte hain, isliye $\Delta H$ (product minus reactant energy) unchanged rehta hai. Dekho Reaction Energy Diagrams. Catalyst sirf kitni tez ko badalta hai, kitna neeche ko kabhi nahi.

KYA: do temperatures ke saath hum two-point Arrhenius form use karte hain. KYUN yeh form: do temperatures par $\ln k$ ghatao; $\ln A$ cancel ho jaata hai, sirf $E_a$ bachta hai. $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ $\ln \frac{8.0\times 10^{-3}}{1.0\times 10^{-3}}=\ln 8=2.0794$ $\frac{1}{340}-\frac{1}{300}=0.0029412-0.0033333=-3.9216\times 10^{-4}{\text{K}}^{-1}$ $2.0794=-\frac{E_a}{8.314}\times (-3.9216\times 10^{-4})$ $E_a=\frac{2.0794\times 8.314}{3.9216\times 10^{-4}}=44,085\text{J/mol}\approx 44.1kJ/mol$

Wahi two-point law, ab $k_2$ ke liye solve karo: $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{360}-\frac{1}{300}\right)$ $\frac{1}{360}-\frac{1}{300}=0.0027778-0.0033333=-5.5556\times 10^{-4}$ $\ln \frac{k_2}{k_1}=-\frac{44,085}{8.314}\times (-5.5556\times 10^{-4})=2.9457$ $\frac{k_2}{k_1}=e^{2.9457}=19.02$ $k_2=1.0\times 10^{-3}\times 19.02=1.90\times 10^{-2}{\mathbf{s}}^{-1}$

KYA: catalysed ko uncatalysed rate constant se divide karo. $A$ wale cancel ho jaate hain. $\frac{k_{cat}}{k_{uncat}}=e^{-(E_{a,cat}-E_{a,uncat})/RT}=e^{-(55,000-80,000)/(8.314\times 310)}$ $=e^{25,000/2577.3}=e^{9.700}=1.63\times 10^4$ 25 kJ/mol ki kami rate ko lagbhag 16,300 guna kar deti hai — exponent ke upar ki thodi si change exponentiate hone ke baad bahut badi ho jaati hai.

$1/T$ aur $\ln k$ mein convert karo:

Endpoints se slope: $m=\frac{-2.3581-(-6.1658)}{0.00285714-0.00344828}=\frac{3.8077}{-5.9113\times 10^{-4}}=-6441\text{K}$ $E_a=-m\times R=6441\times 8.314=53,554\text{J/mol}\approx 53.6kJ/mol$ Straightness check: plotted points (figure) lagbhag perfect line par hain — beech ke points endpoint line par hi baithe hain, isliye single-slope $E_a$ trustworthy hai. Agar curve hota, toh $E_a$ khud temperature-dependent hota (changing mechanism ka sign).

$\ln k=\ln A-\frac{E_a}{RT}$ se, $\ln A$ ke liye solve karo: $\ln A=\ln k+\frac{E_a}{RT}=-3.4674+\frac{53,600}{8.314\times 330}$ $=-3.4674+\frac{53,600}{2743.6}=-3.4674+19.5364=16.069$ $A=e^{16.069}=9.5\times 10^6{\mathbf{s}}^{-1}$ (Order of magnitude $10^6$–$10^7$; units $k$ se match karte hain kyunki yeh first order hai.)

$\frac{k_Q}{k_P}=e^{-(E_{a,Q}-E_{a,P})/RT}=e^{-(48,000-60,000)/(8.314\times 298)}$ $=e^{12,000/2477.6}=e^{4.8434}=126.8$ $k_Q=5.0\times 10^{-4}\times 126.8=6.34\times 10^{-2}{\mathbf{s}}^{-1}$

Reasoning: dono lines ka same intercept $\ln A$ hai (same $A$), isliye woh $1/T=0$ par milti hain (infinite temperature — ek dum baayein). Catalysed line shallower hai ($E_a$ chhota ⇒ slope kam negative). Kyunki yeh zyada dheere utarti hai, isliye kisi bhi real finite temperature par ($1/T>0$) yeh uncatalysed line ke upar rehti hai. Lines $1/T$ badhne par (matlab $T$ ghattne par) fan out hoti jaati hain — catalyst sabse zyada kam temperature par help karta hai. Gap at 320 K: $\Delta (\ln k)=\ln k_{cat}-\ln k_{uncat}=-\frac{E_{a,cat}}{RT}+\frac{E_{a,uncat}}{RT}=\frac{E_{a,uncat}-E_{a,cat}}{RT}$ $=\frac{90,000-50,000}{8.314\times 320}=\frac{40,000}{2660.5}=15.03$ Rate ke terms mein: $k_{cat}/k_{uncat}=e^{15.03}\approx 3.4\times 10^6$ — millions guna speed-up. Dekho Catalysis Mechanisms.

Gap ko $\ln 10$ ke barabar set karo: $\frac{40,000}{8.314T}=2.3026\Rightarrow T=\frac{40,000}{8.314\times 2.3026}=\frac{40,000}{19.144}=2089\mathbf{K}$ Matlab: sirf $\approx 2089\text{K}$ jaise ek pagal zyada temperature par catalyst ka faayda sirf factor of 10 tak aata hai. Ordinary temperatures par faayda astronomically zyada hota hai. Yahi kyun hai ki catalysts sabse zyada matter karte hain un reactions ke liye jo hum room temperature ke paas sasti mein run karna chahte hain — connect karo Industrial Catalysis se.

Step 1 — heating se target rate ratio. 300 K se 360 K tak garam karna (same $E_a$, same $A$): $\ln \frac{k_{360}}{k_{300}}=-\frac{E_a}{R}\left(\frac{1}{360}-\frac{1}{300}\right)=-\frac{100,000}{8.314}\times (-5.5556\times 10^{-4})$ $=12028.6\times 5.5556\times 10^{-4}=6.6826$ Toh heating $k$ ko $e^{6.6826}$ se multiply karti hai. Step 2 — 300 K par catalyst se match karo. Catalyst ko 300 K par wahi factor produce karna hoga: $\ln \frac{k_{cat}}{k_{uncat}}=\frac{E_{a,uncat}-E_{a,cat}}{R\times 300}=6.6826$ $E_{a,uncat}-E_{a,cat}=6.6826\times 8.314\times 300=16,666\text{J/mol}$ $\boxed{\Delta E_a\approx 16.7kJ/mol}$ Matlab: ek catalyst jo $E_a$ sirf 16.7 kJ/mol kam kare at 300 K, wahi kaam karta hai jo 60 K heating karne se hota hai — lekin bina kisi fuel bill ke. Yahi quantitatively woh wajah hai ki industry catalysts ko brute-force heating se zyada prefer karti hai.

Curvature ka matlab: local slope har point par $-E_a/R$ hai. High $T$ par (chhota $1/T$, left side) kam steep slope ka matlab hai ki effective $E_a$ high temperature par chhota hai. KYUN aisa hota hai: ek reaction do competing pathways se chal sakti hai (ya changing rate-determining step se). Kam $T$ par high-barrier path rate constant ke temperature response par dominate karta hai; jaise $T$ badhti hai, ek lower-barrier channel le leta hai — isliye mixture ka apparent $E_a$ ghatta hai. (Transition State Theory mein $A$ ki $T$-dependence bhi halki curvature add kar sakti hai — dekho Transition State Theory.) Single fit ki galti: curved data par ek straight line force karna ek average slope deta hai — ek aisa $E_a$ jo na low-$T$ wala hai na high-$T$ wala, aur dono extremes par $k$ galat predict karta hai. Single $E_a$ quote karne se pehle hamesha curvature check karo (yeh L3.1 ke straightness check se jodta hai).

(a) Reaction energy diagram par, reverse barrier = forward barrier − $\Delta H$ product side se measure kiya hua. $\Delta H=-40$ ke saath (products reactants se 40 kJ/mol neeche): $E_{a,\text{rev}}=E_{a,\text{fwd}}-\Delta H=90-(-40)=130kJ/mol$ (Uncatalysed check: $120-(-40)=160$; catalyst ne forward aur reverse dono ko exactly 30 kJ/mol kam kiya — $120\to 90$ aur $160\to 130$.) (b) Equilibrium constant barriers ke difference se track hota hai: $K\propto \frac{k_{\text{fwd}}}{k_{\text{rev}}}=e^{-(E_{a,\text{fwd}}-E_{a,\text{rev}})/RT}$ Uncatalysed: $E_{a,\text{fwd}}-E_{a,\text{rev}}=120-160=-40$. Catalysed: $90-130=-40$. Bilkul same. Catalyst ne dono barriers se 30 ghataaya, isliye unka difference — aur isliye $K$ — unchanged hai. Yeh rigorous proof hai ki catalysts kinetics badlaate hain, equilibrium kabhi nahi.

Slope $=-mR$? ::: $E_a=-m\times R$ (negative slope $m$ se positive $E_a$ milta hai). Exponential ke andar units? ::: Energy J/mol mein, kyunki $R$ J/(mol·K) mein hai. Same $A$, different $E_a$ — parallel lines? ::: Nahi — same intercept, different slope ⇒ lines $1/T=0$ se fan out hoti hain. Kya catalyst $\Delta H$ ya $K$ badalta hai? ::: Koi nahi; yeh shared barrier kam karta hai, sirf rate badlta hai.