2.8.12 · D3Catalysis — homogeneous, heterogeneous, enzyme catalysis
Yeh page ek drill ground hai. Parent note ne ideas build kiye; yahan hum har tarah ke numbers ko tackle karte hain jo yeh topic throw kar sakta hai — har sign, har extreme, degenerate cases, ek word problem, aur ek exam twist. Padhne se pehle har Forecast line pe apna guess lagao.
Yahan sab kuch un do tools pe tika hai jo tum pehle se jaante ho:
Arrhenius Equation k = A e − E a / R T Activation Energy E a Michaelis–Menten rate law v = K M + [ S ] V ma x [ S ] Rate Law and Order of Reaction logic se aata hai aur Enzymes and Proteins (Biomolecules) pe apply hota hai.
Shuru karne se pehle kuch symbols seedhe kar lete hain:
Definition Wo letters jo hum use karenge
E a activation energy = energy hill ki height (J/mol kJ/mol A Arrhenius Equation mein pre-exponential factor (ya "frequency factor") — yeh roughly batata hai ki molecules kitni baar hill cross karne ki koshish karte hain (collision frequency × orientation). Yeh exponential ko multiply karta hai. Jab hum ek catalysed aur uncatalysed path ko same conditions pe compare karte hain toh A same rakhte hain taaki sirf E a A A f , A b forward aur backward directions ke liye pre-exponential factors (Ex 7 mein use honge).R = 8.314 J mol − 1 K − 1 T kelvin mein (in formulas mein kabhi °C nahi).R T E a / R T "hill kitne budgets unchi hai?" — yeh ek pure number hai jisme koi units nahi.[ S ] V ma x K M [ S ] V ma x aadha deta hai.k a f t er ke saath aur k b e f or e ke bina ke liye, aur yahi before/after naming poore note mein use karte hain.
Har catalysis calculation jo tumse pucha ja sakta hai in cells mein se ek hai. Neeche ke examples us cell ke saath tagged hain jo wo cover karta hai.
Cell
Kya vary karta hai / trap kya hai
Example
A. Sign check Kya E a neeche jaata hai (positive catalyst) ya upar (inhibitor)? Exponent ka sign.
Ex 1, Ex 2
B. Speed-up magnitude E a k a f t er / k b e f or e Ex 1
C. Degenerate: Δ E a = 0 E a 1 Ex 3
D. Temperature limit T → ∞ T → 0 Ex 3
E. Enzyme low [ S ] [ S ] ≪ K M v ∝ [ S ] Ex 4
F. Enzyme high [ S ] [ S ] ≫ K M v → V ma x Ex 4, Ex 5
G. Enzyme at [ S ] = K M bilkul V ma x
Ex 5
H. Word problem Real industrial process, catalysis type + reason chunno.
Ex 6
I. Exam twist Catalysis ke under Equilibrium / Δ H
Ex 7
J. Back-solve Speed-up diya hai, E a
Ex 8
Worked example Ex 1 — positive catalyst, "kitne times faster?" (A, B)Ek catalyst E a 80 60 kJ/mol T = 350 K
Forecast: ek warm temperature pe 20 kJ/mol
Joules mein convert karo aur drop nikalo. Δ E a = E a , b e f or e − E a , a f t er = ( 80 − 60 ) × 1 0 3 = 20000 J/mol Yeh step kyun? Arrhenius exponent ko E a J/mol R T J/mol Thermal budget compute karo. R T = 8.314 × 350 = 2909.9 J/mol Kyun? R T E a / R T Ratio banao. Do Arrhenius expressions ko divide karne se (same A T k b e f or e k a f t er = e ( E a , b e f or e − E a , a f t er ) / R T = e 20000/2909.9 = e 6.873 . A A T sirf hill height ka fark rehta hai — catalyst ko isolate karta hai.Evaluate karo. e 6.873 ≈ 966
Verify: exponent positive hai (hill chhoti ho gayi ⇒ tez ⇒ ratio > 1 ≈ 966× faster .
Worked example Ex 2 — negative catalyst / inhibitor (A, sign flips)Ek inhibitor effective barrier ko 50 70 kJ/mol T = 300 K raise karta hai. Rate kis factor se change hota hai?
Forecast: hill zyada unchi ho gayi — kya tumhe 1
Signed drop. E a , b e f or e − E a , a f t er = 50 − 70 = − 20000 J/mol Sign kyun rakhna hai? Yahi baat hai: ek negative catalyst exponent ko negative banata hai.Budget. R T = 8.314 × 300 = 2494.2 J/mol Ratio. k b e f or e k a f t er = e − 20000/2494.2 = e − 8.019 ≈ 3.3 × 1 0 − 4 A A T A
Verify: ratio < 1 slower ⇒ "inhibitor" ke consistent ✓. Yeh ek speed-up ka reciprocal-style partner hai: hill ko 20 kJ/mol raise karne se rate ~3000× shrink ho jaata hai. Answer ≈ 0.00033× (lagbhag 3000× slower) .
Common mistake Sign confusionAgar tum E a , a f t er − E a , b e f or e slow karta hai. Hamesha exponent ko R T E a , higher − E a , lower speed-up ke liye positive — phir physical meaning attach karo.
Worked example Ex 3 — koi change nahi, aur do temperature extremes (C, D)(a) Ek "catalyst" add kiya jaata hai lekin E a 60 kJ/mol real drop Δ E a = 20 kJ/mol T → ∞ T → 0 +
Forecast: (a) ke liye number guess karo. (b) ke liye, fraction Δ E a / R T T
(a) Δ E a = 0 k b e f or e k a f t er = e 0/ R T = e 0 = 1 Yeh kyun important hai: yeh sanity anchor hai — hill mein koi change nahi matlab rate mein bilkul koi change nahi, kisi bhi T 1 (b) Hot limit. T → ∞ R T Δ E a → 0 e Δ E a / R T → e 0 = 1 Kyun? Jab thermal energy enormous ho, molecules koi bhi hill aasani se cross kar lete hain; shortcut se zyada kuch nahi milta — catalyst ka advantage khatam ho jaata hai.(b) Cold limit. T → 0 + R T Δ E a → + ∞ e Δ E a / R T → ∞ Kyun? Jab thermal energy almost nahi hoti, ek chhoti hill sab kuch hoti hai — catalyst ka relative advantage explode kar jaata hai.
Verify: speed-up vs T 1 300 K 3000 Δ E a = 20
Intuition Catalyst cold mein zyada matter kyun karta haiBlue curve dekho: yeh sabse zyada upar left (cold) pe hai aur right (hot) pe dashed line = 1
Ab hum tool switch karte hain: Rate Law and Order of Reaction result v = K M + [ S ] V ma x [ S ] V ma x = 100 μ M/s K M = 5 μ M
Worked example Ex 4 — low, equal, aur high substrate (E, F, G)(a) [ S ] = 0.5 μ M ≪ K M [ S ] = 5 μ M = K M [ S ] = 500 μ M ≫ K M v
Forecast: kaun sa case roughly [ S ] straight-line dependence deta hai, aur kaun sa plateau karta hai?
(a) Low [ S ] v = 5 + 0.5 100 × 0.5 = 5.5 50 = 9.09 μ M/s Yeh "first order" kyun hai: kyunki [ S ] ≪ K M ≈ K M v ≈ K M V ma x [ S ] proportional hai [ S ] (b) [ S ] = K M v = 5 + 5 100 × 5 = 10 500 = 50 μ M/s Exactly aadha kyun: construction se [ S ] = K M = 2 [ S ] v = V ma x /2 K M (c) High [ S ] v = 5 + 500 100 × 500 = 505 50000 = 99.01 μ M/s Yeh V ma x [ S ] ≫ K M ≈ [ S ] [ S ] v → V ma x = 100 saturation .
Verify (units + trend): teeno μ M/s 9.09 → 50 → 99.01 100
Intuition Upar wala enzyme curve padhnaBlue curve Michaelis–Menten rate hai. Ise left se right follow karo: origin ke paas yeh almost ek straight line ki tarah rise karta hai (orange dot, cell E — rate proportional to [ S ] [ S ] = K M = 5 aadha hai (green dot, cell G). Zyada right mein curve dashed line V ma x = 100 flatten ho jaata hai (red dot, cell F — saturation). Ek picture mein teeno enzyme cells hain: linear start, half-way marker, plateau.
Worked example Ex 5 — saturation trap back-solve karo (F, G)V ma x 90% [ S ]
Forecast: 50% [ S ] = K M = 5 90% [ S ]
Target set karo. Chahiye v = 0.9 V ma x K M + [ S ] V ma x [ S ] = 0.9 V ma x V ma x Cancel karo aur solve karo. K M + [ S ] [ S ] = 0.9 ⇒ [ S ] = 0.9 K M + 0.9 [ S ] ⇒ 0.1 [ S ] = 0.9 K M ⇒ [ S ] = 9 K M Algebra kyun, guessing kyun nahi: relationship ek hyperbola hai, line nahi — intuition fail karti hai.Numbers. [ S ] = 9 × 5 = 45 μ M
Verify: back plug karo: v = 5 + 45 100 × 45 = 50 4500 = 90 μ M/s = 90% V ma x K M 9× chahiye tha, sirf doubling nahi — saturation curve ka top expensive hota hai. Yahi cell F ka exam trap hai.
Worked example Ex 6 — type chunno aur justify karo (H)Contact process mein, sulfur dioxide gas aur oxygen gas solid V 2 O 5 S O 3
Forecast: kya catalyst aur reactants same phase mein hain ya different phases mein?
Phases identify karo. Reactants S O 2 , O 2 gas hain; catalyst V 2 O 5 solid hai. Different phases.
Phase key kyun hai: homogeneous vs heterogeneous ki classification purely is baat se define hoti hai ki catalyst aur reactants ek phase share karte hain ya nahi (dekho Adsorption ).Classify karo. Different phase ⇒ heterogeneous catalysis.Mechanism (surface, chaar explicit steps).
(i) Diffusion — gas molecules S O 2 O 2 travel karna hota hai. Yeh ek transport step hai, chemical nahi; agar gas stagnant ho ya pellet pores clog ho jayein, toh sirf yahi step poori reaction ko throttle kar sakta hai (ek "diffusion-limited" regime). Transport picture Adsorption mein dekho.(ii) Adsorption — molecules V 2 O 5 active sites pe stick karte hain; surface bonds reactant bonds ko stretch aur weaken karte hain, jo exactly E a (iii) Reaction — adsorbed, weakened species surface pe react karke S O 3 (iv) Desorption — S O 3 regenerate ho jaata hai agle cycle ke liye.
Verify: parent ke heterogeneous examples (V 2 O 5 Adsorption sequence se, diffusion (transport) step explicitly ke saath ✓. Compare: Lead-chamber version gas reactants ke saath gas N O same phase — woh homogeneous hai. Same overall reaction (S O 2 → S O 3
Worked example Ex 7 — kya catalyst equilibrium move karta hai? (I)Ek reaction ke forward aur reverse activation energies E a , f = 100 E a , b = 60 kJ/mol dono ko 30 kJ/mol T = 300 K K e q Δ H
Forecast: agar dono directions 30 kJ/mol ratio change hoga?
Reaction enthalpy pehle. Δ H = E a , f − E a , b = 100 − 60 = 40 kJ/mol Yeh formula kyun: reactant side se barrier minus product side se barrier = do valleys ki height difference — dekho Chemical Equilibrium .Catalyst ke baad. Naye barriers 70 30 Δ H = 70 − 30 = 40 kJ/mol identical .
Kyun: dono se same 30 Equilibrium constant. K e q = k f / k b k f = A f e − E a , f / R T k b = A b e − E a , b / R T dono rates ko same factor e 30000/ R T K e q n e w = A b e − ( E a , b − 30000 ) / R T A f e − ( E a , f − 30000 ) / R T = A b A f e − ( E a , f − E a , b ) / R T = K e q o l d . Assumption explicitly: yeh cancellation chahti hai ki catalyst ka same factor dono directions pe ho (sach hai, kyunki catalyst ek shared lowered path offer karta hai) aur ki pre-exponential factors A f , A b e 30000/ R T
Verify: common factor e 30000/ R T = e 12.03 ≈ 1.68 × 1 0 5 k f k b Δ H K e q equilibrium pe pahunchti hai, bas jaldi . "HILL neeche karo, HOUSE nahi."
Worked example Ex 8 — speed-up diya hai,E a T = 310 K 1 0 4 E a
Forecast: humne dekha hai ~20–25 kJ/mol se room temperature ke paas ~10⁴ milta hai — kuch similar expect karo.
Ratio likho aur log lo. k b e f or e k a f t er = e Δ E a / R T = 1 0 4 R T Δ E a = ln ( 1 0 4 ) = 4 ln 10 = 9.210. ln e ( ⋅ ) Δ E a Multiply out karo. Δ E a = 9.210 × R T = 9.210 × 8.314 × 310 R T Δ E a Compute karo. R T = 8.314 × 310 = 2577.3 J/mol Δ E a = 9.210 × 2577.3 = 23737 J/mol ≈ 23.7 kJ/mol
Verify: forward-check — e 23737/2577.3 = e 9.210 = 1.0 × 1 0 4 J/mol
Recall Cover-the-answers self-test (all cells)Positive catalyst 80→60 kJ/mol at 350 K, speed-up? ::: e 20000/2909.9 ≈ 966 e − 8.019 ≈ 3.3 × 1 0 − 4 Δ E a = 0 1 T → ∞ 1 v [ S ] = K M V ma x /2 [ S ] 90% V ma x 9 K M V 2 O 5 K e q 1 0 4 Δ E a = ≈ 23.7
