Rates of Reactions
Rate of a chemical reaction is the rate of formation of products or disappearance of reactants.
A + B ___________> C
Rate of reaction of A = rate of disappearance of A = change in A concentration/change in time=Δ[A]/Δt = -d[A]/dt
The sign delta Δ means change in amount. This becomes d delta for a small amount of change.
Negative sign means loss of substance or decrease in amount of substance with time.
Rate of reaction of B = rate of disappearance of B = Δ[B]/Δt=-d[B]/dt
Rate of FORMATION of C product = rate of appearance of C = Δ[C]/Δt=+d[C]/dt
Positive sign means gain or increase in amount of substance with time.
From the equation above A + B= C, we can deduce that,
-d[A]/dt =-d[B]/dt = +d[C]/dt
EXAMPLE 1
A chemical reaction of 0.02g Bromine liquid completely changed color from dark brown to colorless when reacted with 1.0g 1-hexene in 30 seconds. Calculate the rate of the reaction.
Br2 + CH3(CH2)3CH=CH2 ______________> CH3(CH2)3CHBr-CH2Br
Brown colorless colorless
Rate of reaction = rate of disappearance of Bromine = -Δ[Br2]/Δt
Molar mass of propene =84.16g/mol
Moles of hexene = 1.0g / 84.16 g/mol = 0.0119mol, moles expected product = 0.0119 mol
Density of hexene = 0.678g/ml
Therefore, volume of hexene = 1.0 g x 0.678g/ml = 0.678ml = 6.78 x 10-4 L
Br2 : molar mass = 159.81g/mol
Moles of Br2 = 0.02g/159.81 g/mol = 0.0001252mol, moles of product expected = 0.0001252mol This is smaller than amount from hexene therefore Br2 is LIMITING REACTANT
Since Br2 moles gives the smallest products in the reaction, Br2 is limiting reagent, so we use Br2 concentration to calculate reaction rate.
Ignore the volume of 0.02g Br2 and just use the volume of hexene as total volume.
Concentration of Br2 =[Br2] = 0.0001252mol/6.78 x 10-4 L = 0.1846 mol/L
Rate of disappearance of Br2 = –D[Br2]/Dt = -[0.1846 mol/L] / 30 s = -0.00615 M/s
Negative sign is nothing but just telling the amount was lost.
Hence the rate of REACTION = 0.00615 M/s
Rate of disappearance of hexene = -0.00615M/s
Rate of formation of product = 0.00615M/s
A + 2B _______________> C
Rate of reaction of A = rate of disappearance of A = Δ[A]/Δt = -d[A]/dt
Rate of reaction of B = rate of disappearance of B = (1/2) Δ[B]/Δt=[(1/2)]x(-d[B]/dt)
Rate of FORMATION of C product = rate of appearance of C =Δ [C]/Δt=+d[C]/dt
Therefore, for A + 2B ________________> C
-d[A]/dt = (1/2) x (-d[B]/dt) = +d[C]/dt
EXAMPLE 2
A chemical reaction of 4.05g Bromine liquid completely reacted with 1.0g 1-hexyne in 1minute. Calculate the rate of the reaction.
2Br2 + CH3(CH2)2C=CH ______________> CH3(CH2)2CBr2-CHBr2
Brown colorless colorless
Rate of reaction = rate of disappearance of Bromine = – (1/2) Δ[Br2]/Δt
Molar mass of hexyne =82.14 g/mol
Moles of hexyne = 1.0g / 82.14g/mol = 0.0121743 mol , therefore, Moles of product expected = 0.0121743 mol
Density of hexyne = 0.715g/ml
Therefore, volume of hexyne = 1.0 g x 0.715g/ml = 0.715ml = 7.15 x 10-4 L
Br2 : molar mass = 159.81 g/mol , density = 3.119g/ml
Moles of Br2 = 4.05g/159.81 g/mol = 0.02534mol, therefore, Mole of product expected = ½ (0.02534 mol) product = 0.0127mol
Since hexyne gives the smallest moles of expected products, hexyne is the limiting reagent, we use hexyne concentration to calculate the reaction rate.
Volume of Br2 = 4.05g x 3.119g/ml = 12.632ml = 1.263 x 10-2L
Total volume = 126.32 x 10-4 L + 7.15 x 10-4L = 133.47 x 10-4 L
Concentration of hexyne =[hexyne] = 0.000312872mol / 133.47 x 10-4 L = 0.023441 mol/L
Rate of disappearance of hexyne = – Δ[hexyne]/Δt = – [ 0.023441 mol/L] / 60 s = – 0.000391 M/s
Negative sign is nothing but just indicating that the amount was lost.
Hence the rate of REACTION = 0.000391M/s
Rate of disappearance of Br2 = d[Br2]/dt
-d[hexyne]/dt = rate of reaction = – (1/2) x [d[Br2]/dt]
Therefore, d[Br2]/dt = 2 x rate of reaction = – (2) x [0.000391 M/s] = -0.00078 M/s
RATE LAW
A _______> B
Rate = k[A]x where x is the order of reaction. k is rate constant at temperature T.
FIRST ORDER: reaction rate or speed DOUBLE increases with increase doubling the concentration of reactant A. [x=1, Rate = k[A] ]
SECOND ORDER: reaction rate Quadruples with increasing doubling concentration of reactant A. [x=2, Rate = k[A]2 ]
3rd order: Reaction rate increases EIGHT-fold with double increase in concentration of reactant A. [x=3 , Rate = k[A]3]
Zeroth Order: Reaction rate does not change with increase in concentration of reactant A. [x=0, Rate = k[A]0 ]
However, rate order can be decimal or fraction such as 0.5 or 1.5 and not always a whole number.
The mole concentrations of reactants in reaction equation can be used to derive rate expression but actual rate expression for a reaction can only be obtained from experiments.
e.g. A + B ________> C
Rate = k[A][B] implies reaction is first order in A and also first order in B but second order overall. x=1 for A , x=1 for B but sum total order = 1 + 1 = 2
EXAMPLE 3 :
Experimentally Rate = k[A]x[B]y , x= A reaction order , y = B reaction order . Both x and y can be determined from experimental results.
In a reaction; A + B _____ > C
| Rate M/s | [A] conc /M | [B] conc /M |
| 100 | 2 | 2 |
| 200 | 2 | 4 |
| 300 | 3 | 4 |
| 400 | 4 | 4 |
Find the rate law of the reaction in the above experiment.
100 = k[2]x[2]y ………1
200 = k[2]x[4]y ………..2
Dividing 2 by 1 gives;
200/100 = (4/2)y , find y.
2 = 2y
Ln2 = yLn2 , y = 1
400 = k[4]x[4]y ………..3
200 = k[2]x[4]y …………4
Dividing eqn 3 by eqn 4 gives;
400/200 = (4/2)x , 2=2x , x=1
Therefore, the Rate law is actually Rate = k[A][B]
Find the rate Constant k.
From eqn 2. ,
200 (M/s) = k[2]x[4]y = k[2][4]=k 8,
therefore, k = 200(M/s)/8M2
k= 25/Ms .
You can calculate k value for each stage of the experiment and find the average value.
Reaction Progress;
1st order reaction
x=1, Rate = k[A] = d[A]/dt ,where A = amount of substance A that is lost to form the products.
Ao ______> At, where Ao = original A amount, and At = amount of A at time t.
Therefore, amount or concentration of reactant A lost to form products = [A]= [Ao]-[At]
[ ] means concentration of substance in moles per liter.
[Ao] = initial concentration of A
[At] = concentration of A that is present at time t.
RATE = -d[A]/dt =k[A],
kdt = -d[A]/[A] , Integrating from zero to time t, from [Ao] to [At]
kt = -(ln[At] – ln[Ao])
kt = ln[Ao] – ln[At]
EXAMPLE 4
Calculate the concentration of reactant A after 2 seconds in the first chemical reaction of the rate experiments in EXAMPLE 3 , if reaction is first order in A and the rate constant is 25/s.
ANSWER:
Take initial concentration [Ao]=2M
kt =25/s x 2s = ln[2] – ln[At], find [At]
50 = 0.6932 – ln[At]
50 = 0.6932 – ln[At]
ln[At] = 0.6932 -50
[At] = e(0.6932 -50)
[At] = 3.86 x 10-22 M
Calculate the half-life ( t0.5 ) of the reaction.
Answer:
Half-life (t0.5) = time for half reaction progress to occur = time for when concentration of substance equals half initial concentration, that is
[At] = [Ao]/2 , therefore 2[At] = [Ao]
For a first order reaction, k t0.5= ln[Ao] – ln[At]
Therefore k t0.5= ln(2[At]) – ln([At ])
k t0.5 = ln(2[At]/[At]) = ln2
k t0.5= 0.6932
t0.5 =0.6932/k for first order reaction
k=25/s
t0.5 = 0.6932/(25/s) = 0.02773 s
2nd order Reaction ;
Rate = k[A]2 = d[A]/dt
Therefore, kdt = -(d[A]/[A]2)
kdt = -d(1/[A]) , where [A] =[Ao]-[At ]
therefore, integrating from time zero to time t, and from [Ao] to [At] gives
kt = -(1/[Ao] -1/[At])
kt = 1/[At] – 1/[Ao]
EXAMPLE 5
Calculate the concentration of reactant A after 2s in the first chemical reaction of the rate experiments in EXAMPLE 3, if the reaction is second order overall and the rate constant is 25M/s.
Answer:
Take initial concentration to be 2M and find concentration at time 2s , remember k=25/Ms
kt = 1/[At] – 1/[Ao]
25/Ms x 2s = 1 /[At] – 1/[2M]
50/M = 1 /[At] – 1/[2M]
50/M + 1/2M = 1 / [At]
50.5/M = 1/[At]
[At] = 1/[50.5/M] = 0.0198M
Calculate the half-life t0.5 of the reaction if it is second order overall.
ANSWER:
Half -life (t0.5) = time for half reaction progress to occur = time for when concentration of substance equals half initial concentration, that is [At] = [Ao]/2
Therefore, kt0.5 = 1/[At] – 1/[Ao]
becomes kt0.5 = 2/[Ao] – 1/[Ao] = (2-1)/[Ao] = 1/[Ao]
kt0.5 = 1/[Ao]
t0.5 = 1/([Ao] k) = 1/ [2M x 25/Ms] = 1/ 50/s = 0.02s
