Activation energy from Arrhenius plot; effect of catalyst: Question bank

2.8.8 · D5 · Chemistry › Chemical Kinetics › Activation energy from Arrhenius plot; effect of catalyst

Yeh parent topic ka "gotcha" deck hai. Neeche har item ek aisa tarika target karta hai jisme students Arrhenius equation ya catalyst ke energy diagram ko galat padhte hain. Prompt padho, apna jawab ek reason ke saath zor se bolo, phir reveal karo.

Ab neeche ki picture ko apna reference maano har "slope" aur "intercept" wale item ke liye.

Figure — Activation energy from Arrhenius plot; effect of catalyst

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$\frac{1}{T}$ x-axis par hai, $T$ nahin. Daayein jaana → $1/ T$ badhta hai → temperature girti hai.
Slope hai $- \frac{E _{a}}{R}$ — yeh negative hai, aur zyada steep neeche matlab bada $E_{a}$ .
Catalyst $E_{a}$ badalta hai (kinetics), kabhi $Δ H$ ya $Δ G$ nahin (thermodynamics).

True or false — justify karo

Har prompt ek statement hai. "True" ya "false" aur ek-sentence ka reason do.

Zyada activation energy hone se Arrhenius line ka slope upar ki taraf zyada steep hota hai.

False — slope

- E_{a} / R

hai, toh bada

E_{a}

line ko

1/ T

badhne par zyada steeply neeche giraaता hai; line down-right jaati hai, kabhi upar nahin.

Arrhenius plot par daayein jaane ka matlab reaction garam ho rahi hai.

False — x-axis

1/ T

hai, isliye daayein jaana matlab

1/ T

badh raha hai, yaani temperature ghat rahi hai aur reaction thandi ho rahi hai.

Catalyst sirf forward reaction ki activation energy ko kam karta hai.

False — catalyst ek nayi pathway deta hai jisme lower peak hota hai jise dono directions ko cross karna padta hai, isliye dono forward aur reverse

E_{a}

kam hote hain, aur yahi wajah hai ki

K_{e q}

unchanged rehta hai.

Ek hi

E_{a}

wali do reactions ka rate constant

k

same hoga.

False —

k = A e^{- E_{a} / R T}

pre-exponential factor

A

par bhi depend karta hai; plot par equal slopes alag heights par ho sakti hain (alag intercepts

ln A

Infinite temperature par, rate constant pre-exponential factor

A

ke kareeb pahunch jaata hai.

True — jab

T \to \infty

E_{a} / R T \to 0

isliye

e^{- E_{a} / R T} \to 1

aur

k \to A

; plot par yeh

1/ T \to 0

hai, y-intercept

ln A

Catalyst milane se equilibrium zyada product ki taraf shift ho jaata hai.

False — catalyst forward aur reverse dono ko equally speed karta hai, isliye equilibrium jaldi pohonchta hai lekin same jagah rehta hai;

K_{e q}

sirf

Δ G

par depend karta hai, jise catalyst kabhi nahin chhuta.

A

ki units hamesha

s^{- 1}

hoti hain.

False —

A

ke wahi units hote hain jo $k$ ke hote hain, jo reaction order par depend karte hain; sirf first-order reaction ke liye yeh

s^{- 1}

hota hai.

Agar kisi reaction ka

E_{a} = 0

hai, toh uska rate constant temperature par depend nahin karta.

True —

E_{a} = 0

hone par exponential sabhi

T

ke liye

e^{0} = 1

hai, isliye

k = A

flat rehta hai; Arrhenius line horizontal ho jaati hai (zero slope).

Temperature double karne se rate constant hamesha double ho jaata hai.

False —

k

T

par dependence

e^{- E_{a} / R T}

ke zariye hai, jo exponential aur non-linear hai, isliye

T

double karne par

k

kaafi zyada (ya, saturation ke paas, kaafi kam) se 2 se multiply ho sakta hai.

Catalyzed reaction ka Arrhenius plot par y-intercept chhota hota hai.

False (zyaadatar) — y-intercept

ln A

hai; yeh maante hue ki

A^{'} \approx A

, intercept lagbhag unchanged rehta hai, lekin har real temperature par catalyzed line upar hoti hai kyunki uska slope kam steep hota hai.

Spot the error

Har line mein ek galat statement hai. Correction reveal karo.

E_{a} = m \times R

jahan

m

slope hai."

Slope negative hai (

m = - E_{a} / R

), isliye

E_{a} = - m \times R

; minus sign bhool jaane par negative, physically impossible activation energy milti hai.

"Kyunki

ln k = ln A - E_{a} / R T

ln k

T

ke against plot karne par seedhi line milti hai."

Linear term

E_{a} / (R T)

hai, jo

1/ T

mein linear hai,

T

mein nahin; seedhi line paane ke liye

ln k

1/ T

ke against plot karna zaroori hai.

"Catalyst energy barrier kam karke endothermic reaction ko exothermic bana deta hai."

Barrier kam karna sirf peak ki unchai (

E_{a}

) badalta hai; reactant-to-product energy gap

Δ H

fixed rehta hai, isliye endothermic reaction endothermic hi rehti hai.

"Arrhenius plot par catalyzed aur uncatalyzed lines parallel hoti hain."

Woh parallel hoti hain sirf tab jab dono slopes equal hon, yaani

E_{a, c a t} = E_{a, u n c a t}

— lekin catalyst

E_{a}

kam karta hai, isliye catalyzed slope

- E_{a, c a t} / R

kam steep hota hai aur lines parallel nahin hoti; woh cross karti hain ya diverge hoti hain.

"Kyunki catalyst barrier kam karne ke liye consume ho jaata hai, tumhe zyada dalte rehna padta hai."

Catalyst har cycle mein regenerate hota hai aur overall consume nahin hota, isliye thodi si maatra bahut saari reactant ko turn over kar sakti hai.

"Bada

A

matlab reaction zyada temperature-sensitive hai."

Temperature sensitivity

E_{a}

se set hoti hai (woh exponent mein hota hai);

A

sirf

k

ko uniformly upar-neeche scale karta hai aur intercept shift karta hai, slope nahin.

"Hum Arrhenius equation ka

ln

isliye lete hain kyunki logs numbers ko chhota bana dete hain."

Hum

ln

isliye lete hain taaki exponential ko linear relationship (

y = m x + c

) mein badal sakein aur slope seedhi line se padha ja sake — point shape ka hai, size ka nahin.

Why questions

Sirf label nahin, reasoning do.

Successful collisions ka fraction temperature ke saath linearly ki jagah exponentially kyun badhta hai?

Kyunki Maxwell–Boltzmann distribution un molecules ka fraction jo

E_{a}

se upar hain use exponential tail

e^{- E_{a} / R T}

mein daalti hai, isliye

T

mein chhoti shift bahut saare molecules ko barrier ke paas se bahar dhakeel deti hai.

Modest 25 kJ/mol ki

E_{a}

drop se tens of thousands ka rate boost kyun hota hai?

Rate ratio

e^{- Δ E_{a} / R T}

hai; 300 K par,

R T \approx 2.5

kJ/mol, isliye

Δ E_{a}

lagbhag das "

R T

" units hai aur

e^{10} \approx 22000

— exponential chhote energy changes ko enormously amplify karta hai.

Kisi bhi real reaction ke liye Arrhenius plot ka slope negative kyun hona chahiye?

Real activation energies positive hoti hain (

E_{a} > 0

), aur slope

- E_{a} / R

ke barabar hai; positive

E_{a}

ko positive

R

se divide karne par slope negative aata hai, yaani

k

girti hai jab

1/ T

badhta hai (jab

T

girta hai).

Catalyst

Δ G

change kiye bina reaction tez kyun kar sakta hai?

Δ G

reactants aur products ke beech ki height ka difference hai; catalyst sirf unke beech ke transition-state peak ko neeche laata hai, jo ek kinetics quantity hai, aur dono endpoint energies (aur isliye

Δ G

K_{e q}

) untouched rehti hain.

Slope estimate karne ke liye do adjacent points ki jagah data ke extremes par do points kyun plot karte hain?

Extreme points

1/ T

mein sabse bada range span karte hain, isliye same measurement scatter se "rise over run" slope mein proportionally kam error aata hai (lekin sabhi points par best-fit line aur bhi behtar hai).

Temperature badhane se collision frequency thodi hi badhti hai lekin rate bahut zyada kyun?

Collision frequency molecular speed ke saath scale hoti hai, jo

T

jaisi hai (weak dependence), jabki successful-collision fraction

e^{- E_{a} / R T}

T

mein exponential hai aur rate change par dominate karta hai.

Edge cases

Boundary aur degenerate scenarios jo topic quietly assume kar leta hai.

Agar ek reaction ek temperature range mein alag-alag

E_{a}

wale do mechanisms se chalti ho toh Arrhenius plot kaisa dikhega?

Plot curve (bend) karta hai seedha rehne ki jagah, kyunki lower-

E_{a}

path low

T

par dominate karta hai aur higher-

E_{a}

path high

T

par, aur dono apna alag slope contribute karte hain.

Agar temperature

T \to 0

K ho toh

k

ka kya hoga?

E_{a} / R T \to \infty

, isliye

e^{- E_{a} / R T} \to 0

aur

k \to 0

— reaction effectively ruk jaati hai kyunki kisi bhi molecule ke paas enough energy nahin hoti; plot par

1/ T \to \infty

ln k \to - \infty

bhejet hai.

Kya catalyst thermodynamically unfavourable (

Δ G > 0

) reaction ko completion tak le ja sakta hai?

Nahin — catalyst sirf equilibrium tak pahonchne ko accelerate karta hai; agar

Δ G > 0

toh equilibrium ab bhi reactants favor karta hai, isliye utni hi chhoti amount product milti hai, bas jaldi.

Agar measured

ln A

(y-intercept) negative aaye — kya yeh error hai?

Zaroor nahin —

ln A < 0

ka matlab sirf yeh hai ki

A

apni units mein

1

se chhota hai, jo bilkul theek hai; iska matlab kuch galat nahin hai, sirf pre-exponential factor ek chhota number hai.

Enzyme ke liye substrate saturation ke paas Arrhenius plot seedha rehta hai?

Aksar nahin — saturation par observed rate turnover steps se limited hoti hai (aur high

T

par enzymes denature hote hain), isliye plot flat ho sakta hai ya neeche bend kar sakta hai, simple single-barrier assumption ko violate karta hai.

Agar catalyzed aur uncatalyzed paths ke pre-exponential factors alag hon (

A^{'} \neq = A

), kya catalyzed reaction kabhi slower ho sakti hai?

Principle mein haan kuch temperatures par — agar

A^{'}

bahut chhota hai

A

se, intercept penalty ek narrow

T

window mein lower-

E_{a}

advantage ko outweigh kar sakti hai, lekin genuine catalysts ke liye barrier drop normally jeet jaata hai.