Temperature dependence — Arrhenius equation k = A·e^(−Ea - RT): Worked examples

2.8.7 · D3 · Chemistry › Chemical Kinetics › Temperature dependence — Arrhenius equation k = A·e^(−Ea - R

Yeh page parent Arrhenius note ($k = A\,e^{-E_a/(RT)}$) ka "no surprises" workbook hai. Parent ne equation aur do clean examples dikhaye the. Yahan hum har tarah ka question dhundhte hain jo yeh topic pooch sakta hai — har unknown, har degenerate case, har sneaky exam twist — aur har ek ko scratch se solve karte hain.

Shuru karne se pehle, sirf do tools jo hamesha kaam aate hain unhe re-state karte hain, taaki koi symbol unfamiliar na lage.

The scenario matrix

Har Arrhenius problem mein basically yahi hota hai — "kaun sa quantity missing hai, aur koi input edge par toh nahi?" Yeh poora grid hai. Neeche har example ko us cell ke saath tag kiya gaya hai jo woh fill karta hai.

Cell	Kya unknown hai	Twist / edge case	Example
A	$E_{a}$	do clean $(T, k)$ points	Ex 1
B	$k_{2}$	higher $T$ par predict karo	Ex 2
C	$k_{2}$	lower $T$ par predict karo (cooling, $ln$ negative)	Ex 3
D	$E_{a}$	graph slope se (kai points, error average hoti hai)	Ex 4
E	$A$	pre-exponential factor back out karo	Ex 5
F	$T$	woh temperature dhundo jo target $k$ de	Ex 6
G	degenerate: $E_{a} = 0$	jab koi barrier nahi hota tab kya hota hai	Ex 7
H	limiting: $T \to \infty$	$k$ ki ceiling; aur $T \to 0$ floor	Ex 7
I	real-world word problem	Q10 / food spoilage doubling	Ex 8
J	exam twist	do reactions ka ratio, ya "kitna faster"	Ex 9
K	edge: $E_{a} < 0$	cooling se reaction fast ho jaati hai	Ex 10

Do sign facts pocket mein rakhne wale hain, kyunki cells B/C mein yahi pakad lete hain logon ko. Neeche ki figure inhe draw karti hai: yeh $k$ ko $T$ ke against plot karti hai hamare running reaction ke liye aur heating ke liye burnt-orange arrow (daayein aur upar) aur cooling ke liye teal arrow (baayein aur neeche) dikhaati hai, same reference point se door.

Figure — Temperature dependence — Arrhenius equation k = A·e^(−Ea - RT)

Cell A — do points se $E_{a}$ nikalo

Worked example Example 1 (Cell A)

Ek reaction mein $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $T_{1} = 310 K$ aur $k_{2} = 8.0 \times 1 0^{- 3} s^{- 1}$ at $T_{2} = 330 K$ hai. $E_{a}$ nikalo.

Forecast: $k$ 20 K rise mein 4× badh gaya. Guess: moderate barrier, tens of kJ/mol.

Step 1 — two-temperature form choose karo. Yeh step kyun? $A$ unknown hai aur hume chahiye nahi; two-point form $A$ delete kar deta hai. $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

Step 2 — left side compute karo. Yeh step kyun? Ratio $k_{2} / k_{1} = 4$ hai, aur $ln 4 = 1.3863$ hai.

Step 3 — kelvin mein bracket compute karo. Yeh step kyun? Dono temperatures absolute honi chahiye warna Boltzmann factor meaningless hai. $\frac{1}{310} - \frac{1}{330} = 0.0032258 - 0.0030303 = 1.9550 \times 1 0^{- 4} K^{- 1}$

Step 4 — $E_{a}$ ke liye solve karo. $E_{a} = \frac{R l n ( k _{2} / k _{1} )}{( \frac{1}{T _{1}} - \frac{1}{T _{2}} )} = \frac{8.314 \times 1.3863}{1.9550 \times 1 0 ^{- 4}} \approx 5.90 \times 1 0^{4} J/mol = 59.0 kJ/mol$

Answer: $E_{a} \approx 59.0 kJ/mol$ — moderate barrier, forecast se match karta hai.

Verify: Units: $\frac{J/(mol\cdotK) \times (dimensionless)}{1/ K} = J/mol$ ✓. Plug back: $\frac{59000}{8.314} \times 1.955 \times 1 0^{- 4} = 1.387 \approx ln 4$ ✓.

Cell B — higher temperature par $k$ predict karo

Worked example Example 2 (Cell B)

$E_{a} = 59.0 kJ/mol$ aur $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $310 K$ use karke, $360 K$ par $k$ predict karo.

Forecast: 59 kJ/mol barrier ke saath 50 K heating — roughly order of magnitude jump expect hai.

Step 1 — two-temperature form likhо $k_{2}$ ke liye solve karke. Yeh step kyun? $k_{2}$ hamara unknown hai; baaki sab known hai. $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

Step 2 — bracket. $\frac{1}{310} - \frac{1}{360} = 0.0032258 - 0.0027778 = 4.4803 \times 1 0^{- 4} K^{- 1}$ . Yeh step kyun? Kyunki $T_{2} > T_{1}$ bracket positive hai → $k$ badhega, jaisa sign rule (aur upar figure mein orange arrow) ne promise kiya tha.

Step 3 — $E_{a} / R$ se multiply karo. $\frac{59000}{8.314} \times 4.4803 \times 1 0^{- 4} = 7096.5 \times 4.4803 \times 1 0^{- 4} = 3.1794$

Step 4 — exponentiate aur multiply karo. Yeh step kyun? Humare paas $ln (k_{2} / k_{1})$ tha; log undo karo $e^{(\cdot)}$ se. $k_{2} = k_{1} e^{3.1794} = 2.0 \times 1 0^{- 3} \times 24.03 = 4.81 \times 1 0^{- 2} s^{- 1}$

Answer: $k \approx 4.8 \times 1 0^{- 2} s^{- 1}$ — 50 K rise mein roughly 24× faster.

Verify: Sanity: $k_{2} > k_{1}$ ✓ (heating). Order of magnitude ~ $1 0^{- 2}$ forecast se match karta hai.

Cell C — lower temperature par $k$ predict karo (negative log)

Worked example Example 3 (Cell C)

Same reaction ( $E_{a} = 59.0 kJ/mol$ , $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at $310 K$ ). Fridge mein $277 K$ (4 °C) par $k$ kya hoga?

Forecast: Cooling se cheezein slow hoti hain (is normal $E_{a} > 0$ ke liye), toh $k$ ko $2.0 \times 1 0^{- 3}$ se neeche drop karna chahiye. Log negative aayega — cooling ka tell, upar figure mein teal arrow se match karta hai.

Step 1 — two-temperature form, $T_{2} = 277 K$ ab $T_{1}$ se neeche hai. Yeh step kyun? $T_{1}$ ko known point rakho; sign direction automatically handle kar lega.

Step 2 — bracket. Yeh step kyun? Ab $1/ T_{2} = 1/277 = 0.0036101$ hai jo $1/ T_{1}$ se bada hai, toh bracket negative hai. $\frac{1}{310} - \frac{1}{277} = 0.0032258 - 0.0036101 = - 3.8430 \times 1 0^{- 4} K^{- 1}$

Step 3 — multiply karo. $ln \frac{k _{2}}{k _{1}} = 7096.5 \times (- 3.8430 \times 1 0^{- 4}) = - 2.7272$

Step 4 — exponentiate karo. $k_{2} = 2.0 \times 1 0^{- 3} \times e^{- 2.7272} = 2.0 \times 1 0^{- 3} \times 0.06544 = 1.31 \times 1 0^{- 4} s^{- 1}$

Answer: $k \approx 1.3 \times 1 0^{- 4} s^{- 1}$ — 310 K se roughly 15× slower.

Verify: Negative log ✓ (cooling). $k_{2} < k_{1}$ ✓. Yahi wajah hai refrigeration se food preserve hoti hai — Q10 Temperature Coefficient version ke liye Example 8 dekho.

Cell D — graph slope se $E_{a}$

Sabse reliable lab method: kai temperatures par $k$ measure karo, $ln k$ ko $1/ T$ ke against plot karo, slope padho. Linear form $ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ ek straight line $y = m x + c$ hai jiska slope $m = - E_{a} / R$ hai.

Neeche ki figure exactly yeh Arrhenius plot hai: plum line best fit hai, orange dots paanch noisy measurements hain jo yeh average karta hai, aur teal triangle slope $m = - 6500 K$ mark karta hai jo hum read off karte hain. Intercept label dhyan se dekho far left par, jahan line axis se milti — $1/ T \to 0$ par (yani $T \to \infty$ ) — woh height $ln A$ hai.

Worked example Example 4 (Cell D)

Upar ka plot padhke, best-fit Arrhenius line ka slope $m = - 6500 K$ aur intercept $c = 15.2$ hai. $E_{a}$ aur $A$ nikalo.

Forecast: Slope magnitude 6500 K × 8.314 ≈ 54 kJ/mol. Intercept 15.2 $ln A$ hai, toh $A \approx e^{15} \sim$ kuch million.

Step 1 — slope ko $E_{a}$ se relate karo. Yeh step kyun? Linear form se, $1/ T$ ka coefficient exactly $- E_{a} / R$ hai; figure mein teal triangle yahi slope hai. $m = - \frac{E _{a}}{R} \Rightarrow E_{a} = - m R = 6500 \times 8.314 = 5.40 \times 1 0^{4} J/mol$

Step 2 — intercept ko $A$ se relate karo. Yeh step kyun? $1/ T = 0$ par (infinite temperature) line $ln k = ln A$ ko hit karti hai — figure mein marked intercept. $ln A = 15.2 \Rightarrow A = e^{15.2} = 3.99 \times 1 0^{6} s^{- 1}$

Answer: $E_{a} = 54.0 kJ/mol$ , $A = 4.0 \times 1 0^{6} s^{- 1}$ .

Verify: $- m R$ ki units $K \times J/(mol\cdotK) = J/mol$ hain ✓. Multi-point fit do points se better hai kyunki random scatter (orange dots ka plum line se wobble) average out ho jaata hai — Maxwell-Boltzmann Distribution se connect karo ki individual $k$ measurements kyun fluctuate karti hain.

Cell E — pre-exponential factor $A$ back out karo

Worked example Example 5 (Cell E)

Ek reaction mein $E_{a} = 75.0 kJ/mol$ aur $k = 0.35 s^{- 1}$ at $T = 500 K$ hai. $A$ nikalo.

Forecast: Exponent $E_{a} / R T$ around 18 hoga, toh $e^{- 18}$ bahut chota (~ $1 0^{- 8}$ ) hai; phir bhi $k = 0.35$ milne ke liye, $A$ bahut bada hona chahiye — order $1 0^{7}$ – $1 0^{8}$ .

Step 1 — one-temperature form use karo; $A$ ke liye solve karo. Yeh step kyun? Sirf ek $(T, k)$ point diya hai aur hume $A$ chahiye, toh two-point form yahan useless hai. $k = A e^{- E_{a} / (R T)} \Rightarrow A = k e^{+ E_{a} / (R T)}$

Step 2 — exponent compute karo. Yeh step kyun? $E_{a} / (R T)$ energies ka dimensionless ratio hai; sab kuch J mein rakho. $\frac{E _{a}}{R T} = \frac{75000}{8.314 \times 500} = \frac{75000}{4157} = 18.043$

Step 3 — exponentiate aur multiply karo. $A = 0.35 \times e^{18.043} = 0.35 \times 6.86 \times 1 0^{7} = 2.40 \times 1 0^{7} s^{- 1}$

Answer: $A \approx 2.4 \times 1 0^{7} s^{- 1}$ .

Verify: $A$ same units as $k$ carry karta hai (yahan $s^{- 1}$ , kyunki first-order hai) kyunki $e^{- E_{a} / R T}$ dimensionless hai ✓. Plug back: $2.4 \times 1 0^{7} \times e^{- 18.043} = 2.4 \times 1 0^{7} \times 1.457 \times 1 0^{- 8} = 0.35$ ✓.

Cell F — temperature ke liye solve karo

Worked example Example 6 (Cell F)

$A = 2.4 \times 1 0^{7} s^{- 1}$ aur $E_{a} = 75.0 kJ/mol$ ke saath, kis temperature par $k = 1.0 s^{- 1}$ hoga?

Forecast: Hum $k$ bada chahte hain 500 K par 0.35 se, toh $T$ 500 K se thoda upar chahiye.

Step 1 — one-temperature form, exponential isolate karo. $\frac{k}{A} = e^{- E_{a} / (R T)}$

Step 2 — dono sides ka $ln$ lo. Yeh step kyun? Unknown $T$ ek exponent ke andar phansa hai; natural log woh tool hai jo exponent ko free karta hai. $ln \frac{k}{A} = - \frac{E _{a}}{R T}$

Step 3 — $T$ ke liye solve karo. $T = \frac{- E _{a}}{R l n ( k / A )} = \frac{E _{a}}{R l n ( A / k )}$ $ln \frac{A}{k} = ln \frac{2.4 \times 1 0 ^{7}}{1.0} = ln (2.4 \times 1 0^{7}) = 16.993$ $T = \frac{75000}{8.314 \times 16.993} = \frac{75000}{141.28} = 530.9 K$

Answer: $T \approx 531 K$ — forecast ke mutabiq 500 K se thoda upar.

Verify: Ex 5 ke numbers mein plug karo: 531 K par, $E_{a} / (R T) = 75000/ (8.314 \times 530.9) = 16.99$ , aur $k = 2.4 \times 1 0^{7} e^{- 16.99} = 1.0 s^{- 1}$ ✓.

Cell G & H — degenerate aur limiting cases

Yeh woh questions hain jo expose karte hain ki tum equation ko samajhte ho ya sirf plug in karte ho.

Worked example Example 7 (Cells G & H)

(G) Zero barrier. Agar $E_{a} = 0$ toh $k$ kya hai? (H) Extreme temperatures. $lim_{T \to \infty} k$ aur $lim_{T \to 0^{+}} k$ kya hai?

Forecast: Koi barrier nahi matlab har collision count hogi, toh $k$ bas $A$ ke barabar hona chahiye. Infinite $T$ matlab sab wall cross kar sakte hain, toh $k \to A$ . $T \to 0$ par freeze karo matlab koi nahi cross kar sakta, toh $k \to 0$ .

Step 1 — one-temperature form mein $E_{a} = 0$ set karo. Yeh step kyun? $E_{a} = 0$ se exponent zero ho jaata hai, aur $e^{0} = 1$ hota hai. $k = A e^{- 0/ (R T)} = A \cdot e^{0} = A$ Toh barrier-free reaction apni maximum possible rate par chalti hai, collision-limited rate $A$ . (Sach mein $E_{a} = 0$ rare hai; kuch radical recombinations iske karib aati hain — Catalysis dekho ki catalysts $E_{a}$ ko is ideal tak kaise lower karte hain.)

Step 2 — limit $T \to \infty$ . Yeh step kyun? Jaise $T$ badhta hai, $E_{a} / (R T) \to 0$ hota hai, toh exponent $\to 0$ aur $e^{- E_{a} / R T} \to 1$ . $lim_{T \to \infty} k = A \cdot 1 = A$ $A$ $k$ ki ceiling hai: tum molecules ke collide hone se faster react nahi kar sakte. Yeh Cell D figure mein intercept height hai.

Step 3 — limit $T \to 0^{+}$ . Yeh step kyun? Jaise $T \to 0^{+}$ hota hai, $E_{a} / (R T) \to + \infty$ hota hai, toh $e^{- \infty} \to 0$ hota hai. $lim_{T \to 0^{+}} k = A \cdot 0 = 0$ Absolute zero par, koi molecule barrier cross karne ki energy nahi rakhta — reaction freeze ho jaati hai.

Answer: $E_{a} = 0 \Rightarrow k = A$ ; $k \to A$ jaise $T \to \infty$ (ceiling); $k \to 0$ jaise $T \to 0^{+}$ (floor). Teen edge results har possible $k$ ko $0$ aur $A$ ke beech bracket karte hain.

Verify: $A = 1 0^{13}, E_{a} = 50000$ se numerically check karo: $T = 1 0^{6} K$ par, $e^{- 50000/ (8.314 \times 1 0^{6})} = e^{- 0.006} = 0.994$ , toh $k = 9.94 \times 1 0^{12} \approx A$ ✓. $T = 1 K$ par, $e^{- 50000/8.314} = e^{- 6014} \approx 0$ ✓.

Cell I — real-world word problem (Q10 doubling)

Ek famous rule of thumb: bahut si reaction rates roughly har 10 K rise mein double ho jaati hain. Yeh Q10 Temperature Coefficient hai $= k_{T + 10} / k_{T}$ . Chalte hain derive karte hain ki kaun sa $E_{a}$ $Q_{10} = 2$ banata hai.

Worked example Example 8 (Cell I)

Milk spoilage ka rate roughly $4 °C$ (277 K) aur $14 °C$ (287 K) ke beech double ho jaata hai. Effective $E_{a}$ estimate karo.

Forecast: "Har 10 K mein doubling" room temperature ke paas classic case hai; textbooks $E_{a}$ around 40–60 kJ/mol quote karte hain.

Step 1 — two-temperature form $k_{2} / k_{1} = 2$ ke saath. Yeh step kyun? "Doubling" precisely $ln (k_{2} / k_{1}) = ln 2$ hai.

Step 2 — bracket. Yeh step kyun? Dono temperatures kelvin mein convert karo (already hain) aur reciprocals subtract karo. $\frac{1}{277} - \frac{1}{287} = 0.0036101 - 0.0034843 = 1.2578 \times 1 0^{- 4} K^{- 1}$

Step 3 — $E_{a}$ ke liye solve karo. $E_{a} = \frac{R l n 2}{1.2578 \times 1 0 ^{- 4}} = \frac{8.314 \times 0.6931}{1.2578 \times 1 0 ^{- 4}} = \frac{5.763}{1.2578 \times 1 0 ^{- 4}} = 4.58 \times 1 0^{4} J/mol$

Answer: $E_{a} \approx 45.8 kJ/mol$ — expected band ke andar hai.

Verify: Doubling back predict karo: $\frac{45800}{8.314} \times 1.2578 \times 1 0^{- 4} = 5509.5 \times 1.2578 \times 1 0^{- 4} = 0.693 = ln 2$ ✓. Toh tumhara fridge (Example 3) 310→277 K se ~15× slowdown — yeh wahi physics hai bas zyada push ke saath.

Cell J — exam twist: "kitna zyada fast?"

Worked example Example 9 (Cell J)

Do reactions same 300 K par chal rahi hain. Reaction P ka $E_{a} = 40 kJ/mol$ hai, reaction Q ka $E_{a} = 80 kJ/mol$ hai. Equal $A$ assume karo. P, Q se kitna zyada fast hai?

Forecast: Q ka barrier P ka double hai, toh uska Boltzmann factor bahut chhota hoga — P many orders of magnitude faster honi chahiye.

Step 1 — ratio likho, same $T$ aur same $A$ . Yeh step kyun? Equal $A$ cancel ho jaata hai; sirf exponents differ karte hain. $\frac{k _{P}}{k _{Q}} = \frac{A e ^{- E_{a, P} / (R T)}}{A e ^{- E_{a, Q} / (R T)}} = e^{(E_{a, Q} - E_{a, P}) / (R T)}$

Step 2 — energy difference plug in karo. Yeh step kyun? $E_{a, Q} - E_{a, P} = 40000 J/mol$ woh "extra wall" hai jo Q ko climb karni padti hai. $\frac{E _{a, Q} - E _{a, P}}{R T} = \frac{40000}{8.314 \times 300} = \frac{40000}{2494.2} = 16.037$

Step 3 — exponentiate karo. Yeh step kyun? Humare paas $ln (k_{P} / k_{Q})$ hai; log undo karke raw ratio lo. $\frac{k _{P}}{k _{Q}} = e^{16.037} = 9.2 \times 1 0^{6}$

Answer: P, Q se 300 K par roughly 90 lakh times faster hai.

Verify: Exponents ka difference dimensionless hai ✓. Logs se cross-check: $ln (9.2 \times 1 0^{6}) = 16.03 \approx 16.037$ ✓. Yahi wajah hai ki $E_{a}$ ko catalyst se lower karna itna dramatic speed-up deta hai — aur kyun thermodynamically favourable reaction bhi immeasurably slow ho sakti hai agar $E_{a}$ high ho.

Cell K — genuine exception: negative activation energy

Upar har jagah $E_{a} > 0$ tha aur heating se cheezein fast hoti theen. Lekin kuch real reactions (kuch radical recombinations, aur kai multi-step reactions fast pre-equilibrium ke saath) measured $E_{a} < 0$ dikhate hain: inhe heat karo toh yeh slower ho jaate hain. Arrhenius algebra yeh automatically handle karta hai — bas sign haath se "correct" mat karo.

Worked example Example 10 (Cell K)

Ek reaction ka effective $E_{a} = - 12.0 kJ/mol$ hai aur $k_{1} = 5.0 \times 1 0^{2} s^{- 1}$ at $T_{1} = 300 K$ . $T_{2} = 350 K$ par $k$ predict karo.

Forecast: Negative $E_{a}$ matlab heating se yeh slow hogi, toh $k_{2} < 500$ expect hai. Aur $ln (k_{2} / k_{1})$ negative aana chahiye even though hum heat kar rahe hain — normal sign rule ka ulta.

Step 1 — two-temperature form, koi sign patching nahi. Yeh step kyun? Formula par trust karo; negative $E_{a}$ apna sign khud carry karta hai. $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

Step 2 — bracket (heating, toh still positive). Yeh step kyun? $T_{2} > T_{1}$ hai, toh bracket hamesha ki tarah positive hai; twist $E_{a}$ mein hai, yahan nahi. $\frac{1}{300} - \frac{1}{350} = 0.0033333 - 0.0028571 = 4.7619 \times 1 0^{- 4} K^{- 1}$

Step 3 — $E_{a} / R$ se multiply karo (ab negative). $ln \frac{k _{2}}{k _{1}} = \frac{- 12000}{8.314} \times 4.7619 \times 1 0^{- 4} = - 1443.4 \times 4.7619 \times 1 0^{- 4} = - 0.6873$

Step 4 — exponentiate karo. $k_{2} = 5.0 \times 1 0^{2} \times e^{- 0.6873} = 5.0 \times 1 0^{2} \times 0.5029 = 2.51 \times 1 0^{2} s^{- 1}$

Answer: $k \approx 2.5 \times 1 0^{2} s^{- 1}$ — heating se rate half ho gayi.

Verify: Heating ne negative log diya ✓ (sirf tab possible jab $E_{a} < 0$ ho). $k_{2} < k_{1}$ ✓. Cell-D Arrhenius plot par is reaction ki line "galat" direction mein slope hogi (positive slope $- E_{a} / R > 0$ ) — negative activation energy ka dead giveaway.

Recall Quick self-test

Two-temperature form $A$ jaane bina kaun sa unknown solve karta hai? ::: Ya toh $E_{a}$ ya doosra $k$ , ya doosra $T$ — $A$ ke siwa kuch bhi, kyunki $A$ cancel ho jaata hai. Agar heating ke baad $ln (k_{2} / k_{1})$ negative aaye, iska kya matlab hai? ::: Ya toh subscript swap hua hai, ya reaction genuinely $E_{a} < 0$ hai (Cell K). $E_{a} = 0$ hone par $k$ kya hota hai? ::: $k = A$ , collision-limited maximum. $T \to \infty$ hone par $k$ kya approach karta hai? ::: Ceiling $A$ . $T$ kelvin mein kyun hona chahiye? ::: $E_{a} / (R T)$ ek clean dimensionless energy ratio hona chahiye; Celsius zero ya negative ho sakta hai aur ise tod deta hai. Kya $k$ aur $A$ ki units reaction order ke saath badlti hain? ::: Haan — $A$ hamesha $k$ ki units share karta hai, jo order par depend karti hain (first-order ke liye s⁻¹, second-order ke liye L·mol⁻¹s⁻¹, etc.).