# 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]/d****t =-d[B]/d****t = +d[C]/d****t**

**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.**

**Br _{2} + CH_{3}(CH_{2})_{3}CH=CH_{2 }^{______________}> CH_{3}(CH_{2})_{3}CHBr-CH_{2}Br_{ }**

**Brown colorless colorless**

**Rate of reaction = rate of disappearance of Bromine = -Δ****[Br _{2}]/Δ**

**t**

**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 **

**Br _{2} : molar mass = 159.81g/mol **

**Moles of Br _{2} = 0.02g/159.81 g/mol = 0.0001252mol, moles of product expected = 0.0001252mol This is smaller than amount from hexene therefore Br_{2} is LIMITING REACTANT**

**Since Br _{2} moles gives the smallest products in the reaction, Br_{2} is limiting reagent, so we use Br_{2 }concentration to calculate reaction rate. **

**Ignore the volume of 0.02g Br _{2} and just use the volume of hexene as total volume.**

**Concentration of Br _{2} =[Br_{2}] = 0.0001252mol/6.78 x 10^{-4} L = 0.1846 mol/L**

**Rate of disappearance of Br _{2} = –**

**D**

**[Br**

_{2}]/**D**

**t = -[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]/d****t**

Rate of reaction of B = rate of disappearance of B = (1/2) Δ[B]/Δt**=[(1/2)]x(-d[B]/d****t)**

Rate of FORMATION of C product = rate of appearance of C =Δ **[C]/Δ****t=+d[C]/d****t**

Therefore, for **A + 2B ^{________________}> C**

**-d[A]/d****t = (1/2) x (-d[B]/d****t) ****= +d[C]/d****t**

**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.**

**2Br _{2} + CH_{3}(CH_{2})_{2}C^{=}CH_{ }^{______________}> CH_{3}(CH_{2})_{2}CBr_{2}-CHBr_{2}**

**Brown colorless colorless**

**Rate of reaction = rate of disappearance of Bromine = – (1/****2) Δ[Br**_{2}**]/Δ****t **

**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 **

**Br _{2} : molar mass = 159.81 g/mol , density = 3.119g/ml**

**Moles of Br _{2} = 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 Br _{2} = 4.05g x 3.119g/ml = 12.632ml = 1.263 x 10^{-2}L**

**Total volume = 126.32 x 10 ^{-4} L + 7.15 x 10^{-4}L = 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]/Δ****t = – [ 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 Br _{2} = d[Br_{2}]/dt**

**-d[hexyne]/dt = rate of reaction = – (1/2) x [d[Br _{2}]/dt]**

**Therefore, d[Br _{2}]/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 }]**

**3**^{rd}** 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 = 2 ^{y }**

**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=2^{x }, 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)/8M ^{2} **

**k= 25/Ms . **

**You can calculate k value for each stage of the experiment and find the average value.**

** **

**Reaction Progress;**

**1**^{st} order reaction

^{st}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 ( t**_{0.5 }** ) of the reaction.**

**Answer:**

**Half-life (t _{0.5}) = 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 t _{0.5}= ln[Ao] – ln[At] **

**Therefore k t _{0.5}= ln(2[At]) – ln([At ]) **

**k t _{0.5 }= ln(2[At]/[At]) = ln2 **

**k t _{0.5}= 0.6932**

**t _{0.5 }=0.6932/k for first order reaction**

**k=25/s**

** t _{0.5 } = 0.6932/(25/s) = 0.02773 s **

** **

** **

**2**^{nd} order Reaction ;

^{nd}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 t _{0.5 } of the reaction if it is second order **

**overall.**

**ANSWER: **

**Half -life (t _{0.5}) = time for half reaction progress to occur = time for when concentration of substance equals half initial concentration, that is [At] = [Ao]/2**

**Therefore, kt _{0.5 }= 1/[At] – 1/[Ao] **

**becomes kt _{0.5 }= 2/[Ao] – 1/[Ao] = (2-1)/[Ao] = 1/[Ao]**

**kt _{0.5 } = 1/[Ao]**

**t _{0.5 } = 1/([Ao] k) = 1/ [2M x 25/Ms] = 1/ 50/s = 0.02s**