Understanding differential rate equations means connecting how fast a reaction proceeds (rate) to the current concentration of reactants. Each order of reaction gives a different mathematical relationship, and deriving them from scratch reveals why concentration affects rate the way it does.
Intuition Why Different Orders ExistThink of chemical reactions like cars on a highway. In a zero-order reaction, the highway has one narrow toll booth—no matter how many cars pile up (concentration), only the same fixed number can pass through per minute (rate is constant). In first-order , every car drives independently—double the cars, double the throughput. In second-order , cars must collide to merge—if you double the cars, collisions quadruple because each car has twice as many partners to hit.
Definition Rate Law (General Form)For a reaction aA → products \text{aA} \to \text{products} aA → products differential rate equation is:
Rate = − 1 a d [ A ] d t = k [ A ] n \text{Rate} = -\frac{1}{a}\frac{d[\text{A}]}{dt} = k[\text{A}]^n Rate = − a 1 d t d [ A ] = k [ A ] n
k k k rate constant (units depend on n n n n n n order of reaction (experimentally determined)The negative sign: concentration decreases over time
Starting point : Rate depends on concentration raised to power zero.
− d [ A ] d t = k [ A ] 0 = k -\frac{d[\text{A}]}{dt} = k[\text{A}]^0 = k − d t d [ A ] = k [ A ] 0 = k
WHY this step? Zero power means concentration doesn't matter—rate is always k k k
Separate variables :
d [ A ] = − k d t d[\text{A}] = -k \, dt d [ A ] = − k d t
Integrate both sides from t = 0 t=0 t = 0 [ A ] 0 [\text{A}]_0 [ A ] 0 t t t [ A ] t [\text{A}]_t [ A ] t
∫ [ A ] 0 [ A ] t d [ A ] = − k ∫ 0 t d t \int_{[\text{A}]_0}^{[\text{A}]_t} d[\text{A}] = -k \int_0^t dt ∫ [ A ] 0 [ A ] t d [ A ] = − k ∫ 0 t d t
[ A ] t − [ A ] 0 = − k t [A]_t - [A]_0 = -kt [ A ] t − [ A ] 0 = − k t
Worked example Worked Example: Surface-Catalyzed DecompositionAmmonia decomposes on a platinum surface: 2NH 3 (g) → N 2 (g) + 3H 2 (g) \text{2NH}_3\text{(g)} \rightarrow \text{N}_2\text{(g)} + \text{3H}_2\text{(g)} 2NH 3 (g) → N 2 (g) + 3H 2 (g)
Given: [ NH 3 ] 0 = 0.10 M [\text{NH}_3]_0 = 0.10 \, \text{M} [ NH 3 ] 0 = 0.10 M k = 2.5 × 10 − 4 M s − 1 k = 2.5 \times 10^{-4} \, \text{M s}^{-1} k = 2.5 × 1 0 − 4 M s − 1
Find : Concentration after 200 s
Solution :
[ NH 3 ] t = 0.10 − ( 2.5 × 10 − 4 ) ( 200 ) [\text{NH}_3]_t = 0.10 - (2.5 \times 10^{-4})(200) [ NH 3 ] t = 0.10 − ( 2.5 × 1 0 − 4 ) ( 200 ) = 0.10 − 0.05 = 0.05 M = 0.10 - 0.05 = 0.05 \, \text{M} = 0.10 − 0.05 = 0.05 M Why this step? Direct substitution into [ A ] t = [ A ] 0 − k t [\text{A}]_t = [\text{A}]_0 - kt [ A ] t = [ A ] 0 − k t NH 3 \text{NH}_3 NH 3
Starting point : Rate proportional to concentration.
− d [ A ] d t = k [ A ] -\frac{d[\text{A}]}{dt} = k[\text{A}] − d t d [ A ] = k [ A ]
Separate variables :
d [ A ] [ A ] = − k d t \frac{d[\text{A}]}{[\text{A}]} = -k \, dt [ A ] d [ A ] = − k d t
WHY this step? We need [ A ] [\text{A}] [ A ] t t t
Integrate :
∫ [ A ] 0 [ A ] t d [ A ] [ A ] = − k ∫ 0 t d t \int_{[\text{A}]_0}^{[\text{A}]_t} \frac{d[\text{A}]}{[\text{A}]} = -k \int_0^t dt ∫ [ A ] 0 [ A ] t [ A ] d [ A ] = − k ∫ 0 t d t
ln [ A ] t − ln [ A ] 0 = − k t \ln[\text{A}]_t - \ln[\text{A}]_0 = -kt ln [ A ] t − ln [ A ] 0 = − k t
ln [ A ] t [ A ] 0 = − k t \ln\frac{[\text{A}]_t}{[\text{A}]_0} = -kt ln [ A ] 0 [ A ] t = − k t
Worked example Worked Example: Radioactive Decay14 C \mathrm{^{14}C} 14 C t 1 / 2 = 5730 years t_{1/2} = 5730\, \text{years} t 1/2 = 5730 years
Find : k k k
Solution :
Step 1 : Find k k k k = 0.693 5730 = 1.21 × 10 − 4 yr − 1 k = \frac{0.693}{5730} = 1.21 \times 10^{-4} \, \text{yr}^{-1} k = 5730 0.693 = 1.21 × 1 0 − 4 yr − 1
Why this step? Half-life formula rearranged.
Step 2 : Use integrated law
ln [ A ] t [ A ] 0 = − ( 1.21 × 10 − 4 ) ( 10000 ) = − 1.21 \ln\frac{[\text{A}]_t}{[\text{A}]_0} = -(1.21 \times 10^{-4})(10000) = -1.21 ln [ A ] 0 [ A ] t = − ( 1.21 × 1 0 − 4 ) ( 10000 ) = − 1.21
[ A ] t [ A ] 0 = e − 1.21 = 0.298 \frac{[\text{A}]_t}{[\text{A}]_0} = e^{-1.21} = 0.298 [ A ] 0 [ A ] t = e − 1.21 = 0.298
Why this step? Exponentiating both sides inverts the natural log. About 30% remains after 10,000 years (note: t = 10,000 t = 10{,}000 t = 10 , 000 k t = 1.21 × 10 − 4 × 10,000 = 1.21 kt = 1.21 \times 10^{-4} \times 10{,}000 = 1.21 k t = 1.21 × 1 0 − 4 × 10 , 000 = 1.21
Starting point : Rate proportional to concentration squared.
− d [ A ] d t = k [ A ] 2 -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^2 − d t d [ A ] = k [ A ] 2
WHY squared? Often means two molecules of A must collide. Probability of collision scales as [ A ] × [ A ] [\text{A}] \times [\text{A}] [ A ] × [ A ]
Separate variables :
d [ A ] [ A ] 2 = − k d t \frac{d[\mathrm{A}]}{[\mathrm{A}]^2} = -k \, dt [ A ] 2 d [ A ] = − k d t
Rewrite left side:
[ A ] − 2 d [ A ] = − k d t [\text{A}]^{-2} d[\text{A}] = -k \, dt [ A ] − 2 d [ A ] = − k d t
Integrate :
∫ [ A ] 0 [ A ] t [ A ] − 2 d [ A ] = − k ∫ 0 t d t \int_{[\text{A}]_0}^{[\text{A}]_t} [\text{A}]^{-2} d[\text{A}] = -k \int_0^t dt ∫ [ A ] 0 [ A ] t [ A ] − 2 d [ A ] = − k ∫ 0 t d t
[ − [ A ] − 1 ] [ A ] 0 [ A ] t = − k t \left[ -[\text{A}]^{-1} \right]_{[\text{A}]_0}^{[\text{A}]_t} = -kt [ − [ A ] − 1 ] [ A ] 0 [ A ] t = − k t
− 1 [ A ] t + 1 [ A ] 0 = − k t -\frac{1}{[\text{A}]_t} + \frac{1}{[\text{A}]_0} = -kt − [ A ] t 1 + [ A ] 0 1 = − k t
1 [ A ] t − 1 [ A ] 0 = k t \frac{1}{[\text{A}]_t} - \frac{1}{[\text{A}]_0} = kt [ A ] t 1 − [ A ] 0 1 = k t
Worked example Worked Example: Dimerization2 NO 2 ( g ) → N 2 O 4 ( g ) 2\text{NO}_2(g) \rightarrow \text{N}_2\text{O}_4(g) 2 NO 2 ( g ) → N 2 O 4 ( g ) NO 2 \text{NO}_2 NO 2
Given: [ NO 2 ] 0 = 0.0100 M [\text{NO}_2]_0 = 0.0100 \, \text{M} [ NO 2 ] 0 = 0.0100 M k = 0.543 M − 1 s − 1 k = 0.543 \, \text{M}^{-1}\text{s}^{-1} k = 0.543 M − 1 s − 1 t = 300 s t = 300 \, \text{s} t = 300 s
Find : [ NO 2 ] t [\text{NO}_2]_t [ NO 2 ] t
Solution :
1 [ NO 2 ] t = 1 0.0100 + ( 0.543 ) ( 300 ) \frac{1}{[\text{NO}_2]_t} = \frac{1}{0.0100} + (0.543)(300) [ NO 2 ] t 1 = 0.0100 1 + ( 0.543 ) ( 300 ) Why this step? Direct substitution into second-order law.
= 100 + 162.9 = 262.9 M − 1 = 100 + 162.9 = 262.9 \, \text{M}^{-1} = 100 + 162.9 = 262.9 M − 1
[ NO 2 ] t = 1 262.9 = 3.80 × 10 − 3 M [\text{NO}_2]_t = \frac{1}{262.9} = 3.80 \times 10^{-3} \, \text{M} [ NO 2 ] t = 262.9 1 = 3.80 × 1 0 − 3 M
Why this step? Taking reciprocal gives concentration. Concentration dropped to ~38% of initial because two NO 2 \text{NO}_2 NO 2
Order
Differential
Integrated
Linear Plot
Slope
Units of k k k
Half-life
0
− d [ A ] d t = k -\frac{d[\text{A}]}{dt} = k − d t d [ A ] = k [ A ] t = [ A ] 0 − k t [\text{A}]_t = [\text{A}]_0 - kt [ A ] t = [ A ] 0 − k t [ A ] [\text{A}] [ A ] t t t − k -k − k M s − 1 \text{M s}^{-1} M s − 1 [ A ] 0 2 k \frac{[\text{A}]_0}{2k} 2 k [ A ] 0
1
− d [ A ] d t = k [ A ] -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}] − d t d [ A ] = k [ A ] ln [ A ] t = ln [ A ] 0 − k t \ln[\text{A}]_t = \ln[\text{A}]_0 - kt ln [ A ] t = ln [ A ] 0 − k t ln [ A ] \ln[\text{A}] ln [ A ] t t t − k -k − k s − 1 \text{s}^{-1} s − 1 0.693 k \frac{0.693}{k} k 0.693
2
− d [ A ] d t = k [ A ] 2 -\frac{d[\text{A}]}{dt} = k[\text{A}]^2 − d t d [ A ] = k [ A ] 2 1 [ A ] t = 1 [ A ] 0 + k t \frac{1}{[\text{A}]_t} = \frac{1}{[\text{A}]_0} + kt [ A ] t 1 = [ A ] 0 1 + k t 1 [ A ] \frac{1}{[\text{A}]} [ A ] 1 t t t k k k M − 1 s − 1 \text{M}^{-1}\text{s}^{-1} M − 1 s − 1 1 k [ A ] 0 \frac{1}{k[\text{A}]_0} k [ A ] 0 1
Common mistake Common Error: Mixing Up Which Graph to UseWrong thinking : "I'll just plot [ A ] [{\rm A}] [ A ] t t t
Why it feels right : Simplest plot, and works for zero-order.
The fix : Each order linearizes with a different y-axis transformation:
Zero-order: [ A ] [\text{A}] [ A ] t t t
First-order: ln [ A ] \ln[\text{A}] ln [ A ] t t t
Second-order: 1 [ A ] \frac{1}{[\text{A}]} [ A ] 1 t t t How to remember : The integrated form tells you what to plot—the left side is your y-axis!
Common mistake Common Error: Forgetting That Units ofk k k Wrong thinking : "k k k
Why it feels right : First-order is most common, and there k k k s − 1 \text{s}^{-1} s − 1
The fix : Units must make the rate equation dimensionally consistent. Rate always has units M s − 1 \text{M s}^{-1} M s − 1 k [ A ] n must equal M s − 1 k[\text{A}]^n \text{ must equal } \text{M s}^{-1} k [ A ] n must equal M s − 1 k = M s − 1 [ M ] n = M 1 − n s − 1 k = \frac{\text{M s}^{-1}}{[\text{M}]^n} = \text{M}^{1-n}\text{s}^{-1} k = [ M ] n M s − 1 = M 1 − n s − 1
Recall Feynman Explain-to-a-12-Year-OldImagine you're playing a video game where enemies spawn and you have to shoot them.
Zero-order : Your gun fires exactly 10 bullets per second no matter how many enemies are on screen. If 5 enemies appear, you shoot 10 times. If 500 appear, still 10 times per second. The "rate" doesn't care about how many targets exist.
First-order : Now your gun is automatic and tracks enemies—if there are 10 enemies, it fires 10 shots per second. If 20 enemies, 20 shots. The rate is directly proportional to how many are there.
Second-order : You need to hit two enemies at once with a special laser that only fires when enemies line up. If you have 10 enemies, there might be 45 possible pairs. If you have 20 enemies, there are 190 pairs! The number of successful hits grows way faster because every enemy can pair with every other one. That's why the rate depends on concentration squared .
Mnemonic Order Memory Aid"Zero, One, Two = Flat, Ln, Divide"
Zero : Plot stays flat over time (constant rate), use [ A ] [\text{A}] [ A ] One : Use natural Ln of concentrationTwo : Divide (take reciprocal 1 [ A ] \frac{1}{[\text{A}]} [ A ] 1
Half-life mnemonic : "Zero Depends, One Never, Two Inverts"
Zero-order t 1 / 2 t_{1/2} t 1/2 depends on [ A ] 0 [\text{A}]_0 [ A ] 0
First-order never depends on [ A ] 0 [\text{A}]_0 [ A ] 0
Second-order is inverted (larger [ A ] 0 [\text{A}]_0 [ A ] 0 t 1 / 2 t_{1/2} t 1/2
#flashcards/chemistry
What is the integrated rate law for a zero-order reaction? [ A ] t = [ A ] 0 − k t [\text{A}]_t = [\text{A}]_0 - kt [ A ] t = [ A ] 0 − k t . Linear plot of
[ A ] [\text{A}] [ A ] vs
t t t with slope
− k -k − k .
For a first-order reaction, how does half-life depend on initial concentration? It doesn't.
t 1 / 2 = 0.693 k t_{1/2} = \frac{0.693}{k} t 1/2 = k 0.693 is constant regardless of
[ A ] 0 [\text{A}]_0 [ A ] 0 .
What are the units of the rate constant for a second-order reaction? M − 1 s − 1 \text{M}^{-1}\text{s}^{-1} M − 1 s − 1 or
L mol − 1 s − 1 \text{L mol}^{-1}\text{s}^{-1} L mol − 1 s − 1 .
How do you determine reaction order graphically? Plot
[ A ] [\text{A}] [ A ] vs
t t t (zero-order),
ln [ A ] \ln[\text{A}] ln [ A ] vs
t t t (first-order), or
1 [ A ] \frac{1}{[\text{A}]} [ A ] 1 vs
t t t (second-order). Whichever gives a straight line indicates the order.
Derive the first-order integrated rate law starting from − d [ A ] d t = k [ A ] -\frac{d[\text{A}]}{dt} = k[\text{A}] − d t d [ A ] = k [ A ] Separate variables:
d [ A ] [ A ] = − k d t \frac{d[\text{A}]}{[\text{A}]} = -k dt [ A ] d [ A ] = − k d t . Integrate:
∫ d [ A ] [ A ] = − k ∫ d t \int \frac{d[\text{A}]}{[\text{A}]} = -k \int dt ∫ [ A ] d [ A ] = − k ∫ d t , giving
ln [ A ] t − ln [ A ] 0 = − k t \ln[\text{A}]_t - \ln[\text{A}]_0 = -kt ln [ A ] t − ln [ A ] 0 = − k t , or
ln [ A ] t = ln [ A ] 0 − k t \ln[\text{A}]_t = \ln[\text{A}]_0 - kt ln [ A ] t = ln [ A ] 0 − k t .
Why does a second-order reaction have rate proportional to [ A ] 2 [\text{A}]^2 [ A ] 2 Because two molecules of A must collide. The collision frequency is proportional to the product of their concentrations:
[ A ] × [ A ] = [ A ] 2 [\text{A}] \times [\text{A}] = [\text{A}]^2 [ A ] × [ A ] = [ A ] 2 .
For a zero-order reaction, how does half-life change if you double the initial concentration? It doubles.
t 1 / 2 = [ A ] 0 2 k t_{1/2} = \frac{[\text{A}]_0}{2k} t 1/2 = 2 k [ A ] 0 , so doubling
[ A ] 0 [\text{A}]_0 [ A ] 0 doubles
t 1 / 2 t_{1/2} t 1/2 .
What is the relationship between rate constant units and reaction order n n n Units of
k k k are
M 1 − n s − 1 \text{M}^{1-n}\text{s}^{-1} M 1 − n s − 1 . For
n = 0 n=0 n = 0 :
M s − 1 \text{M s}^{-1} M s − 1 ;
n = 1 n=1 n = 1 :
s − 1 \text{s}^{-1} s − 1 ;
n = 2 n=2 n = 2 :
M − 1 s − 1 \text{M}^{-1}\text{s}^{-1} M − 1 s − 1 .
General Rate Law k A to n
1 over A = 1 over A0 + kt
Intuition Hinglish mein samjhoDifferential rate equations matlab ki reaction ka speed aur reactant ki concentration ke bech ka mathematical relationship hai. Socho ki tumhare pas ek bucket hai jisme pani hai—pani kitni tezi se kam ho raha hai ye depend karta hai ki reaction kaun se order ki hai.
Zero-order matlab jaise koi tap constant rate se pani de raha ho—bucket mein kitna bhi paani ho, rate same rahega. Formula simple hai: concentration linearly decrease hota hai time ke sath ([ A ] t = [ A ] 0 − k t [\text{A}]_t = [\text{A}]_0 - kt [ A ] t = [ A ] 0 − k t First-order mein jitna zyada paani hai, ut
