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Reaction Rates / Kinetics
• During chemical reaction, reactants are consumed, products are formed
• Amount of reactants decreases ↓ , Amount of products increases ↑
• Rate of reaction follows stoichiometric principles
Reaction A → B
• For every ONE A breakdown = ONE B will form
• Rate of decomposition A = Rate of formation of B
dt
Bd
dt
Ad ][][
2NO 2 → N2O4
• Two moles NO2 decompose = One mole of N2O4 form • NO2 used up is twice as fast as N2O4 produced
dt
Nd
dt
Nd ]0[1
2
]0[1 422
dt
HId
dt
Id
dt
Hd
2
][][1][1 22
H 2 + I2 → 2HI
• One mole H2 decompose = TWO moles of HI form • Rate of H2 and I2 decomposition are the same but only half the rate of HI formation
Rate of reaction can be defined as a change of property over time
X → Y (Reactants) • X decrease ↓ over time • X consumed ↓
(Products) • Y increase ↑over time • X formed ↑
Rate of Decrease of X
• Decrease ↓ Concentration X /time
• Decrease ↓ Volume X /Time
• Decrease ↓ Absorbance X /Time
Rate of Increase of Y
• Increase ↑ Concentration Y /time
• Increase ↑ Volume Y /Time
• Increase ↑ Absorbance Y /Time
Graphical Representation of Order of reactions, ZERO, FIRST and SECOND order
• Conc double x2 – Rate constant • Rate Vs Conc – Constant • Conc Vs Time – Linear
• Conc double x2 – Rate double x2 • Rate triple x3 – Rate triple x3 • Conc Vs Time – Linear
• Conc double x2 – Rate quadruple x 4 • Rate triple x3 – Rate increase x 9
• Rate = k[A]0
•Rate is independent of [A] • Rate = k[A]1
•Rate - 1st order respect to [A] • Rate = k[A]2
•Rate - 2nd order respect to [A]
•Unit for k (rate constant) •Rate = k[A]0
•Rate = k • k = Ms-1
•Unit for k (rate constant) •Rate = k[A]1
•Rate = kA • k = s-1
•Unit for k (rate constant) •Rate = k[A]2
•Rate = kA2
• k = M-1s-1
ZERO ORDER FIRST ORDER SECOND ORDER
1 → 1/2 → 1/4 → 1/8 1 → 1/2 → 1/4 → 1/8
1 → 1/2 → 1/4 → 1/8
10m 10m 10m 10m 5m 2.5m 10m 20m 40m
Graphical Representation of Half Life for ZERO, FIRST and SECOND order
ZERO ORDER FIRST ORDER SECOND ORDER
Half Life is directly proportional to Conc Half Life is independent of Conc (constant) Half Life is inversely proportional to Conc
Graphical Representation of Order of reactions, ZERO, FIRST and SECOND order
Reaction Law / Rate Expression
For a reaction: aA + bB → cC + dD
• Stoichiometry equation : Shows the mole ratio of reactants and products
• Rate equation : Equation relates rate of reaction with concentration of reactants
: How concentration of reactants affect the rate
Reaction equation = k[A]x[B]y x = rate order with respect to [A] y = rate order with respect to [B] (x +y) = overall order k = rate constant Rate order must be determined experimentally , NOT derived from stoichiometry coefficients
Order of reaction can be determined using THREE methods
Initial Rate method (Multiple Single Runs)
Conc Vs Time Method (Half Life)
Conc Vs Time Method (Whole Curve/Tangent Method)
• Multiple Single Runs • Varying and Keeping certain Conc fixed • Wasteful as multiple runs needed
• Monitoring the decrease in Conc of a single reactants • Using Half Life to determine the order
• Monitoring the decrease in Conc of a single reactants • Using gradients of tangents at different reactant conc
1st order
Zero order
2nd order
• Conc x2 – rate x2 - 1st order • Conc x2 – rate x4 – 2nd order • Conc x2 – rate x0 – zero order
Convert Conc Vs Time to Rate vs Conc
• Rate Vs Conc – straight line – 1st Order • Initial Rate is taken at time O by drawing a tangent at time O
Half Life is directly proportional to Conc
Half Life is independent of Conc (constant)
Half Life is inversely proportional to Conc
Using Conc Vs Time and Conc Vs Rate Method to determine Order of reaction
Reaction between 2A → B + C
Plot a graph of Conc A Vs Time to determine the order, initial rate and rate constant, k
Reaction between 2N205 → 4N02 + 02
Plot a graph of Rate Vs Conc to determine the order and rate constant, k
Conc Vs Time Method • Half Life for A is constant = 80s – 1st order respect to [A]
• Formula for 1st order half life t1/2 = 0.693/k = 0.693/80 = 8.66 x 10-3 s-1
Conc Vs Rate Method • Straight Line – 1st order respect to [N205]
• Rate = k[N205 ], k = gradient = 7.86 x 10-6 s-1
Using Initial rate and Half Life to determine order of reaction
Reaction on hydrolysis of ester by OH- given by Ester + OH- → X + Y
Reaction was done using two different OH- concentration
Run 1 – [OH- ] – 0.20M Run 2 – [OH- ] – 0.40M
Conc ester Vs Time was plotted. Determine the order and initial rate of reaction
Determine rate order for OH- (fix conc ester)
Let Rate = k[OH-]x [ester] y
Determine rate order for ester (Using Half Life )
Using run 2 : Conc ester vs Time
Half Life for Ester t1/2 = 11.5min (constant)
1st order with respect to ester
Calculate rate constant, k • Rate = k[OH-]1 [ester]1
For run 2 :
• Initial rate = 8.00 x10-5, [OH-]= 0.4M, [ester] = 0.001M
• Rate = k[OH-]1 [ester]1
• 8.00 x 10-5 = k[0.4]1[0.001]1 = k = 0.200M-1min-1
Half life : 0.001 → 0.0005 → 0.00025 is (11.5min) • Initial Rate Run 1 – Gradient at time 0 = (0.006 – 0.0006)/(10-0) = 4.00 x 10-5
• Initial Rate Run 2 – Gradient time 0 = (0.001 – 0.0002)/(10-0) = 8.00x10-5
1st order with respect to OH -
Reaction between RBr + OH- → ROH + Br-
Reaction was done using TWO different OH- concentration
Run 1 – [OH- ] – 0.10M Run 2 – [OH- ] – 0.15M
Conc RBr Vs Time was plotted. Determine the order and initial rate of reaction
Determine rate order for OH- (fix conc RBr)
Let Rate = k[OH-]x [RBr] y
Determine rate order for RBr (using half life)
Using run 2 : Conc vs Time
Half Life for RBr t1/2 = 78min (constant)
1st order with respect to RBr
Calculate rate constant, k • Rate = k[OH-]1 [RBr]1
• For run 1 : Initial rate = 5.25 x10-5, [OH-] = 0.10M, [RBr] = 0.01M
• Rate = k[OH-]1 [RBr]1
• 5.25 x10-5 = k[0.10]1[0.01]1 = k = 0.0525M-1min-1
Half life : 0.01 → 0.005 → 0.0025 Initial Rate Run 1 – Gradient time 0 = 5.25 x10-5
Initial Rate Run 2 – Gradient time 0 = 8.00 x10-5
Using Initial rate and Half Life to determine order of reaction
1st order with respect to OH -
Rate = k[OH-]1 [RBr]1
Reaction between CH3CO2C2H5 + H2O → CH3CO2H + C2H5OH
Reaction was done using TWO different HCI concentration
Run 1 – [HCI] – 0.10M Run 2 – [HCI ] – 0.20M
Conc CH3CO2C2H5 Vs Time was plotted. Determine the order and initial rate of reaction
Determine rate order for HCI (fix conc CH3CO2C2H5 )
Let Rate = k[HCI]x [CH3CO2C2H5] y
Half life is 31min (constant) Initial rate Run 1 – Gradient time 0 = 1.90 x10-3
Initial rate Run 2 – Gradient time 0 = 3.80 x10-3
Rate = k[HCI]1[CH3CO2C2H5]1
Using Initial rate and Half Life to determine order of reaction
Determine rate order for CH3CO2C2H5 (using half life)
Half life for run 2 -> 0.200 → 0.100 → 0.050
Half Life for CH3CO2C2H5 t1/2 = 31min (constant)
1st order with respect to CH3CO2C2H5
1st order with respect to HCI
Using Initial Rate Method to determine Order of reaction
Reaction between A + B → AB
Determine rate order for A (fix conc B )
Let Rate = k[A]x[B] y
Rate = k[A]2[B]1
Determine rate order for B (fix conc A )
Let Rate = k[A]x[B] y
Ex 1
2nd order with respect to A 1st order with respect to B
Reaction between F2 + 2CIO2 → 2FCIO2
Determine rate order for CIO2 (fix conc F2 )
Let Rate = k[F2]x [CIO2]
y
Determine rate order for F2 (fix conc CIO2)
Let Rate = k[F2]x [CIO2]
y
For expt 1 : Initial rate = 1.2 x 10-3 , [F2] = 0.10M, [CIO2] = 0.01M
Rate = k[F2]1[CIO2]
1
1.2 x 10-3 = k[0.10]1[0.01]1, k = 1.2M-1s-1
Using Initial Rate Method to determine Order of reaction
Ex 2
1st order with respect to CIO2 1st order with respect to F2
Rate = k [CIO2] 1[F2]1 To calculate k
Reaction between 2CIO2 + 2OH- → CIO3- + CIO2
- + H2O
Determine rate order for CIO2 (fix conc OH- )
Let Rate = k[CIO2]x[OH-]y
For expt 1 : Initial rate = 8.00 x 10-3 , [CIO2] = 0.025M, [OH-] = 0.046M
Rate = k[CIO2]2[OH-] 1
8.00 x 10-3 = k[0.025]1[0.046]1, k = 278.3M-1s-1
Determine rate order for OH- (fix conc CIO2 )
Let Rate = k[CIO2]x[OH-]y
Using Initial Rate Method to determine Order of reaction
2nd order with respect to CIO2 1st order with respect to OH-
Rate = k[CIO2]2[OH-] 1 To calculate k
Ex 3
Reaction between Br2 + 2NO → 2NOBr
Determine rate order for Br2 (fix conc NO )
Let Rate = k[Br2]x[NO]y
Determine rate order for NO (fix conc Br2 )
Let Rate = k[Br2]x[NO]y
For expt 1 : Initial rate = 12Ms-1, [Br2] = 0.10M, [NO] = 0.10M
Rate = k[Br2]1[NO]2
12 = k[0.10]1[0.10]2 , k = 12,000M-1min-1
Using Initial Rate Method to determine Order of reaction
To calculate initial rate when 0.1M Br2 and 0.3M NO are used
1st order with respect to Br2 2nd order with respect to NO
Rate = k[Br2]1[NO]2 To calculate k
Rate = k[Br2]1[NO]2
Rate = 12,000 x 0.1 x 0.1 2
Rate = 108Ms-1
Ex 4
• C3H8 + 5O2 → 3C02 + 4H2O
• 2H2 + 02 → 2H20
When rate of O2 decrease ↓ is at 0.23Ms-1, what is the rate of H2O formation/ increases ↑
When rate of C3H8 decrease ↓ is at 0.30Ms-1, what is the rate of 02 decrease ↓
Questions on Reaction Rates / Kinetics
EX 5
EX 6