Pseudo-first-order kinetics: Question bank — Hinglish deep dive

2.8.5 · D5 · Chemistry › Chemical Kinetics › Pseudo-first-order kinetics

Shuru karne se pehle, teen notation pieces hain jo kisi bhi question se pehle earn karne zaroori hain:

Order aur molecularity ko alag rakho aur aadhe traps gayab ho jaate hain; $k^{'}$ vs $k$ ke units ko alag rakho aur baaki aadhe bhi.

Poore page ko pehle ek figure mein picture karo. Neeche ke do panels woh mental image hain jinhe har question secretly refer karta hai — ek genuinely second-order run $ln [A]$ -vs- $t$ plot par bend karta hai, jabki wohi reaction $B$ se flood hone par seedhi ho jaati hai.

Aur $B$ ko constant treat karne ki wajah purely visual hai: flooded reservoir muskil se dip karta hai jab $A$ khatam hoti hai. Agla figure exactly dikhata hai ki "excess" kitna bada hona chahiye pehle ki ye claim safe ho.

Do chhoti derivations jinhein hum baar baar lean karenge:

True or false — justify

Pseudo-first-order kinetics reaction ki true molecularity change karta hai.

False — molecularity ek step mein colliding particles count karta hai aur mechanism se fix hota hai; ek reactant flood karna sirf ye change karta hai ki hamara measured rate law kaisa dikhta hai, physical collision nahi.

ln [A]

t

ka pseudo-first-order plot straight hona prove karta hai ki reaction genuinely first-order overall hai.

False — ye sirf un conditions mein

A

mein first-order prove karta hai. Straightness

[B]

ko constant rakh kar engineer ki gayi hai; true overall order abhi bhi second hai (dekho Method of Isolation).

Pseudo-first-order rate constant

k^{'}

aur true rate constant

k

ke units same hote hain.

False —

k^{'}

ka unit

s^{- 1}

hai (first-order) jabki

k

ka unit

M^{- 1} s^{- 1}

hai (second-order); ye excess concentration

[B]_{0}

ke units se differ karte hain, kyunki

k^{'} = k [B]_{0}

(dekho Integrated Rate Laws).

Agar tum

[B]_{0}

double karo (abhi bhi large excess mein), toh measured

k^{'}

double ho jaata hai — assuming reaction

B

mein first-order hai.

True —

Rate = k [A] [B]

ke saath humhare paas

k^{'} = k [B]_{0}^{1}

hai, toh

k^{'}

linearly scale karta hai; lekin agar

B

mein order

n

hota, toh

[B]_{0}

double karne se

k^{'}

2^{n}

se multiply hota.

Agar tum

[B]_{0}

double karo, toh A ki half-life change hoti hai.

True — Half-life hai

t_{1/2} = ln 2/ k^{'} = ln 2/ (k [B]_{0})

, toh

[B]_{0}

double karne se half-life half ho jaati hai bhale hi "first-order half-life is constant" ek fixed run ke liye abhi bhi hold karta ho.

Ek fixed pseudo-first-order run ke liye, A ki half-life

[A]_{0}

par independent hoti hai.

True — ye first-order behaviour ki signature hai; ek run ke andar

k^{'}

constant hai, toh

t_{1/2} = ln 2/ k^{'}

depend nahi karta ki A kitni le ke shuru kiya.

Aqueous hydrolysis mein water ek spectator hai aur true rate law mein appear nahi karta.

False — water ek genuine reactant hai; ye sirf

\approx 55.5 M

par hai aur muskil se change hota hai, toh ye gayab hone ki jagah

k^{'}

ke andar chhup jaata hai.

Ek reaction pseudo-second-order ho sakti hai.

True — ek third-order reaction

k [A] [B] [C]

ke liye, dono

B

aur

C

flood karne se ye

k^{''} [A]

tak collapse ho jaati hai jahan

k^{''} = k [B]_{0} [C]_{0}

; lekin teeno mein se sirf ek flood karo aur genuine second-order dependence bachti hai, toh "pseudo-lower-order" ek general idea hai.

[A]_{0}

(limiting reactant) double karo jabki B ka same excess rakho —

k^{'}

change hota hai.

False —

k^{'} = k [B]_{0}

sirf excess reactant par depend karta hai; limiting reactant ki start amount change karna sirf run kitne time tak chalega ye change karta hai,

k^{'}

nahi.

Ek reaction jo flooded reactant

B

mein order

2

hai, phir bhi ek straight

ln [A]

-vs-

t

line deta hai jab

A

first-order hai.

True — flooding

[B]_{0}^{2}

k^{'} = k [B]_{0}^{2}

ke andar constant fix kar deta hai, toh

A

-dependence unchanged rehti hai; plot straight hai, lekin

k^{'}

[B]_{0}

ke square ke roop mein scale karta hai.

Spot the error

Ek student

k^{'} = 0.05 s^{- 1}

measure karta hai

[B]_{0} = 1.0 M

ke saath aur reaction ke true rate constant ko

0.05 s^{- 1}

report karta hai.

Woh excess divide karna bhool gaya:

k = k^{'} / [B]_{0} = 0.05/1.0 = 0.05 M^{- 1} s^{- 1}

— same number, lekin units second-order mein change hote hain, yahi toh poora point hai.

Ek student B ka 10× excess use karta hai, A ka 50% consume karta hai, aur claim karta hai ki

[B]

constant tha.

Jab A ka aadha react ho jaata hai, B

10 [A]_{0}

9.5 [A]_{0}

tak drop ho jaata hai — ek 5% change — jo "straight" line ko bend karne ke liye kaafi bada hai; 10× reliable pseudo-first-order ke liye bahut chhota hai.

Ester hydrolysis run diya, ek student likhta hai

k = k^{'} \times [H_{2} O] = 2.5 \times 1 0^{- 4} \times 55.5

Relation hai

k = k^{'} / [B]_{0}

k^{'} \times [B]_{0}

nahi; true

k

recover karne ke liye tum flooded concentration remove karte ho, toh

k = 2.5 \times 1 0^{- 4} /55.5 = 4.5 \times 1 0^{- 6} M^{- 1} s^{- 1}

Ek student curved

ln [A]

-vs-

t

plot dekhta hai aur conclude karta hai ki reaction first-order hai lekin changing

k

ke saath.

Rate "constant" ek run ke dauran fixed temperature par change nahi ho sakta; curvature ka matlab hai ki assumed order galat hai — yahan dono concentrations vary karte hain, toh ye genuinely second-order hai, drifting

k

wala first nahi.

Ek student kehta hai: "Kyunki

k^{'} = k [B]_{0}

[B]_{0}

raise karne se reaction fast hoti hai, toh true rate constant

k

increase hua."

k

untouched hai — ye sirf temperature aur Activation Energy par depend karta hai (Collision Theory ke zariye); sirf apparent constant

k^{'}

bada hua kyunki wo

[B]_{0}

carry karta hai.

Ek student third-order reaction ko sirf

B

se flood karta hai aur likhta hai

k^{''} = k [B]_{0} [C]_{0}

Tum

[C]_{0}

fold in nahi kar sakte — tumne

C

flood nahi kiya, toh

[C]

abhi bhi vary karta hai; sahi collapse hai

Rate = k [B]_{0} [A] [C]

, ek pseudo-second-order law, pseudo-first nahi.

Ek student

k^{'}

determine karta hai

[A] = [A]_{0} e^{- k^{'} t}

se lekin

t

minutes mein plug in karta hai jabki

k^{'}

s^{- 1}

mein hai.

Units match karni chahiye: exponentiate karne se pehle time ko seconds mein convert karo (ya

k^{'}

min^{- 1}

mein), warna exponent meaningless hai.

Why questions

Ek reactant flood karne se curved second-order plot straight first-order line mein kyun badal jaata hai?

[B] \approx [B]_{0}

hold karna doosre varying concentration ko remove kar deta hai, toh

Rate = k [B]_{0} [A] = k^{'} [A]

— mathematically first-order ka form — aur tabhi

ln [A]

linearly girta hai (upar ke left→right panels dekho).

Hum roughly 50–100× excess kyun require karte hain sirf "zyada" ki jagah?

Pure run ke dauran B mein fractional change approximately

[A]_{0} / [B]_{0}

hota hai; 100× par ye ~1% error hai, acceptable, jabki 10× par ye ~5–10% tak pahunch sakta hai, jo

k^{'}

ko visibly distort karta hai.

Flooded reactant par exponent

n

apparent order mein kyun appear nahi karta?

Kyunki

Rate = k [A]^{m} [B]_{0}^{n} = k^{'} [A]^{m}

saara

[B]_{0}^{n}

andar

k^{'} = k [B]_{0}^{n}

ke hide kar deta hai; slope abhi bhi

m

report karta hai, jabki

n

tabhi show hota hai jab tum

[B]_{0}

vary karo aur dekho

k^{'}

kaise scale karta hai.

Pseudo-first-order water mein reactions ke liye natural regime kyun hai?

Pure water

\approx 55.5 M

par hota hai, typical solute concentrations ko hazaaron times dwarf karta hai, toh koi bhi reaction jo water consume karti hai automatically flooded hai aur solute mein first-order behave karti hai.

Hum

[B]

k^{'}

ke andar "hide" karne ki koshish kyun karte hain poora second-order equation solve karne ki jagah?

Full second-order integrated law dono concentrations mix karta hai aur fit karna awkward hai; pseudo trick ek variable isolate karta hai, ek clean straight line deta hai aur ek easily measured slope.

Low substrate par enzyme catalysis pseudo-first-order in substrate kyun ban jaati hai?

Jab

[S] ≪ K_{M}

, Michaelis–Menten rate

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

tak simplify ho jaata hai, ek constant times

[S]

— exactly pseudo-first-order form, kyunki enzyme effectively relative excess mein hai.

Method of Isolation har ek reactant ke liye baari baari deliberate excess kyun use karta hai?

Saare reactants ko ek ko chhod kar flood karke, rate sirf uss ek par depend karta hai; har reactant ke liye repeat karne se uska individual order reveal hota hai, jisse tum poora rate law piece by piece rebuild kar sakte ho.

Temperature raise karne se

k^{'}

kyun change hota hai fixed

[B]_{0}

par bhi?

k^{'} = k [B]_{0}

aur true

k

temperature ke saath Arrhenius/Activation Energy dependence ke zariye badhta hai, toh

k^{'}

badhta hai bhale hi flooded concentration unchanged ho.

Edge cases

Pseudo-first-order approximation ka kya hota hai jab

[A]_{0}

[B]_{0}

ke kareeb aane lagta hai?

Excess ratio 1 ki taraf collapse karta hai,

[B]

ko constant treat nahi kiya ja sakta, aur behaviour true second-order mein revert ho jaata hai — straight line curve ho jaati hai (figure ka right panel apni straightness khota hai).

Kya hota hai agar reaction appreciably reversible ho,

A + B ⇌ P

Reverse rate

k_{- 1} [P]

ko ignore nahi kiya ja sakta;

B

flood hone ke bawajood integrated law ban jaata hai

ln ([A] - [A]_{e q}) = ln ([A]_{0} - [A]_{e q}) - (k^{'} + k_{- 1}) t

, toh plot

[A] - [A]_{e q}

ke against straight hai aur iska slope sum

k^{'} + k_{- 1}

hai, sirf

k^{'}

nahi.

Reversible correction

[A]

directly use karne ki jagah

[A]_{e q}

subtract kyun karta hai?

Kyunki equilibrium par net rate zero hai,

[A] = 0

nahi; driving quantity equilibrium se distance

[A] - [A]_{e q}

hai, aur woh combination ek clean single exponential ki tarah decay karta hai jahan plain

ln [A]

curve karta.

Kya hoga agar

[B]_{0}

enormous ho lekin B bhi ek product ho?

Agar B regenerate hota hai ya muskil se consumed hota hai toh regardless constant rehta hai, toh pseudo-first-order form abhi bhi hold karta hai — key

[B]

ki constancy hai, ye nahi ki ye technically consume hua ya nahi.

Ek pseudo-first-order reaction ka "order" kya hai agar tum koi bhi reactant flood karne se refuse karo?

Bilkul bhi pseudo behaviour nahi hoga — tum genuine order measure karte ho (jaise overall second), kyunki pseudo-first-order ek condition hai jo tum impose karte ho, na ki koi intrinsic property.

Irreversible pseudo-first-order run ke liye

t \to \infty

ki limit mein

[A]

kya karta hai?

[A] = [A]_{0} e^{- k^{'} t} \to 0

; A exponentially fully consumed ho jaata hai, jabki flooded B

[B]_{0} - [A]_{0} \approx [B]_{0}

par settle hota hai, practically abhi bhi unchanged. (Reversible run ke liye ye

[A]_{e q} > 0

par level off karta hai.)

k^{'}

ka kya hota hai agar

[B]_{0} \to 0

(excess bilkul remove kar diya)?

k^{'} = k [B]_{0} \to 0

, matlab apparent rate vanish ho jaati hai — correctly reflect karta hai ki B ke bina reaction proceed nahi kar sakti; pseudo picture degenerate ho jaata hai.

Kya ek pseudo-first-order reaction mein pH dependence hoti hai jab acid-catalyzed ho (jaise sucrose inversion)?

Haan — catalyst

[H^{+}]

bhi effective constant mein fold ho jaata hai, toh

k^{'}

pH par depend karta hai bhale hi plot sucrose mein first-order rahe; buffer pH change karna

k^{'}

shift karta hai.

Agar tum accidentally "pseudo" run sirf 3× excess mein B ke saath shuru karo, toh

ln [A]

plot perfectly straight hoga, perfectly curved hoga, ya slightly bent hoga?

Slightly bent — B appreciably change karta hai jab A consumed hoti hai, toh line mein gentle curvature hogi na ki clean slope, aur jo bhi

k^{'}

tum extract karo woh sirf ek rough average hai.

Recall Ek-line self-test

"Pseudo" ka matlab hai ::: apparent/false — reaction truly higher-order hai; humne ise sirf first-order dikhaya ek concentration constant rakh kar. True $k$ recover karne ke liye $k^{'}$ se tum ::: excess concentration se divide karo, $k = k^{'} / [B]_{0}^{n}$ (jahan $n$ flooded reactant mein order hai), jo units bhi genuine rate constant ke units mein wapas fix karta hai.