2.8.6 · D5 · HinglishChemical Kinetics
Question bank — Methods to determine order — initial rates, integrated method, half-life method
2.8.6 · D5· Chemistry › Chemical Kinetics › Methods to determine order — initial rates, integrated metho
True or false — justify
For a reaction , the order in must be
False. Order experimental hota hai; coefficient exponent ke barabar tab hi hota hai jab step elementary ho, jo aap overall reaction ke liye assume nahi kar sakte.
The three methods (initial rates, integrated, half-life) can give different orders for the same reaction
False. Ye teeno ek hi underlying rate law ke teen alag fingerprints hain, isliye inhe agree karna hi chahiye; disagreement matlab measurement ya fitting error hai.
A straight line in the vs plot proves the reaction is first order
True — sirf hi ko time ke saath linearly girne deta hai, isliye wo particular plot ka straight hona first order ki unique signature hai.
If vs is straight, then vs must be curved
True. Ek data set sirf ek plot par linear ho sakta hai; agar wala straight hai, toh raw plot ko curve karna chahiye (wo exponentially decay karta hai, linearly nahi).
Every reaction has a constant half-life
False. Sirf first order ka , se independent hota hai; zero-order half-life shrink hoti hai aur second-order half-life badhti hai jaise reaction aage badhti hai.
The rate constant has the same units for all orders
False. Kyunki , rate ke units () match karne ke liye ko units carry karne padte hain, jo ke saath change hote hain.
In the integrated method, a negative slope always means a positive
aur ke liye True hai, jahan linear slope hota hai; concentration term girta hai, isliye slope negative hai aur . ke liye plotted quantity badhti hai, isliye slope deta hai.
Spot the error
"I measured the rate at s in both experiments and used the initial-rate formula."
Formula ko initial rates chahiye. s par products build up ho chuke hain, uncertain hai, aur reverse/side reactions rate ko distort kar deti hain — isolation ka logic collapse ho jaata hai.
"Doubling doubled the rate, so ."
Formula hai . Double concentration ke liye rate ka double hona first order hai, second nahi; student ne ek extra factor of 2 invent kar liya.
", so I'll use this to find for my zero-order reaction."
sirf first-order half-life hai. Zero order ke liye hota hai, isliye galat formula use karne se galat milega aur concentration dependence bilkul chhup jaayegi.
"To find the order in I doubled both and and saw the rate go ×8."
Ek saath do concentrations change karne se dono exponents mix ho jaate hain (). Isolation method demand karta hai ki baaki har reactant fixed rakho taaki sirf ka exponent dikhe.
"The slope of my vs line is , so ."
Rate constants kabhi negative nahi hote. Slope ke barabar hota hai, isliye ; minus sign sirf ye record karta hai ki concentration gir rahi hai.
" vs was straight with slope ."
Second order ke liye plot time ke saath badhta hai, isliye uska slope hota hai, nahi. Wahan negative slope physically impossible hoga.
"My data fit all three plots equally well, so the reaction is zero, first, and second order."
Sirf ek hi plot truly straight ho sakta hai; "teeno fit hain" ka matlab hai ki tumhara concentration range ya time window itna chhota hai ki unhe alag nahi kar sakte. Run extend karo — jaise aur girta hai, curves zyada diverge hote hain.
Why questions
Why do we take the logarithm in the initial-rate formula?
Order ke exponent mein hota hai. lene se neeche multiplier ban jaata hai, ek exponent-solving problem ko simple division mein badal deta hai.
Why must all other reactant concentrations be held constant when isolating the order in one reactant?
Taaki rate ko sirf wo reactant change kare jise hum study kar rahe hain; baaki kisi mein bhi change unknown factors add kar dega aur measured ratio ambiguous ho jaayega.
Why does the integrated method plot three different quantities (, , ) rather than just ?
Kyunki har order ki ek alag function predict karta hai jo time mein linear hoti hai. Test karna ki kaunsi transformed quantity data ko straighten karti hai, yahi order reveal karta hai.
Why is a first-order half-life independent of starting concentration?
Kyunki mein koi nahi hai — fractional decay per unit time constant hai, isliye halving hamesha utna hi clock time leta hai chahe shuru mein kitna bhi ho.
Why can't you determine order just by inspecting the balanced equation?
Overall equation mass balance reflect karta hai, mechanism nahi. Actual rate slow (rate-determining) step ke molecularity par depend karta hai, jiska overall coefficients se match zaroori nahi.
Why is the general rule useless exactly at ?
par exponent hota hai, isliye — dependence vanish ho jaati hai. Yahi reason hai ki first order ka half-life constant hota hai, lekin iska matlab ye bhi hai ki wahan exponent se extract nahi kar sakte aur flat behaviour ko directly recognize karna padta hai.
Why do initial rates avoid the problem of reverse reactions?
par practically koi product nahi hota, isliye reaction ko backwards chalane ke liye kuch nahi hota; measured rate purely wo forward process hai jise hum characterise karna chahte hain.
Edge cases
What does the half-life method report if doubling doubles ?
Tab hoga, jo zero-order fingerprint hai (), isliye reaction zero order hai.
For a zero-order reaction, what happens to the rate as approaches zero?
Rate (constant) rehti hai jab tak khatam nahi ho jaata, phir abruptly ruk jaata hai — zero order ka matlab hai ki rate par depend nahi karta jab tak koi bhi bacha ho.
Can the order be a fraction or negative, and would these methods still work?
Haan. Initial-rate log formula koi bhi real return karta hai (jaise ya ); fractional/negative orders complex mechanisms indicate karte hain, lekin isolation aur log methods inhe bina kisi modification ke handle karte hain.
If a plot of vs is a straight line all the way to , what is the order and what is the slope?
Zero order, aur slope hai; concentration ek constant rate par girta hai chahe kitna bhi bacha ho.
Two experiments give the same initial rate despite different — what order is that?
Us reactant mein zero order: kyunki rate par depend nahi karta, change karne se rate unchanged rehta hai, jo deta hai.
Why does the second-order half-life grow as the reaction proceeds?
Kyunki mein current concentration ko new starting point ki tarah use kiya jaata hai; har successive half-life ek chhote se shuru hoti hai, isliye bada hota hai aur halving mein zyada time lagta hai.
For a first-order reaction, what is the relationship between the third half-life and the first?
Dono equal hain. First-order half-life constant hoti hai, isliye 1st, 2nd, aur 3rd teeno same time lete hain — yahi reason hai ki radioactive decay (Radioactive Decay) ka ek fixed half-life hota hai.
If measured "order" from initial rates differs from the order that straightens the integrated plot, what should you suspect?
Koi hidden complication — accumulating product, temperature drift (Arrhenius Equation), ya run ke dauran mechanism change. Genuine kinetics dono methods ko agree karne par majboor karti hai.
Recall Is page ko close karne se pehle ek-line self-test
Answers cover karo aur re-derive karo: (1) har order ke liye kaunsa plot straight hota hai, (2) har order ke liye ka par kya dependence hai, (3) kyun "initial" aur "ek reactant ko isolate karo" non-negotiable hain. Agar inme se koi bhi shaky lage, toh Integrated Rate Equations aur Half-Life of Reactions revisit karo.