Worked examples — Activation energy from Arrhenius plot; effect of catalyst

2.8.8 · D3 · Chemistry › Chemical Kinetics › Activation energy from Arrhenius plot; effect of catalyst

Yeh page ek cases ki drill hai. Parent note (parent topic) ne tumhe Arrhenius Equation aur plot diya tha. Yahan hum ise har kone mein push karte hain: normal data, two-point tricks, catalyst factors, "temperature-that-does-what" twist, aur woh degenerate cases jahan equation almost toot jaati hai.

Neeche jo bhi hai, sirf wahi tools use hote hain jo tumhare paas already hain: multiplication, division, natural logarithm $\ln$, aur uska inverse $e^x$. Agar inme se koi bhi abhi bhi jaadu jaisa lagta hai, pehle do boxes padh lo.

Recall Reminder:

$\ln$ aur $e^x$ actually kya karte hain $e^x$ ek "growth machine" hai: ise number $x$ do, yeh batata hai koi cheez kitne guna badi ho gayi. $e^0=1$ (koi change nahi), $e^1\approx 2.718$, $e^{10}\approx 22026$ (bahut bada). $\ln$ woh sawaal hai jo ise undo karta hai: "$\ln y$" poochta hai "e ki kaunsi power $y$ deti hai?" Toh $\ln (e^x)=x$. Yahi ek reason hai kyun hum kabhi Arrhenius equation ka $\ln$ lete hain — taaki $E_a$ ko exponent se neeche kheench sakein jahan hum use ek straight line se padh sakein.

---

Scenario matrix

Neeche har alag tarah ka problem hai jo yeh topic pooch sakta hai. Har worked example ek cell tag ke saath hai.

#	Case class	Kya alag hai	Example
C1	Full data set → slope	Bahut saare $(T,k)$ points, best-fit line	Ex 1
C2	Two-point shortcut	Sirf do temperatures diye, koi plot nahi	Ex 2
C3	$A$ find karo (intercept)	Pre-exponential factor ke liye pucha	Ex 3
C4	Same $A$, do $E_a$ compare karo	Rate constants ka ratio	Ex 4
C5	Catalyst rate-enhancement	$\Delta E_a<0$, factor $\gg 1$	Ex 5
C6	Degenerate: $E_a=0$	Barrier-free reaction, limiting case	Ex 6
C7	Limit: $T\to \infty$ aur $T\to 0$	Extremes par kya hota hai	Ex 7
C8	Real-world word problem	Food spoilage, "kitna time?"	Ex 8
C9	Exam twist: $T$ find karo	Unknown temperature ke liye solve karo	Ex 9

Hum C1–C9 neeche cover karte hain. Do sign-traps (positive vs negative exponent) almost har ek mein aate hain, toh "Why this step?" notes dhyan se dekho.

---

Example 1 — C1: Full data set, best-fit slope se $E_a$ nikalo

Forecast: Guess karo — kya $E_a$ tens of kJ/mol mein aayega ya hundreds mein? (Hint: $k$ roughly har 20 K par triple ho jaata hai. Yeh ek "gentle" temperature response hai, toh bet karo tens par.)

Step 1 — raw table ko plot-coordinates mein convert karo. Hum chahte hain $\frac{1}{T}$ aur $\ln k$.

Why this step? Arrhenius equation tabhi straight line hai jab variables $x=\frac{1}{T}$ aur $y=\ln k$ hon. Seedha $k$ vs $T$ plot karo toh ek curve aati hai jisme se clean slope nahi padh sakte (figure ka left panel); $x=1/T,y=\ln k$ switch karo toh points ek line par aate hain (right panel).

Figure — same data, do views. Left: raw $k$ vs $T$, ek bending curve jisme koi readable slope nahi. Right: Arrhenius plot $\ln k$ vs $1/T$; chaar lavender dots line up hain, aur mint line unka best fit hai. Line ka slope $-E_a/R$ hai.

$T$	$1/T$ (${\text{K}}^{-1}$)	$\ln k$
300	0.003333	$-5.404$
320	0.003125	$-4.160$
340	0.002941	$-3.062$
360	0.002778	$-2.083$

Step 2 — sabhi chaar points se line fit karo, sirf do se nahi. Jab do se zyada measurements hon toh sahi method least squares hai: woh single straight line $y=mx+c$ dhundho jiska slope $m$ charon dots ke total squared vertical gap ko as small as possible banaye. Closed-form slope hai

$m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-{\left(\sum x\right)}^2},n=4.$

Charon $(x=1/T,y=\ln k)$ pairs is mein feed karne par

$m\approx -5972\text{K}$

milta hai.

Why this step? Real data mein scatter hota hai; two-point slope tumhare chaar measurements mein se do throw away kar deta hai aur endpoints par blindly trust karta hai. Least squares har point use karta hai, isliye ek galat reading poore answer ko hijack nahi kar sakti. (Yahan data almost perfectly linear hai, toh neeche diya quick endpoint slope best-fit ka $0.2\%$ ke andar aata hai.)

Step 2b — exam shortcut (sirf endpoints). Agar tumhare paas fit function wala calculator nahi hai, to do extreme points use karo:

$m=\frac{\ln k_4-\ln k_1}{\frac{1}{T_4}-\frac{1}{T_1}}=\frac{-2.083-(-5.404)}{0.002778-0.003333}=\frac{3.321}{-0.000555}=-5984\text{K}$

Why this step? Slope = rise-over-run. Run negative hai (jab $T$ badhta hai, $1/T$ girta hai) aur rise positive hai — genuinely downward-sloping line. Note karo $-5984$ vs best-fit $-5972$: teen figures tak agreement, jo batata hai data sach mein linear hai.

Step 3 — slope ko $E_a$ mein convert karo.

$E_a=-m\times R=-(-5984)(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1})=49748\text{J/mol}\approx 49.7\text{kJ/mol}$

Why this step? Slope $=-E_a/R$, toh $E_a=-mR$. Do minus signs multiply hokar positive $E_a$ dete hain — activation energy kabhi negative nahi hoti.

Verify: Units: $\text{K}\times \text{J}{\text{mol}}^{-1}{\text{K}}^{-1}=\text{J/mol}$ ✓. Magnitude "tens of kJ/mol" band mein hai jaise forecast kiya tha ✓. Sanity: ek mid-range small-molecule reaction 40–120 kJ/mol hoti hai, toh 49.7 chemically believable hai.

---

Example 2 — C2: Two-point shortcut (koi graph nahi)

Forecast: $k$ 40 K rise ke liye 8× jump kiya. Ex 1 se bada jump → guess karo bada $E_a$.

Step 1 — Arrhenius do baar likho aur divide karo.

$\frac{k_2}{k_1}=\frac{A{e}^{-E_a/R{T}_2}}{A{e}^{-E_a/R{T}_1}}=e^{-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$

Why this step? Hum $A$ nahi jaante. Do equations divide karne par unknown $A$ exactly cancel ho jaata hai, aur ek equation mein ek unknown $E_a$ bacha rehta hai.

Step 2 — $\ln$ lo taaki $E_a$ free ho.

$\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Why this step? $\ln$ woh tool hai jo $E_a$ ko exponent se bahar nikalti hai (recall box dekho). Maine bracket ka sign flip kiya taaki front sign $+$ ho jaye — cleaner hai, kyunki $\frac{1}{T_1}>\frac{1}{T_2}$ bracket ko positive banata hai.

Step 3 — numbers plug in karo.

$\ln \frac{2.0\times 10^{-3}}{2.5\times 10^{-4}}=\ln 8=2.079$ $\frac{1}{310}-\frac{1}{350}=0.0032258-0.0028571=3.687\times 10^{-4}{\text{K}}^{-1}$ $E_a=\frac{R\ln (k_2/k_1)}{\frac{1}{T_1}-\frac{1}{T_2}}=\frac{8.314\times 2.079}{3.687\times 10^{-4}}=46880\text{J/mol}\approx 46.9\text{kJ/mol}$

Why this step? Humne Step 2 equation ko $E_a$ ke liye solve kiya (dono sides $R$ se multiply karo, bracket se divide karo) aur wo do numbers substitute kiye jo humne compute kiye. $R$ J/mol·K mein rehta hai, aur $E_a$ J/mol mein nikalta hai — 46.9 kJ/mol quote karne ke liye 1000 se divide karo.

Verify: Positive $E_a$ ✓. Ex 1 jaisi band (~50 kJ/mol) ✓. Bracket positive hai aur $\ln 8$ positive hai, toh $E_a>0$ automatically — ek built-in sign check.

---

Example 3 — C3: Pre-exponential factor $A$ recover karo

Forecast: $A$ woh rate constant hai jo reaction ka hoga agar barrier disappear ho jaye — toh $A\gg k$. Ek bada number guess karo.

Step 1 — linear form se shuru karo aur $\ln A$ ke liye solve karo.

$\ln k=\ln A-\frac{E_a}{RT}\Rightarrow \ln A=\ln k+\frac{E_a}{RT}$

Why this step? $\ln A$ Arrhenius line ka y-intercept hai — $\ln k$ ki value jab $\frac{1}{T}=0$ ho. Hum ise reconstruct karte hain woh bit wapas jod ke jo barrier ne subtract kiya tha.

Step 2 — plug in karo. ($E_a$ already J/mol mein hai, toh yahan ×1000 ki zaroorat nahi.)

$\frac{E_a}{RT}=\frac{49748}{8.314\times 300}=19.94$ $\ln A=\ln (0.0045)+19.94=-5.404+19.94=14.53$

Step 3 — log undo karo.

$A=e^{14.53}=2.04\times 10^6{\text{s}}^{-1}$

Why this step? $e^x$, $\ln$ ko undo karta hai; hum $A$ khud chahte hain, $\ln A$ nahi.

Verify: $A\approx 2\times 10^6\gg k=0.0045$ ✓ (jaise forecast kiya). $A$ ke units $k$ se match karte hain (${\text{s}}^{-1}$, first-order reaction) ✓.

---

Example 4 — C4: Do reactions, same $A$, alag $E_a$

Forecast: B ka barrier 25 kJ/mol unccha hai. Unccha barrier → slower. Guess karo $k_B$ hazaron guna chota hai.

Step 1 — ratio, taaki $A$ cancel ho.

$\frac{k_B}{k_A}=e^{-(E_{a,B}-E_{a,A})/RT}$

Why this step? Same $A$ ⇒ divide karne par woh khatam ho jaata hai (Ex 2 jaisi hi trick).

Step 2 — J/mol mein convert karo, phir exponent evaluate karo.

$E_{a,B}-E_{a,A}=(75-50)\frac{\text{kJ}}{\text{mol}}=25\frac{\text{kJ}}{\text{mol}}=25000\frac{\text{J}}{\text{mol}}$ $\frac{25000}{8.314\times 298}=10.09,\frac{k_B}{k_A}=e^{-10.09}=4.15\times 10^{-5}$

×1000 aur minus kyun? ×1000 J-based $R$ se match karta hai (units box dekho). B ka exponent $-E_{a,B}/RT$ A se zyada negative hai, toh ratio 1 se kam hai — B slower hai. Sign encode karta hai "kiska barrier unccha hai."

Step 3 — $k_B$ get karne ke liye multiply karo.

$k_B=k_A\times \frac{k_B}{k_A}=2.0\times 10^{-3}\times 4.15\times 10^{-5}=8.3\times 10^{-8}{\text{s}}^{-1}$

Why this step? Hum $k_A$ jaante hain aur humne abhi ratio $k_B/k_A$ find kiya; jaane hue $k_A$ ko us ratio se multiply karke target $k_B$ isolate ho jaata hai. Units: ${\text{s}}^{-1}\times (\text{dimensionless ratio})={\text{s}}^{-1}$ ✓.

Verify: $1/(4.15\times 10^{-5})\approx 24000$, toh B ~24,000× slower hai ✓ — "hazaron," jaise forecast kiya. Extra 25 kJ/mol ka cost $\approx 2.4\times 10^4$ factor: $E_a$ ke liye exponential ki brutal sensitivity.

---

Example 5 — C5: Catalyst rate-enhancement factor

Figure — energy landscape, catalyzed vs not. Coral hump uncatalyzed path hai (ek tall barrier $E_a$ bada). Mint path catalyzed route hai: do chhote humps. Crucially dono mein reactant aur product levels identical hain — lavender $\Delta H$ arrow unchanged hai. Catalyst sirf peak lower karta hai, endpoints ko kabhi nahi.

Forecast: Barrier 25 kJ/mol drop karta hai — Ex 4 ke gap jaisi same size drop, toh expect karo similar ~20,000× — lekin ab yeh hamare liye kaam karta hai.

Step 1 — ratio likho.

$\frac{k_{cat}}{k_{uncat}}=e^{-\Delta E_a/RT},\Delta E_a=E_{a,cat}-E_{a,uncat}$

Why this step? Catalyst sirf exponent (barrier height) change karta hai. $\Delta E_a$ negative hoga kyunki catalyst path lower hai — yahi toh catalyst ka point hai.

Step 2 — J/mol mein convert karo aur evaluate karo.

$\Delta E_a=(60-85)\frac{\text{kJ}}{\text{mol}}=-25\frac{\text{kJ}}{\text{mol}}=-25000\frac{\text{J}}{\text{mol}}$ $-\frac{\Delta E_a}{RT}=-\frac{-25000}{8.314\times 300}=+10.02,\frac{k_{cat}}{k_{uncat}}=e^{10.02}=22400$

Why this step? ×1000 $R$ ke joules se match karta hai. Do minus signs (aage ka $-$, negative $\Delta E_a$) milkar ek positive exponent banate hain → factor $\gg 1$ → speed-up. Agar catalyst se kabhi factor $<1$ aaye, toh tumne koi sign flip kiya hai.

Verify: 1-hour ($3600$ s) uncatalyzed reaction ab $3600/22400\approx 0.16$ s leti hai ✓. Note karo figure mein: reactant aur product energies dono paths mein identical hain — catalyst $\Delta H$ move nahi karta, sirf peak (dekho Reaction Energy Diagrams aur Catalysis Mechanisms).

---

Example 6 — C6: Degenerate case, $E_a=0$

Forecast: Climb karne ke liye koi barrier nahi → guess karo ki har collision count hota hai, aur temperature zyada matter nahi karta.

Step 1 — $k=A{e}^{-E_a/RT}$ mein $E_a=0$ set karo.

$k=A{e}^{-0/RT}=A{e}^0=A\cdot 1=A$

Why this step? $RT$ times zero over anything zero hai, aur $e^0=1$. Exponential "gate" poori tarah se open hai.

Step 2 — temperature dependence padho.

Kyunki $k=A$ (ek constant), is idealisation mein $k$ temperature-independent hai.

Why this step? Arrhenius plot par slope hai $-E_a/R=0$: ek perfectly flat horizontal line. Koi barrier nahi ⇒ koi steepness nahi.

Verify: Collision Theory se consistent: jab koi energy requirement nahi, "successful" collisions ka fraction 1 hai, toh rate sirf is par depend karta hai ki molecules kitni baar milte hain — $A$ se capture hota hai. (Real barrier-free reactions mein $A$ ke through mild $T$-dependence hoti hai, lekin Arrhenius exponent kuch contribute nahi karta.)

---

Example 7 — C7: Do limits, $T\to \infty$ aur $T\to 0$

Forecast: Itna hot → har molecule bar clear kar le. Itna cold → koi nahi karta. Guess karo $k\to A$ (hot) aur $k\to 0$ (cold).

Step 1 — hot limit $T\to \infty$.

$\frac{E_a}{RT}\to \frac{E_a}{R\cdot \infty }=0\Rightarrow k\to A{e}^0=A$

Why this step? Ek fixed $E_a$ ko ever-growing $RT$ se divide karne par 0 ho jaata hai; gate poori tarah se khul jaata hai. Plot par, $\frac{1}{T}\to 0$ (far left), aur $\ln k\to \ln A$ — isliye intercept $\ln A$ hai.

Step 2 — cold limit $T\to 0^{+}$.

$\frac{E_a}{RT}\to \frac{E_a}{R\cdot 0^{+}}=+\infty \Rightarrow k\to A{e}^{-\infty }=0$

Why this step? Chota $RT$ exponent ko enormously negative banata hai; $e^{-\infty }=0$. Plot par, $\frac{1}{T}\to \infty$ (far right), $\ln k\to -\infty$ — line bottom se neeche gir jaati hai.

Verify: Dono forecast aur Maxwell-Boltzmann Distribution picture se match karte hain: high $T$ par high-energy tail moti hoti hai (almost sab $E_a$ clear kar lete hain); $T\to 0$ par distribution zero energy ki taraf collapse ho jaata hai aur essentially kisi ke paas $\ge E_a$ nahi hoti. $k$, $0$ aur $A$ ke beech bounded hai sabhi $T>0$ ke liye ✓.

---

Example 8 — C8: Real-world word problem (food spoilage)

Forecast: Garam → faster spoiling → 7 din se bahut kam. Guess karo zyada se zyada do din.

Step 1 — "shelf life" ko rate se jodo. Spoiling same visible endpoint tak matlab same amount of reaction, toh time $\propto 1/k$:

$\frac{t_{warm}}{t_{cold}}=\frac{k_{cold}}{k_{warm}}=e^{-\frac{E_a}{R}\left(\frac{1}{T_{cold}}-\frac{1}{T_{warm}}\right)}$

Why this step? Zyada chalna matlab slower reaction; time aur rate reciprocals hain. Ratio $A$ cancel karta hai, toh hum ise kabhi need nahi karte.

Step 2 — $E_a$ ko J/mol mein convert karo aur exponent evaluate karo.

$E_a=75\frac{\text{kJ}}{\text{mol}}=75000\frac{\text{J}}{\text{mol}}$ $\frac{1}{277}-\frac{1}{298}=0.0036101-0.0033557=2.544\times 10^{-4}{\text{K}}^{-1}$ $\frac{E_a}{R}\times 2.544\times 10^{-4}=\frac{75000}{8.314}\times 2.544\times 10^{-4}=9021\times 2.544\times 10^{-4}=2.295$

Why this step? ×1000 units ko $R$ ke J/mol·K ke saath consistent rakhta hai; phir $E_a/R$ (units K) times bracket (units ${\text{K}}^{-1}$) ek pure number $2.295$ deta hai, exactly wahi jo exponent hona chahiye.

Step 3 — shelf life assemble karo. Cold side slower hai, toh $k_{cold}/k_{warm}=e^{-2.295}=0.1008$:

$t_{warm}=t_{cold}\times 0.1008=7\text{days}\times 0.1008=0.706\text{days}\approx 16.9\text{hours}$

Why this step? Hum $t_{warm}$ chahte the; Step 1 se, $t_{warm}=t_{cold}\times (k_{cold}/k_{warm})$, toh jaani hui 7-day cold life ko (small) rate ratio se multiply karo. $e^{-2.295}$ mein minus sign ratio ko 1 se kam banata hai, yaani ek choti warm shelf life — physically sensible direction.

Verify: Counter par ek din se kam vs fridge mein ek hafta ✓ — lived experience se match karta hai aur "much shorter" forecast se bhi. ~21 K rise ne ~10× rate change diya: rozana ka "rule of thumb" ki reactions roughly har 10 °C par double ya zyada ho jaati hain.

---

Example 9 — C9: Exam twist — temperature ke liye solve karo

Forecast: Bade $A$ ke saath lekin bade barrier ke saath, tumhe gate kholne ke liye kaafi thermal energy chahiye — likely kuch hundred K.

Step 1 — Arrhenius ka $\ln$ lo aur $T$ isolate karo.

$\ln k=\ln A-\frac{E_a}{RT}\Rightarrow \frac{E_a}{RT}=\ln A-\ln k=\ln \frac{A}{k}$

Why this step? Unknown $T$ exponent ke andar trapped hai. $\ln$ ise free karta hai, aur ab yeh ek clean fraction mein baitha hai.

Step 2 — $T$ ke liye rearrange karo aur $E_a$ ko J/mol mein convert karo.

$T=\frac{E_a}{R\ln (A/k)},E_a=100\frac{\text{kJ}}{\text{mol}}=100000\frac{\text{J}}{\text{mol}}$

Why this step? Hum Step 1 ko dono sides $\frac{RT}{\ln (A/k)}$ se multiply karte hain taaki $T$ akela ho jaye; ×1000 phir se $R$ ko joules mein match karta hai taaki units kelvin mein cancel ho jayein.

Step 3 — plug in karo.

$\ln \frac{A}{k}=\ln \frac{1.0\times 10^{13}}{1.0}=\ln (10^{13})=29.93$ $T=\frac{100000}{8.314\times 29.93}=\frac{100000}{248.9}=401.8\text{K}$

Why this step? Seedha substitution; note karo units $\frac{\text{J/mol}}{(\text{J}{\text{mol}}^{-1}{\text{K}}^{-1})\times (\text{pure number})}=\text{K}$ ✓.

Verify: $T\approx 402$ K $=129$ °C — "few hundred K," jaise forecast kiya ✓. Back-substitute: $\frac{E_a}{RT}=\frac{100000}{8.314\times 401.8}=29.94$, aur $A{e}^{-29.94}=10^{13}\times e^{-29.94}\approx 1.0{\text{s}}^{-1}$ ✓ (dekho Rate Constant Temperature Dependence).

---

Recall Self-test (right side cover karo)

Arrhenius plot ka slope kya equals karta hai? ::: $-E_a/R$ Chaar data points ke saath, konsa fitting method unhe sab use karta hai? ::: Least-squares (best-fit) line kJ/mol value ko $E_a/RT$ mein plug karne se pehle kya karna zaroori hai? ::: Multiply by 1000 to get J/mol (to match $R$ in J/mol·K) Sirf do $(T,k)$ points se $E_a$ find karne ke liye, kya cancel hota hai? ::: Pre-exponential factor $A$ 300 K par $E_a$ 25 kJ/mol drop karne wala catalyst reaction ko roughly kitna fast karta hai? ::: $\sim$22,000× Jab $T\to \infty$, $k\to$ ? ::: $A$ (intercept, gate fully open) Jab $E_a=0$, Arrhenius plot line kaisi dikhti hai? ::: A flat horizontal line ($k=A$, koi $T$-dependence nahi) Kya catalyst $K_{eq}$ change karta hai? ::: No — only how fast equilibrium is reached