Yeh page ek self-test ladder hai. Har problem apna difficulty level batata hai, aur poora solution ek collapsible callout ke andar chhupa hua hai — pehle khud try karo, phir reveal karo. Yahan use kiye gaye har symbol ko parent note mein build kiya gaya hai; agar koi symbol bhool gaye ho, wahan pehle dekho.
Shuru karne se pehle, ek picture jo do rate ideas ko dimag mein fix kar de.
Teen moves jo tum is poore page par repeat karte rahoge:
Slope = change in timechange in concentration.
Sign flip — ek reactant ke liye aage minus lagao taaki rate positive aaye.
Normalize — reaction rate r paane ke liye species rate ko uske stoichiometric coefficient se divide karo.
Hum kya karte hain: ek reactant ke liye average rate ki definition apply karte hain.
ravg=−ΔtΔ[A]=−60−00.50−0.80Minus kyun:Δ[A]=0.50−0.80=−0.30M negative hai (A disappear ho raha hai). Definition mein minus ise positive speed mein flip kar deta hai.
ravg=−60−0.30=5.0×10−3M/sAnswer:5.0×10−3M/s.
Recall Solution 1.2
Hum kya karte hain: pehchante hain ki konsi line konsi hai (figure s01 dekho).
Chord (amber) poore interval par average rate deta hai. Tangent (white) curve ko exactly ek point par touch karta hai aur wahan instantaneous rate deta hai.
Answer: ==tangent line at t=40 s==.
(a) N₂O₅ ki Consumption rate — slope phir sign flip.
−ΔtΔ[N2O5]=−500.070−0.100=−50−0.030=6.0×10−4M/s(b) Reaction rate — N₂O₅ ke coefficient 2 se normalize karo.
r=21(6.0×10−4)=3.0×10−4M/s2 se divide kyun: ek single reaction event mein do N₂O₅ molecules khatam hote hain, toh reaction "clock" utni hi aadhi speed se tick karta hai jitna N₂O₅ disappear hota hai.
Answer: (a) 6.0×10−4 M/s; (b) 3.0×10−4 M/s.
Recall Solution 2.2
Hum kya use karte hain: linked-rate relation
r=41dtd[NO2]=11dtd[O2].NO₂ (coefficient 4, product hai toh minus ki zaroorat nahi):
dtd[NO2]=4r=4×3.0×10−4=1.2×10−3M/sO₂ (coefficient 1):
dtd[O2]=1⋅r=3.0×10−4M/sMultiply kyun: har reaction event mein 4 NO₂ appear hote hain lekin sirf 1 O₂ — toh NO₂ chaar guna tezi se build up hota hai.
Answer: NO₂ 1.2×10−3 M/s par form hota hai; O₂ 3.0×10−4 M/s par.
Recall Solution 2.3
Relation:r=−dtd[H2]=21dtd[HI] (H₂ coefficient 1, HI coefficient 2).
−dtd[H2]=21(8.0×10−4)=4.0×10−4M/sAadha kyun: 1 H₂ use hone par 2 HI form hote hain, toh HI utni hi tezi se appear hota hai jitna H₂ disappear hota hai, matlab do guna.
Answer:H24.0×10−4 M/s par consume hota hai.
H₂O₂ ki Average consumption rate:−ΔtΔ[H2O2]=−12000.40−1.00=12000.60=5.0×10−4M/sAverage reaction rate (coefficient 2 se divide karo):
ravg=21(5.0×10−4)=2.5×10−4M/sInitial instantaneous reaction rate (given consumption 1.2×10−3, 2 se divide karo):
rinst(0)=21(1.2×10−3)=6.0×10−4M/sRatio:2.5×10−46.0×10−4=2.4.
Instant bada kyun hai: reaction slow hoti hai jab [H2O2] girta hai (collide karne ke liye kam molecules — Collision theory dekho). t=0 par concentration sabse zyada hoti hai, toh tangent sabse steep hota hai. 0–1200 s ka average tez early moments aur slow late moments ko blend kar deta hai, aur peak initial rate se neeche aa jaata hai. s01 dekho: left par tangent chord se zyada steep hai.
Answer:ravg=2.5×10−4 M/s, rinst(0)=6.0×10−4 M/s; initial instant 2.4× bada hai.
Recall Solution 3.2
Yahan derivative kyun: instantaneous rate = tangent slope = −dtd[A]. Kyunki [A] ek smooth formula ke roop mein diya gaya hai, hum slope measure karne ki jagah exactly differentiate kar sakte hain.
ekt ka derivative kekt hota hai, toh
dtd[A]=0.80⋅(−0.010)e−0.010t=−8.0×10−3e−0.010t.t=0 par:e0=1,
rinst(0)=−dtd[A]=8.0×10−3M/s.t=100 par:e−1≈0.3679,
rinst(100)=8.0×10−3×0.3679=2.94×10−3M/s.0–100 s par Average: pehle endpoint concentrations.
[A]0=0.80 M; [A]100=0.80e−1=0.2943 M.
ravg=−1000.2943−0.80=1000.5057=5.06×10−3M/s.Sanity check:ravg (5.06×10−3) fast initial (8.0×10−3) aur slow later (2.94×10−3) instantaneous rates ke beech mein hai — exactly jaisa ek chord slope ko steepest aur shallowest tangents ke beech hona chahiye.
Answer:rinst(0)=8.0×10−3, rinst(100)=2.94×10−3, ravg=5.06×10−3 M/s.
Master relation:r=−dtd[N2]=−31dtd[H2]=21dtd[NH3].(a) NH₃ se (coefficient 2):
r=21(6.0×10−4)=3.0×10−4M/s.(b) N₂ (coefficient 1): −dtd[N2]=r=3.0×10−4M/s.(c) H₂ (coefficient 3): −dtd[H2]=3r=9.0×10−4M/s.H₂ sabse fast kyun: ek reaction event mein teen H₂ use hote hain, toh H₂ N₂ se teen guna tezi se khatam hota hai.
Common error: yeh claim karna ki teeno species same 6.0×10−4 M/s par change karti hain. Aisa nahi hota — sirf normalized rate r shared hoti hai.
Answer:r=3.0×10−4; N₂ 3.0×10−4 par; H₂ 9.0×10−4 M/s par.
Recall Solution 4.2
Idea: average rate × time interval = concentration change (r=−Δ[A]/Δt se rearrange karo, coefficient 1 hai toh r consumption rate ke barabar hai).
Pehla interval (0→100 s):Δ[A]=−rΔt=−(4.0×10−3)(100)=−0.40M.[A]100=1.00−0.40=0.60M.Doosra interval (100→300 s):Δ[A]=−(1.5×10−3)(200)=−0.30M.[A]300=0.60−0.30=0.30M.Speed up ya slow down? Average rate 4.0×10−3 se 1.5×10−3 M/s tak giri: reaction slow down ho rahi hai jaise A use hota hai — yeh standard behaviour Collision theory se explain hota hai.
Answer:[A]100=0.60 M, [A]300=0.30 M; slowing down.
2A+B→3C ke liye Master relation:
r=−21dtd[A]=−11dtd[B]=31dtd[C].(a) C se (coefficient 3):
r=31(9.0×10−4)=3.0×10−4M/s.(b) Har species solve karo, signs rakhte hue (A aur B reactants hain → negative):
dtd[A]=−2r=−2(3.0×10−4)=−6.0×10−4M/s,dtd[B]=−1⋅r=−3.0×10−4M/s.(c) Student ki galti: unhone negative sign drop kar diya. Reactant A consume ho raha hai, toh d[A]/dtnegative hona chahiye. Magnitude 6.0×10−4 sahi hai, lekin sign nahi.
Sahi value:dtd[A]=−6.0×10−4M/s.
Answer: (a) r=3.0×10−4; (b) d[A]/dt=−6.0×10−4, d[B]/dt=−3.0×10−4; (c) minus sign missing hai, sahi hai −6.0×10−4 M/s.
Recall Solution 5.2
(a) t=0 par Instantaneous rate.[A]=[A]0(1+k[A]0t)−1 ko differentiate karo. Chain rule use karke (u−1 ka derivative −u−2u′ hai jahan u=1+k[A]0t, u′=k[A]0):
dtd[A]=−[A]0⋅(1+k[A]0t)2k[A]0=−(1+k[A]0t)2k[A]02.t=0 par denominator 1 hai:
rinst(0)=−dtd[A]0=k[A]02=0.20×(0.50)2=0.050M/s.(b) 0–10 s par Average rate. Endpoint concentrations:
[A]0=0.50 M.
1+k[A]0t=1+0.20×0.50×10=1+1.0=2.0, toh [A]10=0.50/2.0=0.25 M.
ravg=−100.25−0.50=100.25=0.025M/s.(c) Initial rates kyun bharosemand hain: bilkul shuroo ka instantaneous rate (0.050 M/s) exactly k[A]02 hai — known starting concentration mein ek clean expression, koi product build up hone se pehle ya reverse reactions shuru hone se pehle. Average (0.025 M/s) already 10 s ke slowdown se blur ho chuka hai. t=0 par rate measure karna (initial-rate method) true starting speed ko isolate karta hai aur rate law dhundhne ka basis hai.
Answer: (a) 0.050 M/s; (b) 0.025 M/s; (c) initial rate =k[A]02, baad ke slowdown se unblurred.
Recall Quick self-check questions
2A→B ke liye reaction rate kya hai agar [A]0.04 M/s par girta hai? ::: r=21(0.04)=0.02 M/s.
Average ya instantaneous: konsa tangent line use karta hai? ::: Instantaneous.
Average ya instantaneous: konsa chord (do endpoints) use karta hai? ::: Average.
Ek reactant ke liye, definition mein minus sign kyun hota hai? ::: Kyunki Δ[A] negative hota hai aur rate positive report ki jaati hai; minus ise flip kar deta hai.
N2+3H2→2NH3 mein, konsi species ki concentration sabse tezi se change hoti hai? ::: H2 (coefficient 3).
