Catalysis — homogeneous, heterogeneous, enzyme catalysis: Exercises

2.8.12 · D4 · Chemistry › Chemical Kinetics › Catalysis — homogeneous, heterogeneous, enzyme catalysis

Un do engines ka ek reminder jo hum baar baar use karenge — dono seedhe parent note se aate hain.

In tools ki roots ke liye Arrhenius Equation aur Activation Energy dekho.

Ek student char substances list karta hai jo reactions mein add ki gayi hain: (a) Haber process mein Mo, (b) $Pt$ pe deposited As, (c) ester ko hydrolyse karta hua aqueous $H^{+}$ , (d) Contact process mein $V_{2} O_{5}$ . Har ek ke liye batao ki wo promoter, poison, homogeneous catalyst, ya heterogeneous catalyst hai.

Recall Solution

(a) Mo — ek promoter hai. Yeh khud catalyst nahi hai; yeh Fe catalyst ki activity boost karta hai.
(b) Pt pe As — ek poison hai. Arsenic platinum ke active sites se chipak jaata hai aur uski catalytic activity barbad kar deta hai.
(c) aqueous $H^{+}$ — ek homogeneous catalyst hai. Ester, paani, aur $H^{+}$ teeno same liquid phase mein hain (yahan Adsorption ki zaroorat nahi — koi surface nahi).
(d) $V_{2} O_{5}$ — ek heterogeneous catalyst hai. Ek solid jo gaseous $S O_{2} + O_{2}$ pe kaam karta hai: alag phase, surface reaction.

L1.2

Batao har ek True hai ya False, ek-line reason ke saath:

Catalyst reaction mein consume ho jaata hai.
Catalyst ek endothermic reaction ko exothermic bana sakta hai.
Ek negative catalyst effective activation energy badhata hai.

Recall Solution

False. Catalyst regenerated hota hai — yeh amount mein unchanged nikal aata hai.
False. Catalyst sirf hill (transition-state peak) ko neeche karta hai, valleys ko kabhi nahi. $Δ H = H_{p r o d} - H_{r e a c t}$ fixed hota hai, isliye exo/endo character kabhi flip nahi hota. ("Hill neeche karo, ghar nahi.")
True. Ek negative catalyst (inhibitor) easy path block kar deta hai, toh molecules ko ek zyada uunche effective $E_{a}$ pe chadhna padta hai → slower.

Level 2 — Application

L2.1

Ek catalyst ek reaction ki activation energy $E_{a, u n} = 95 kJ/mol$ se ghatakar $E_{a, c a t} = 60 kJ/mol$ kar deta hai $T = 310 K$ pe. Rate constant kitne factor se increase hota hai?

Recall Solution

Hum kya karenge: speed-up ratio use karenge, kyunki yeh catalyst ke effect ko isolate karta hai (same $A$ , same $T$ cancel ho jaate hain). Step 1 — $Δ E_{a}$ joules mein nikalo. $Δ E_{a} = E_{a, u n} - E_{a, c a t} = 95 - 60 = 35 kJ/mol = 35000 J/mol$ . Step 2 — $R T$ compute karo (woh "thermal energy scale" jiske against hum hill measure karte hain): $R T = 8.314 \times 310 = 2577.3 J/mol .$ Step 3 — divide karo ( $R T$ ki units mein drop): $\frac{Δ E _{a}}{R T} = \frac{35000}{2577.3} = 13.58.$ Step 4 — exponentiate karo: $\frac{k _{c a t}}{k _{u n}} = e^{13.58} \approx 7.9 \times 1 0^{5} .$ Reaction lagbhag 790,000× faster chalti hai — 35 kJ ka cut, exponential se magnify hokar.

L2.2

Ek enzyme reaction ke liye, $V_{ma x} = 4.0 μ M s^{- 1}$ aur $K_{M} = 2.0 mM$ hai. Rate $v$ nikalo jab substrate concentration hai (a) $[S] = 2.0 mM$ , (b) $[S] = 18.0 mM$ .

Recall Solution

Tool: parent note se Michaelis–Menten law, $v = \frac{V _{ma x} [ S ]}{K _{M} + [ S ]} .$ (a) $[S] = K_{M} = 2.0 mM$ . Jab substrate $K_{M}$ ke barabar ho, denominator $K_{M} + K_{M} = 2 K_{M}$ hota hai, toh $v = \frac{V _{ma x} K _{M}}{2 K _{M}} = \frac{V _{ma x}}{2} = 2.0 μ M s^{- 1} .$ Yahi reason hai ki $K_{M}$ ko "woh $[S]$ jo $V_{ma x}$ ka aadha deta hai" define kiya jaata hai. (b) $[S] = 18.0 mM$ . $v = \frac{4.0 \times 18.0}{2.0 + 18.0} = \frac{72.0}{20.0} = 3.6 μ M s^{- 1} .$ Gaur karo hum already $V_{ma x}$ ke $90%$ pe hain — saturation approach kar rahe hain.

Level 3 — Analysis

L3.1

Neeche diya energy profile (Figure s01) ek reaction ko catalyst ke saath aur bina catalyst ke dikhata hai. Jawab do:

Kaun sa arrow $Δ H$ hai, aur kya catalyst use change karta hai?
Catalyst ne forward $E_{a}$ kitna lower kiya? Backward $E_{a}$ ?
Apne (2) ke answer ka use karke argue karo ki $K_{e q}$ unchanged hai.

Figure — Catalysis — homogeneous, heterogeneous, enzyme catalysis

Recall Solution

(1) $Δ H$ reactant valley aur product valley ke beech ka vertical gap hai (pink arrow). Catalyst sirf peak ko neeche kheenchta hai; valleys apni jagah rehti hain, isliye $Δ H$ unchanged hai. (2) Forward $E_{a}$ = (peak) − (reactant level). Uncatalysed peak $100 kJ/mol$ pe hai, catalysed peak $60 kJ/mol$ pe, reactants $0$ pe. Toh forward $E_{a}$ $100$ se $60$ tak girta hai: $40 kJ/mol$ ki fall. Product valley $- 20 kJ/mol$ pe hai. Backward $E_{a}$ = (peak) − (product level):

uncatalysed: $100 - (- 20) = 120 kJ/mol$ ,
catalysed: $60 - (- 20) = 80 kJ/mol$ . Backward $E_{a}$ bhi $40 kJ/mol$ girta hai — utni hi maatra. (3) Kyunki $k_{f}$ aur $k_{b}$ dono identical factor $e^{40000/ R T}$ se scale hote hain, $K_{e q} = \frac{k _{f}}{k _{b}} \Rightarrow K_{e q}^{c a t} = \frac{k _{f} \cdot e ^{40000/ R T}}{k _{b} \cdot e ^{40000/ R T}} = \frac{k _{f}}{k _{b}} = K_{e q} .$ Equal factors cancel ho jaate hain. Chemical Equilibrium dekho.

L3.2

Do catalysts ko $T = 300 K$ pe test kiya jaata hai. Catalyst X deta hai $E_{a} = 50 kJ/mol$ ; catalyst Y deta hai $E_{a} = 45 kJ/mol$ . Y, X se kitni baar faster hai? Comment karo ki kya 5 kJ ka difference "bada" hai.

Recall Solution

Dono ek hi uncatalysed baseline share karte hain, toh inhe directly compare karo: $\frac{k _{Y}}{k _{X}} = e^{(50000 - 45000) / (8.314 \times 300)} = e^{5000/2494.2} = e^{2.005} \approx 7.4.$ Sirf $5 kJ/mol$ Y ko lagbhag 7.4× faster bana deta hai. Kyunki $E_{a}$ exponential ke andar hai, hill height mein "chote" differences bhi bade speed differences mein amplify ho jaate hain — isliye catalyst screening har kJ ke peeche bhaagti hai.

Level 4 — Synthesis

L4.1

Ek reaction mein $A = 1.0 \times 1 0^{13} s^{- 1}$ hai. Uncatalysed $E_{a} = 90 kJ/mol$ ; ek catalyst $E_{a} = 55 kJ/mol$ deta hai. $T = 298 K$ pe:

$k_{u n}$ aur $k_{c a t}$ compute karo (units $s^{- 1}$ ).
Agar ek batch uncatalysed mein 5 ghante leta hai, toh catalysed time estimate karo.

Recall Solution

$R T$ set up karo: $R T = 8.314 \times 298 = 2477.6 J/mol$ . (1) Uncatalysed: $k_{u n} = 1 0^{13} e^{- 90000/2477.6} = 1 0^{13} e^{- 36.32} = 1 0^{13} \times 1.68 \times 1 0^{- 16} \approx 1.68 \times 1 0^{- 3} s^{- 1} .$ Catalysed: $k_{c a t} = 1 0^{13} e^{- 55000/2477.6} = 1 0^{13} e^{- 22.20} = 1 0^{13} \times 2.27 \times 1 0^{- 10} \approx 2.27 \times 1 0^{3} s^{- 1} .$ (2) Ek fixed reaction ke liye, ek diye gaye conversion tak pahunchne ka time $k$ ke inversely proportional hota hai (faster $k$ → shorter time). Toh $t_{c a t} = t_{u n} \times \frac{k _{u n}}{k _{c a t}} = 5 h \times \frac{1.68 \times 1 0 ^{- 3}}{2.27 \times 1 0 ^{3}} = 5 \times 7.40 \times 1 0^{- 7} h .$ Yeh $3.70 \times 1 0^{- 6} h \approx 0.0133 s$ hai. Paanch ghante ka kaam lagbhag 13 milliseconds mein ho jaata hai — $35 kJ$ cut ki taakat. (Yahan $t \propto 1/ k$ kyun hold karta hai fixed order ke liye, Rate Law and Order of Reaction dekho.)

L4.2

Ek enzyme Michaelis–Menten follow karta hai $K_{M} = 5.0 mM$ ke saath. Kaun se substrate concentration $[S]$ pe rate $V_{ma x}$ ke $80%$ tak pahunchti hai?

Recall Solution

Set up: hum chahte hain $v = 0.80 V_{ma x}$ . Law mein plug karo aur $V_{ma x}$ cancel karo: $0.80 = \frac{[ S ]}{K _{M} + [ S ]} .$ Solve karo (cross-multiply): $0.80 (K_{M} + [S]) = [S] \Rightarrow 0.80 K_{M} = [S] (1 - 0.80) = 0.20 [S] .$ $[S] = \frac{0.80}{0.20} K_{M} = 4 K_{M} = 4 \times 5.0 = 20.0 mM .$ $50%$ se (at $[S] = K_{M}$ ) $80%$ tak jaane ke liye chaar baar $K_{M}$ lagta hai — curve flatten ho jaata hai, isliye $V_{ma x}$ ka har extra percent disproportionately zyada substrate maangta hai. Yeh saturation action mein hai; Enzymes and Proteins (Biomolecules) dekho.

Level 5 — Mastery

L5.1

$T = 350 K$ pe ek reaction $5.0 \times 1 0^{4}$ factor se speed up hoti hai jab ek catalyst add kiya jaata hai. Maan lo prefactor $A$ unchanged hai, activation energy mein drop $Δ E_{a}$ (in $kJ/mol$ ) nikalo.

Recall Solution

Kaun sa tool aur kyun? Hum ratio jaante hain aur energy drop chahiye — toh speed-up formula ko invert karo. Exponential ka inverse natural logarithm $ln$ hai (yeh jawab deta hai " $e$ kis power pe yeh ratio deta hai?"). Start: $\frac{k _{c a t}}{k _{u n}} = e^{Δ E_{a} / R T} \Rightarrow ln (\frac{k _{c a t}}{k _{u n}}) = \frac{Δ E _{a}}{R T} .$ Log lo: $ln (5.0 \times 1 0^{4}) = 10.82$ . $R T$ se multiply karo: $R T = 8.314 \times 350 = 2909.9 J/mol$ , toh $Δ E_{a} = 10.82 \times 2909.9 = 3.148 \times 1 0^{4} J/mol \approx 31.5 kJ/mol .$ $50, 000 \times$ boost ke liye sirf $\sim 31.5 kJ/mol$ ka cut kafi tha — phir se exponential heavy lifting kar raha hai.

L5.2

Ek enzyme ka $k_{2} = 300 s^{- 1}$ (turnover step) aur total enzyme $[E]_{0} = 0.010 mM$ hai, saath mein $K_{M} = 4.0 mM$ hai.

$V_{ma x}$ nikalo.
$[S] = 1.0 mM$ pe $v$ nikalo.
Woh $V_{ma x}$ ka kitna fraction hai, aur yeh site occupancy ke baare mein kya batata hai?

Recall Solution

(1) Parent ki definition $V_{ma x} = k_{2} [E]_{0}$ se: $V_{ma x} = 300 s^{- 1} \times 0.010 mM = 3.0 mM s^{- 1} .$ (2) Michaelis–Menten apply karo: $v = \frac{V _{ma x} [ S ]}{K _{M} + [ S ]} = \frac{3.0 \times 1.0}{4.0 + 1.0} = \frac{3.0}{5.0} = 0.60 mM s^{- 1} .$ (3) Fraction $= v / V_{ma x} = 0.60/3.0 = 0.20$ , yaani $V_{ma x}$ ka $20%$ . Kyunki fraction $[S] / (K_{M} + [S])$ bhi enzyme ka woh fraction hai jo $E S$ ke roop mein bound hai, kisi bhi instant mein sirf lagbhag $20%$ active sites occupied hain — hum curve ke low- $[S]$ , near-first-order part pe hain, saturation se bahut door. Figure s02 se compare karo.

Recall Master checklist (answers cover karo)

$E_{a}$ ko joules mein convert kiya $R T$ se divide karne se pehle? ::: Hamesha — $R$ J mein hai.
$E_{a}$ ko $e^{E_{a} / R T}$ se bahar nikalane ka inverse operation? ::: Natural logarithm $ln$ .
Time vs rate constant relationship (fixed order)? ::: $t \propto 1/ k$ — bada $k$ , shorter time.
$[S]$ jo half $V_{ma x}$ deta hai? ::: $[S] = K_{M}$ .
$[S]$ jo $V_{ma x}$ ka $80%$ deta hai? ::: $[S] = 4 K_{M}$ .
Kya catalyst energy diagram ki do valleys change karta hai? ::: Nahi — sirf peak.

Exercises — Catalysis — homogeneous, heterogeneous, enzyme catalysis

Level 1 — Recognition

L1.1

L1.2

Level 2 — Application

L2.1

L2.2

Level 3 — Analysis

L3.1

L3.2

Level 4 — Synthesis

L4.1

L4.2

Level 5 — Mastery

L5.1

L5.2