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

2.8.7 · D4 · Chemistry › Chemical Kinetics › Temperature dependence — Arrhenius equation k = A·e^(−Ea - R

Parent note dekho jahan se ye aate hain: the topic note (Hindi-English explanation). Supporting ideas yahan hain: Activation Energy and Transition State Theory, Rate Laws and Rate Constants, Maxwell-Boltzmann Distribution, Catalysis, Thermodynamics vs Kinetics, aur Q10 Temperature Coefficient.

State: Equation $k = A e^{- E_{a} / (R T)}$ mein, charon quantities $k$ , $A$ , $E_{a}$ , $T$ ko naam do, $E_{a}$ , $R$ , $T$ ke SI units batao, aur ek sentence mein explain karo ki exponent $- E_{a} / (R T)$ ek pure number (bina units) kyun hona chahiye.

Recall Solution

$k$ = rate constant (units reaction order par depend karte hain).
$A$ = pre-exponential / frequency factor, $k$ jaise hi units.
$E_{a}$ = activation energy — reactants se transition state tak ka energy barrier. Units: $J mol^{- 1}$ .
$T$ = absolute temperature, units $K$ .
$R = 8.314 J mol^{- 1} K^{- 1}$ .

Exponent hai $\frac{E _{a}}{R T}$ . Iske units hain: $\frac{J mol ^{- 1}}{( J mol ^{- 1} K ^{- 1} ) \cdot K} = \frac{J mol ^{- 1}}{J mol ^{- 1}} = dimensionless.$ Ek exponent hamesha dimensionless hona chahiye — tum $e$ ko "5 joules" tak raise nahi kar sakte. Isliye hi $R$ (per mole, per kelvin) $E_{a}$ (per mole) aur $T$ (kelvin mein) ka sahi companion hai.

Exercise 1.2

State: Do reactions X aur Y ka $A$ same hai. Reaction X ka $E_{a} = 30 kJ/mol$ hai; reaction Y ka $E_{a} = 90 kJ/mol$ hai. Same temperature par, kaun faster hai, aur words mein kyun?

Recall Solution

Jitna bada $E_{a}$ , utna hi exponent $- E_{a} / (R T)$ zyada negative hoga, isliye $e^{- E_{a} / (R T)}$ chhota hoga. Equal $A$ ke saath, chhota Boltzmann factor matlab chhota $k$ . Isliye X ( $E_{a} = 30$ ) faster hai. Same temperature par 90 kJ/mol ki wall ko utni molecules cross nahi kar sakti jitni 30 kJ/mol ki wall ko kar sakti hain.

Level 2 — Application

Exercise 2.1

State: Ek reaction ka $E_{a} = 50 kJ/mol$ aur $A = 2.0 \times 1 0^{13} s^{- 1}$ hai. $T = 300 K$ par $k$ nikalo.

Recall Solution

Step 1 — kya/kyun: Seedha $k = A e^{- E_{a} / (R T)}$ mein plug karo (single form, kyunki humein $A$ pata hai). Step 2 — exponent ( $E_{a}$ ko joules mein rakho taaki $R$ se match kare): $\frac{E _{a}}{R T} = \frac{50000}{8.314 \times 300} = \frac{50000}{2494.2} = 20.05.$ Step 3 — Boltzmann factor: $e^{- 20.05} = 1.96 \times 1 0^{- 9}$ . Step 4 — multiply karo: $k = 2.0 \times 1 0^{13} \times 1.96 \times 1 0^{- 9} = 3.9 \times 1 0^{4} s^{- 1} .$ Answer: $k \approx 3.9 \times 1 0^{4} s^{- 1}$ .

Exercise 2.2

State: Pehle convert karo, phir compute karo. Ek reaction ka $E_{a} = 75 kJ/mol$ , $A = 5.0 \times 1 0^{11} s^{- 1}$ hai, $2 7^{\circ} C$ par measure kiya gaya. $k$ nikalo.

Recall Solution

Step 1 — temperature convert karo: $T = 27 + 273.15 = 300.15 K \approx 300 K$ . Step 2 — exponent: $\frac{E _{a}}{R T} = \frac{75000}{8.314 \times 300} = 30.07.$ Step 3: $e^{- 30.07} = 8.7 \times 1 0^{- 14}$ . Step 4: $k = 5.0 \times 1 0^{11} \times 8.7 \times 1 0^{- 14} = 4.4 \times 1 0^{- 2} s^{- 1} .$ Answer: $k \approx 0.044 s^{- 1}$ .

Level 3 — Analysis

Exercise 3.1

State: Ek rate constant $k_{1} = 0.001 s^{- 1}$ hai $T_{1} = 300 K$ par aur $k_{2} = 0.010 s^{- 1}$ hai $T_{2} = 350 K$ par. $E_{a}$ nikalo.

Recall Solution

Step 1 — two-point form kyun: humein $A$ nahi pata, isliye wo form use karo jo use cancel kar de. $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) .$ Step 2 — left side: $ln (0.010/0.001) = ln 10 = 2.3026$ . Step 3 — $1/ T$ bracket: $\frac{1}{300} - \frac{1}{350} = 0.0033333 - 0.0028571 = 4.762 \times 1 0^{- 4} K^{- 1} .$ Step 4 — solve karo: $E_{a} = \frac{R l n ( k _{2} / k _{1} )}{\frac{1}{T _{1}} - \frac{1}{T _{2}}} = \frac{8.314 \times 2.3026}{4.762 \times 1 0 ^{- 4}} = 4.02 \times 1 0^{4} J/mol .$ Answer: $E_{a} \approx 40.2 kJ/mol$ — ek moderate barrier, kai organic reactions mein typical.

Exercise 3.2

State: Paanch measurements ko $ln k$ (vertical) vs $1/ T$ (horizontal) plot kiya gaya hai. Best-fit straight line ka slope $- 5800 K$ hai aur intercept $+ 12.6$ hai. $E_{a}$ aur $A$ nikalo.

Recall Solution

$ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ se compare karo: slope $= - E_{a} / R$ , intercept $= ln A$ . Figure dekho — line ki steepness $E_{a}$ carry karti hai, aur $1/ T = 0$ par iska height $A$ carry karta hai.

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

Figure s01 — Arrhenius plot: yellow dots paanch measured $(1/ T, ln k)$ points hain; blue line best fit hai. Red arrow "run" along $1/ T$ mark karta hai aur green arrow "rise" in $ln k$ ; inका ratio slope $= - E_{a} / R = - 5800 K$ hai. Line ko $1/ T = 0$ tak extend karne par intercept $ln A = 12.6$ milta hai. (Agar image load na ho: $ln k$ upar, $1/ T$ across plot karo, straight fit line khicho — iska downward slope $E_{a}$ encode karta hai, left-edge height $A$ encode karta hai.)

Step 1 — activation energy: $E_{a} = - slope \times R = 5800 \times 8.314 = 4.82 \times 1 0^{4} J/mol \approx 48.2 kJ/mol .$ Step 2 — pre-exponential factor: $ln A = 12.6 \Rightarrow A = e^{12.6} = 2.96 \times 1 0^{5} s^{- 1} .$ Answer: $E_{a} \approx 48.2 kJ/mol$ , $A \approx 3.0 \times 1 0^{5} s^{- 1}$ .

Level 4 — Synthesis

Exercise 4.1

State: Us reaction ke liye jiska $E_{a} = 40.2 kJ/mol$ aur $k = 0.001 s^{- 1}$ hai $300 K$ par, $400 K$ par $k$ predict karo.

Recall Solution

Step 1 — two-point form, unknown $k_{2}$ hai, $T_{1} = 300$ , $T_{2} = 400$ ke saath: $ln \frac{k _{2}}{0.001} = \frac{E _{a}}{R} (\frac{1}{300} - \frac{1}{400}) .$ Step 2 — factor $E_{a} / R$ : $\frac{40200}{8.314} = 4835 K$ . Step 3 — bracket: $\frac{1}{300} - \frac{1}{400} = 0.0033333 - 0.0025000 = 8.333 \times 1 0^{- 4}$ . Step 4 — combine karo: $ln (k_{2} /0.001) = 4835 \times 8.333 \times 1 0^{- 4} = 4.029$ . Step 5 — exponentiate karo: $k_{2} /0.001 = e^{4.029} = 56.2$ , isliye $k_{2} = 0.001 \times 56.2 = 0.056 s^{- 1} .$ Answer: $k \approx 0.056 s^{- 1}$ — $+ 100 K$ ke liye $\sim 56 \times$ ka jump. Yahi extreme sensitivity Arrhenius behaviour ka poora point hai.

Exercise 4.2

State: Ek catalyst $E_{a}$ ko $75 kJ/mol$ se $50 kJ/mol$ tak kam kar deta hai. Maan lo $A$ unchanged hai, $298 K$ par $k$ kitne factor se badhta hai?

Recall Solution

Step 1 — ratio kyun: $A$ cancel ho jaata hai kyunki dono ke liye same hai, isliye $\frac{k _{cat}}{k _{uncat}} = \frac{A e ^{- E_{a, cat} / (R T)}}{A e ^{- E_{a, unc} / (R T)}} = e^{(E_{a, unc} - E_{a, cat}) / (R T)} .$ Step 2 — energy gap: $E_{a, unc} - E_{a, cat} = 75000 - 50000 = 25000 J/mol$ . Step 3 — $R T$ se divide karo: $\frac{25000}{8.314 \times 298} = 10.09$ . Step 4 — exponentiate karo: $e^{10.09} = 2.4 \times 1 0^{4}$ . Answer: catalyst reaction ko $\approx 2.4 \times 1 0^{4}$ (lagbhag 24000×) factor se fast kar deta hai. Barrier mein 25 kJ/mol ka dent bahut bada hota hai kyunki ye ek exponential ke andar baitha hai.

Level 5 — Mastery

Exercise 5.1

State: Ek reaction ka rate constant double ho jaata hai jab temperature $300 K$ se $310 K$ tak badhti hai (ye "doubling per 10 K" $Q_{10}$ rule hai, dekho Q10 Temperature Coefficient). $E_{a}$ nikalo.

Recall Solution

Step 1 — set up karo: $k_{2} / k_{1} = 2$ , $T_{1} = 300$ , $T_{2} = 310$ . $ln 2 = \frac{E _{a}}{R} (\frac{1}{300} - \frac{1}{310}) .$ Step 2 — bracket: $\frac{1}{300} - \frac{1}{310} = 0.0033333 - 0.0032258 = 1.0753 \times 1 0^{- 4}$ . Step 3 — solve karo: $E_{a} = \frac{R l n 2}{1.0753 \times 1 0 ^{- 4}} = \frac{8.314 \times 0.6931}{1.0753 \times 1 0 ^{- 4}} = 5.36 \times 1 0^{4} J/mol .$ Answer: $E_{a} \approx 53.6 kJ/mol$ . Toh famous "rate doubles per 10 °C" room temperature ke paas ek specific barrier correspond karta hai — ye koi universal law nahi hai.

Exercise 5.2

State: Do reactions ka $A$ equal hai. Reaction P: $E_{a} = 40 kJ/mol$ . Reaction Q: $E_{a} = 80 kJ/mol$ . Kis temperature par (agar koi ho) dono ka $k$ same hoga? Agar kabhi nahi, explain karo ki kaun hamesha faster hai aur $T$ badhne par dono ke beech gap kaise change hota hai. (Figure use karo.)

Recall Solution

Step 1 — ratio banao (equal $A$ cancel ho jaata hai): $\frac{k _{P}}{k _{Q}} = e^{(E_{a, Q} - E_{a, P}) / (R T)} = e^{40000/ (R T)} .$ Step 2 — equal kab? Humein chahiye hoga $k_{P} / k_{Q} = 1$ , yaani $e^{40000/ (R T)} = 1$ , yaani $40000/ (R T) = 0$ . Iske liye $T \to \infty$ chahiye. Kisi bhi finite $T$ par exponent positive hai, isliye $k_{P} / k_{Q} > 1$ . Answer: ye finite $T$ par kabhi exactly equal nahi hote; P (lower $E_{a}$ ) hamesha faster hai. Step 3 — gap kaise shrink karta hai: jaise jaise $T$ badhta hai, $40000/ (R T)$ 0 ki taraf shrink hota hai, isliye $k_{P} / k_{Q} \to 1$ . Figure dekho: do Arrhenius lines ki different slopes hain (Q ke liye steeper) aur dono sirf $1/ T = 0$ (infinite $T$ ) par milte hain. High $T$ par dono barriers "easy" hote hain, isliye low-barrier advantage fade ho jaata hai.

Figure s02 — do Arrhenius lines equal intercept ke saath ( $ln A = 20$ ): green hai P ( $E_{a} = 40$ kJ/mol, gentler slope) aur red hai Q ( $E_{a} = 80$ kJ/mol, steeper slope). White dashed segment finite temperature par vertical gap dikhata hai — P hamesha Q ke upar rehta hai. Yellow dot sirf ek meeting point mark karta hai, $1/ T = 0$ par (infinite $T$ ). (Agar image load na ho: do straight lines draw karo jo left edge $1/ T = 0$ par saath shuru hoon; steeper wali high- $E_{a}$ reaction hai, aur wo har real temperature par doosri ke neeche rehti hai.)

Shrinking gap ka numeric check: $300 K$ par, $k_{P} / k_{Q} = e^{40000/ (8.314 \cdot 300)} = e^{16.04} = 9.2 \times 1 0^{6}$ ; $1000 K$ par, $k_{P} / k_{Q} = e^{40000/ (8.314 \cdot 1000)} = e^{4.81} = 123$ . Abhi bhi faster hai, lekin margin millions se hundreds par aa gaya.

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

Level 1 — Recognition

Exercise 1.1

Exercise 1.2

Level 2 — Application

Exercise 2.1

Exercise 2.2

Level 3 — Analysis

Exercise 3.1

Exercise 3.2

Level 4 — Synthesis

Exercise 4.1

Exercise 4.2

Level 5 — Mastery

Exercise 5.1

Exercise 5.2