Temperature dependence — Arrhenius equation k = A·e^(−Ea - RT): Question bank

2.8.7 · D5 · Chemistry › Chemical Kinetics › Temperature dependence — Arrhenius equation k = A·e^(−Ea - R

Shuru karne se pehle, ek reminder symbols ka taaki kuch bhi blindly use na ho:

$k$ = rate constant (reaction kitni fast chalti hai — dekho Rate Laws and Rate Constants).
$A$ = pre-exponential factor (collision frequency × orientation).
$E_{a}$ = activation energy (barrier ki height — dekho Activation Energy and Transition State Theory).
$R$ = gas constant $8.314 J/(mol\cdotK)$ ; $T$ = absolute temperature Kelvin mein.
Woh piece $e^{- E_{a} / (R T)}$ Boltzmann factor hai, yaani collisions ka woh fraction jo react karne ke liye enough energetic hain (from the Maxwell-Boltzmann Distribution).

True or false — justify

Doubling the temperature always doubles the rate constant.

False.

k

T

par dependence exponential hai

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

ke through, linear nahi — thodi si

T

ki rise

k

ko kai-guna kar sakti hai, aur yeh factor har reaction aur har starting

T

ke liye alag hota hai.

A reaction with a larger

E_{a}

is more sensitive to temperature changes.

True.

ln k

1/ T

ki slope

- E_{a} / R

hoti hai, isliye bada

E_{a}

matlab

k

zyada steeply climb karta hai jab

T

badhta hai — high-barrier reactions dramatically respond karte hain heating se.

T \to \infty

, the rate constant

k

approaches

A

True. Jab

T

bahut bada ho,

E_{a} / (R T) \to 0

isliye

e^{- E_{a} / (R T)} \to e^{0} = 1

, aur

k \to A

reh jaata hai — maximum rate agar energy kabhi limiting factor na hoti, sirf collision frequency aur orientation matter karte.

The Arrhenius equation predicts

k = 0

at absolute zero.

True (as a limit).

T \to 0

par,

E_{a} / (R T) \to + \infty

isliye

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

— koi bhi molecule barrier clear karne ke liye enough energy nahi rakhta, toh reaction effectively ruk jaati hai.

A catalyst increases

A

in the Arrhenius equation.

False (mostly). Ek catalyst mainly

E_{a}

ko kam karta hai ek naya pathway offer karke, jo

k

ko bahut zyada powerfully badhata hai kisi bhi

A

mein change se zyada.

ln k

is a straight-line function of

T

False.

ln k

seedhi line hai

1/ T

ke against,

T

ke against nahi. Linear form hai

ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}

, isliye horizontal axis

1/ T

hona chahiye.

The pre-exponential factor

A

has the same units as

k

True. Kyunki

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

dimensionless hai,

k = A \cdot (dimensionless)

force karta hai ki

A

ke paas

k

ke units hon, jo reaction order par depend karte hain.

For an exothermic reaction the activation energy is negative.

False.

E_{a}

woh barrier hai reactants se transition state tak aur essentially hamesha positive hoti hai; overall energy change

Δ H < 0

ek alag quantity hai — dekho Thermodynamics vs Kinetics.

If two reactions have the same

E_{a}

they will have the same rate constant.

False. Ye

A

par bhi depend karta hai; do reactions ek hi barrier height share kar sakti hain phir bhi collision frequency aur orientation mein differ kar sakti hain, alag

k

dete hue.

The Arrhenius equation applies only to gas-phase reactions.

False. Yeh broadly empirical hai aur solution aur surface reactions ko bhi describe karta hai, jahan bhi ek activation barrier rate ko govern karta ho.

Orientation ek geometric requirement hai jo energy se independent hai, isliye yeh collision frequency ko multiply karta hai (dono temperature-weak hain); exponential sirf energy criterion ke liye reserved hai.

Why does a catalyst speed a reaction without being consumed?

Yeh ek alternative route provide karta hai kam

E_{a}

ke saath, Boltzmann factor

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

ko badhata hai; har cycle mein regenerate hokar, yeh unchanged bahar aata hai — dekho Catalysis.

Edge cases

What does the Arrhenius equation predict when

E_{a} = 0

Boltzmann factor

e^{0} = 1

ban jaata hai, isliye

k = A

aur rate temperature-independent hoti hai — har collision enough energetic hai, isliye sirf collision frequency matter karti hai.

What happens to

k

E_{a}

is extremely large (very high barrier)?

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

negligible ho jaata hai, isliye

k

tiny hai aur reaction effectively frozen hai ordinary temperatures par jab tak

T

itna nahi badhaya jaata ki tail populate ho sake.

Can the two-point formula ever give

E_{a} < 0

Tabhi jab

k

increasing

T

ke saath decrease kare; genuine elementary reactions

E_{a} > 0

deti hain, isliye negative result usually ek composite mechanism ya measurement error indicate karta hai.

If a reaction rate barely changes with temperature, what does that imply about

E_{a}

Iska matlab hai bahut chhota

E_{a}

(ek low ya nearly flat barrier); jab thodi energy chahiye, warming se kuch naye successful collisions add hote hain, isliye

k

roughly constant rehta hai — related hai ek chhote Q10 Temperature Coefficient se.

What would a perfectly horizontal Arrhenius plot (

ln k

1/ T

) mean?

Zero slope ka matlab

- E_{a} / R = 0

, yaani

E_{a} = 0

: rate constant bilkul temperature par depend nahi karta, ek idealized barrier-free process.

Recall Quick self-check

The slope of $ln k$ vs $1/ T$ is ::: $- E_{a} / R$ (negative, kyunki $E_{a} > 0$ ). As $T \to \infty$ , $k \to$ ::: $A$ , kyunki Boltzmann factor 1 ki taraf jaata hai. Temperature in the Arrhenius equation must be in ::: Kelvin (absolute). A catalyst mainly changes ::: $E_{a}$ (use kam karta hai), fundamental thermodynamics nahi.

Question bank — Temperature dependence — Arrhenius equation k = A·e^(−Ea - RT)

True or false — justify

Spot the error

Why questions

Edge cases