Catalysis — homogeneous, heterogeneous, enzyme catalysis: Question bank

2.8.12 · D5 · Chemistry › Chemical Kinetics › Catalysis — homogeneous, heterogeneous, enzyme catalysis

Definition Is page par use hone wale har symbol ki definition (pehle yeh padho)

$E_{a}$ = activation energy — energy ki hill ki height jitni ek reaction ko climb karni padti hai.
$Δ H$ = enthalpy change — product valley aur reactant valley ke beech ka height difference ( $H_{p r o d} - H_{r e a c t}$ ).
$Δ G$ = Gibbs free-energy change — thermodynamic quantity jiska sign spontaneity decide karta hai: $Δ G < 0$ matlab reaction apne aap chal sakti hai. $Δ H$ ki tarah, yeh sirf valleys pe depend karta hai, hill pe nahi.
$k$ = ek rate constant — $rate = k [\dots]$ mein woh number jo batata hai ki ek given reaction kitni fast chalti hai. Do special ones:
- $k_{f}$ = forward rate constant, $k_{b}$ = backward rate constant, aur $K_{e q} = k_{f} / k_{b}$ equilibrium constant hai jo batata hai ki reaction kahan settle hogi.
Jab hum usi reaction ko catalyst ke saath aur bina catalyst ke compare karte hain to uske forward $k$ ko do tarike se tag karte hain: $k_{c a t}$ = rate constant catalyst ke saath, $k_{u n}$ = rate constant bina catalyst ke (uncatalysed). To $k_{c a t}$ aur $k_{u n}$ bas wahi same $k$ hai jo do alag roads pe measure ki gayi hai.
$T$ = temperature, $R$ = gas constant, $A$ = $k = A e^{- E_{a} / R T}$ mein pre-exponential (frequency) factor — yeh count karta hai ki molecules kitni baar sahi orientation ke saath collide karte hain.
Enzyme symbols: $[S]$ = substrate concentration; $[E]_{0}$ = total enzyme concentration; $v$ = reaction rate (velocity); $k_{2}$ = step $E S \to E + P$ ka rate constant; $V_{ma x} = k_{2} [E]_{0}$ = maximum possible rate jab har enzyme busy ho.

True or false — justify

Ek catalyst us reaction mein consume ho jaata hai jise woh speed up karta hai.

False. Woh regenerate hota hai — mechanism mein enter karta hai aur wahan se nikalta hai, isliye end mein uski total amount unchanged rehti hai, chahe midway mein woh chemically active raha ho.

Ek catalyst sirf forward reaction ki activation energy ko lower karta hai.

False. Woh hill ko hi lower karta hai, jo dono directions ke liye shared hai; backward

E_{a}

exactly utni hi amount se girta hai, aur yahi precise wajah hai ki

K_{e q}

fixed rehta hai.

Ek catalyst kisi non-spontaneous (

Δ G > 0

) reaction ko spontaneous bana deta hai.

False. Spontaneity

Δ G

se set hoti hai, jo ek thermodynamic quantity hai aur valleys se fix hoti hai; catalyst sirf hill (kinetics) ko change karta hai, isliye ek "nahi chalegi" reaction ab bhi nahi chalegi — bas faster nowhere.

Ek negative catalyst (inhibitor) reverse reaction ko speed up karta hai.

False. Yeh reaction ko slow karta hai path ke effective

E_{a}

ko block ya raise karke; yeh selectively reverse direction ko accelerate nahi karta — yeh dono taraf hill ke paar jaane mein hindrance karta hai.

Catalyst add karne se equilibrium par product ki yield increase hoti hai.

False. Equilibrium par yield

K_{e q}

se govern hoti hai, jise catalyst touch nahi kar sakta; tumhe wahi final amount milti hai, bas jaldi milti hai.

Heterogeneous catalysis mein reactant pressure ko double karne se rate hamesha double hoti hai.

False. Jab surface ke saare active sites occupied ho jaate hain to rate saturate ho jaati hai; us point ke baad extra reactant bas apni baari ka wait karta hai, isliye rate plateau kar jaati hai, double nahi hoti.

Enzymes kisi bhi temperature par equally well kaam karte hain kyunki catalysts unchanged rehte hain.

False. Enzymes proteins hain; temperature ko unke optimum se zyada badhane par woh denature ho jaate hain (folded shape unravel ho jaata hai), active site destroy ho jaati hai aur catalysis bhi khatam ho jaati hai.

Ek promoter reaction ke liye khud ek catalyst hai.

False. Ek promoter (jaise Haber mein Mo) ka akele koi ya bahut kam catalytic action hota hai; woh true catalyst ki activity ko enhance karta hai, isliye woh ek helper hai, catalyst nahi.

Michaelis constant

K_{M}

ke units rate ke units jaisi hain.

False.

K_{M}

substrate concentration

[S]

ke barabar hoti hai jis par

v = V_{ma x} /2

hoti hai, isliye iske units concentration ke hain (jaise mol/L), rate ke nahi.

Spot the error

"Reaction

2 S O_{2} + O_{2} V_{2} O_{5} 2 S O_{3}

homogeneous catalysis hai kyunki equation mein sab kuch gas region mein hai."

Error:

V_{2} O_{5}

ek solid hai jabki reactants gases hain — alag phases — isliye yeh surface par heterogeneous catalysis hai.

"Lead chamber mein NO jo

2 S O_{2} + O_{2} \to 2 S O_{3}

catalyse karta hai, woh heterogeneous hai kyunki kaafi saare substances hain."

Error: "alag substances" ka matlab "alag phases" nahi hota. NO(g) aur gas reactants sab same (gas) phase mein hain, isliye yeh homogeneous catalysis hai.

"Kyunki catalyst

E_{a}

ko lower karta hai, energy diagram mein product valley neeche khich jaati hai."

Error: Sirf peak (transition state) lower hoti hai. Reactant aur product valleys fixed rehti hain, isliye

Δ H = H_{p r o d} - H_{r e a c t}

unchanged rehta hai — hill girta hai, ghar nahi.

E + S ⇌ E S \to E + P

mein, enzyme

E

product banane mein use up ho jaata hai."

Error:

E

final step ke right side par phir se aata hai; woh regenerate hota hai aur agle substrate ko bind karne ke liye free rehta hai, yahi wajah hai ki enzyme ki thodi si matra substrate ki vast amounts process kar leti hai.

"Ek enzyme reactions ki ek badi range catalyse karta hai kyunki woh itna achha catalyst hai."

Error: Enzymes apni specificity ke liye famous hain — lock-and-key active site essentially ek hi substrate shape ke saath fit hoti hai, isliye ek enzyme roughly ek reaction drive karta hai.

"Kyunki

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

hai,

E_{a}

ko half karne se sirf exponent half hota hai, isliye rate barely change hoti hai."

Error:

E_{a}

ek negative exponential ke andar hota hai, isliye exponent mein change

k

mein exponentially amplify hota hai — ek modest

E_{a}

cut rate ko hazaaron-guna multiply kar sakta hai.

"Poison aur inhibitor ek hi cheez hain."

Error (subtle): ek poison ek catalyst ki activity destroy karta hai (jaise Pt par As), jabki ek negative catalyst/inhibitor directly reaction ko slow karta hai; woh alag targets par act karte hain, chahe dono observed rate ko reduce karte hain.

Why questions

Speed-up ratio

k_{c a t} / k_{u n}

derive karte waqt dono

k_{c a t}

(catalysed forward

k

) aur

k_{u n}

(uncatalysed forward

k

) ke liye same pre-exponential factor

A

kyun use kiya jaata hai?

Kyunki

A

collision frequency aur orientation count karta hai, jo temperature aur molecular geometry se set hote hain — hill ki height se nahi;

A

ko fixed rakhne par

E_{a}

single difference reh jaata hai, isliye division par

A

's cancel ho jaate hain aur catalyst ka true effect

e^{(E_{a, u n} - E_{a, c a t}) / R T}

ke roop mein isolate ho jaata hai.

Kya yeh exactly sach hai ki

A

catalyst se unchanged rehta hai?

Perfectly nahi — ek catalyst collision geometry ko thoda alter kar sakta hai — lekin

A

mein change

E_{a}

ko lower karne se hone wale exponential change ke comparison mein tiny hota hai, isliye

A

ko equal treat karna ek excellent, deliberately simplifying, approximation hai.

Solid catalyst par adsorption activation energy ko kyun lower karta hai?

Jab ek molecule active site par adsorb hota hai, uske bonds stretch aur weaken ho jaate hain, isliye unhe break karne ke liye kam extra energy chahiye — hill effectively surface par hi lower ho jaati hai.

High substrate concentration

[S]

par enzyme rate curve

V_{ma x}

tak kyun flatten hoti hai?

High

[S]

par practically har enzyme

E S

complex mein locked hoti hai; zyada substrate add karne se zyada enzyme nahi banta, isliye throughput ceiling

V_{ma x} = k_{2} [E]_{0}

par hit karta hai (

k_{2}

E S \to E + P

rate constant,

[E]_{0}

= total enzyme).

Ek catalyst second law of thermodynamics ko kabhi violate kyun nahi kar sakta?

Yeh sirf equilibrium tak pahunchne ki speed badhata hai,

K_{e q}

Δ G

shift kiye bina; yeh route change karta hai, destination nahi, isliye koi free energy create nahi hoti.

Heterogeneous catalysis mein desorption sirf ek nicety nahi, balki ek necessary step kyun hai?

Agar products kabhi active site nahi chhodte, to site blocked rehti; desorption site ko free karta hai, catalyst ko regenerate karta hai taaki woh next reactant molecule process kar sake.

Ek precise sense mein catalyst forward aur backward reactions ko "equally" speed up kyun karta hai?

Dono directions same lowered peak cross karti hain; forward

E_{a}

mein drop, reverse

E_{a}

mein drop ke barabar hai, isliye dono rate constants (

k_{f}

forward,

k_{b}

backward) same factor se multiply hote hain aur unka ratio

K_{e q} = k_{f} / k_{b}

untouched rehta hai.

Enzymes ke liye steady-state assumption

d [E S] / d t \approx 0

reasonable kyun hai?

E S

ek fleeting intermediate hai jo almost utni hi fast form aur consume hota hai jitni fast woh appear hota hai, isliye interest ke timescale par uski concentration barely change hoti hai, jisse formation rate breakdown rate ke barabar ho jaati hai.

Edge cases

Jab

T \to \infty

ho to speed-up ratio

k_{c a t} / k_{u n}

ka kya hoga?

Exponent

(E_{a, u n} - E_{a, c a t}) / R T \to 0

ho jaata hai, isliye ratio

\to e^{0} = 1

ho jaata hai: bahut high temperature par thermal energy

E_{a}

difference ko dwarf kar deti hai, aur catalyst ka relative advantage shrink hokar zero ke paas aa jaata hai.

Jab

[S]

(substrate concentration)

K_{M}

ke comparison mein bahut chhoti ho to

[S]

mein reaction order kya hoga?

Jab

[S] ≪ K_{M}

, denominator

K_{M} + [S] \approx K_{M}

ho jaata hai, isliye

v \approx (V_{ma x} / K_{M}) [S]

— rate substrate mein first order hai.

Jab

[S]

K_{M}

ke comparison mein bahut badi ho to

[S]

mein reaction order kya hoga?

Jab

[S] ≫ K_{M}

v \to V_{ma x}

, ek constant jo

[S]

se independent hai — reaction substrate mein zero order hai (saturation).

Agar ek "catalyst" final products mein us amount mein appear kare jo start se alag hai, kya woh catalyst hai?

Nahi — regeneration (utni hi amount out jitni in) definition ka part hai; agar amount change hoti hai to woh ek reactant tha ya consume hua, true catalyst nahi.

Ek reaction jo already equilibrium par hai, usme catalyst add karne par kya hota hai?

Kuch bhi observable nahi hota composition mein: forward aur reverse rates equally badhti hain aur equal rehti hain, isliye system exactly usi equilibrium position par rehta hai.

Ek enzyme ko strongly acidic solution mein rakha jaata hai jo uske optimum pH se bahut door hai — uske behaviour ka prediction karo.

Extreme pH protein ki folded structure ko disrupt karta hai (denaturation), active site deform ho jaati hai, isliye catalytic activity collapse ho jaati hai, chahe enzymes otherwise "unchanged" catalysts hain.

Ek solid catalyst ko bahut fine powder mein grind kiya jaata hai — kya change hota hai aur kya nahi?

Surface area badhta hai, isliye zyada active sites hote hain aur rate badhti hai; lekin

Δ H

K_{e q}

, aur final equilibrium position unchanged rehte hain — zyada sites ka matlab faster hai, further nahi.

Recall Har trap ki ek-line summary

Ek catalyst hill (kinetics) ko change karta hai, kabhi valleys (thermodynamics) ko nahi: woh dono directions ke liye $E_{a}$ lower karta hai, regenerate hota hai, $K_{e q}$ / $Δ G$ / $Δ H$ shift nahi kar sakta, aur — surfaces par ya enzymes mein — finite active sites se cap hota hai, jisse saturation milti hai.