Chemical Kinetics : Calculation of the Order

Methods for Determination of Order of a reaction

(1)     Substitution method in integrated rate equation: (Hit and Trial method)

(i)      The method can be used with various sets of a, x, and t with integrated rate equations.

(ii)     The value of k is determined and checked for all sets of  and t.

(iii)    If the value of k is constant, the used equation gives the order of reaction.

(iv)    If all the reactants are at the same molar concentration, the kinetic equations are :

k =  \frac { 2.303 }{ t }   log10  \frac { a }{ (a-x) }   (For first order reactions)

k = \quad \frac { 1 }{ t } \left[ \frac { 1 }{ a } -\frac { 1 }{ a-x } \right]   (For second order reactions)

k = \quad \frac { 1 }{ 2t } \left[ \frac { 1 }{ \left( a-x \right) ^{ 2 } } -\frac { 1 }{ a^{ 2 } } \right]   (For third order reactions)

 

(2)     Half life method :  This method is employed only when the rate law involved only one concentration term.

t1/2 ∞ a1–n ; t1/2 = ka 1–n ; logt1/2 = log k + (1 – n) log a, a plotted graph of logt1/2 vs log a  gives a straight line with slope (1–n), determining the slope we can find the order n. If half life at different concentration is given then.

(t1/2)1 ∝  \frac { 1 }{ a_{ 1 }^{ n-1 } } .  (t1/2)2  ∝  \frac { 1 }{ a_{ 2 }^{ n-1 } } ; \frac { (t_{ 1/2 })_{ 1 } }{ (t_{ 1/2 })_{ 2 } } = \left( \frac { a_{ 2 } }{ a_{ 1 } } \right) ^{ n-1 }       

log10(t1/2)1 – log10(t1/2)2 = (n – 1) [log10 a2 – log10 a1]

n =1 +  \frac { log_{ 10 }(t_{ 1/2 })_{ 1 }-log_{ 10 }(t_{ 1/2 })_{ 2 } }{ (log_{ 10 }a_{ 2 }-log_{ 10 }a_{ 1 }) }

This relation can be used to determine order of reaction ‘n’

Plots of half-lives Vs concentrations (t1/2 µ a1–n)

(3)     Graphical method :  A graphical method based on the respective rate laws, can also be used.

(i)      If the plot of log (a – x) Vs t is a straight line, the reaction follows first order.

(ii)     If the plot of  \frac { 1 }{ (a-x) }   Vs t is a straight line, the reaction follows second order.

(iii)    If the plot of  \frac { 1 }{ (a-x)^{ 2 } } Vs t is a straight line, the reaction follows third order.

(iv)    In general, for a reaction of nth order, a graph of  \frac { 1 }{ (a-x)^{ n-1 } }   Vs t must be a straight line.

 

Graphical determination of order of the reaction

Order
 Equation Straight line plot Slope Intercept on Y-axis
Y- axis X-axis
Zero x = k0t x t k0 0
First  log10 (a-x) = \frac { -k_{ 1 }t }{ 2.303 } + log10 a Log10(a–x)    \frac { -k }{ 2.303 } log10a
Second (a–x)–1 = k2t + a-1 (a–x)–1   k2 a–1
nth (a–x)1–n = (n – 1)knt + a1–n (a–x)1–n   (n–1)kn a1–n

 

Plots from integrated rate equations

 

Plots of rate Vs concentrations [Rate = k(conc.)n ]

(4)     Van’t Haff differential Method :  The rate of reaction varies as the nth power of the concentration Where n is the order of the reaction. Thus for two different initial concentrations C1 and C2 equation, can be written in the form,  \frac { -dC_{ 1 } }{ dt } =kC_{ 1 }^{ n } and \frac { -dC_{ 2 } }{ dt } =kC_{ 2 }^{ n }

            Taking logarithms, log10  \left( \frac { -dC_{ 1 } }{ dt } \right) = log10 k + nlog10 C1        …..(i)

            and log10 \left( \frac { -dC_{ 2 } }{ dt } \right)   = log10 k + n log10 C2                                   …..(ii)

            Subtracting equation (ii) from (i),

             n = \frac { log_{ 10 }\left( \frac { -dC_{ 1 } }{ dt } \right) -log_{ 10 }\left( \frac { -dC_{ 2 } }{ dt } \right) }{ log_{ 10 }C_{ 1 }-log_{ 10 }C_{ 2 } }                                            …..(iii)

\frac { -dC_{ 1 } }{ dt }   and  \frac { -dC_{ 2 } }{ dt }   are determined from concentration Vs time graphs and the value of  can be determined.

 

(5)     Ostwald’s isolation method (Initial rate method) : This method can be used irrespective of the number of reactants involved e.g., consider the reaction, n1A + n2B + n3C → Products.

This method consists in finding the initial rate of the reaction taking known concentrations of the different reactants (A, B, C). Now the concentration of one of the reactants is changed (say that of A) taking the concentrations of other reactants (B and C) same as before. The initial rate of the reaction is determined again. This gives the rate expression with respect to A and hence the order with respect to A. The experiment is repeated by changing the concentrations of B and taking the same concentrations of A and C and finally changing the concentration of C and taking the same concentration of A and B. These will give rate expressions with respect to B and C and hence the orders with respect to B and C respectively. Combining the different rate expressions, the overall rate expression and hence the overall order can be obtained.

Suppose it is observed as follows:

(i)      Keeping the concentrations of B and C constant, if concentration of A is doubled, the rate of reaction becomes four times. This means that,           Rate α [A]2 i.e., order with respect to A is 2

(ii)     Keeping the concentrations of A and C constant, if concentration of B is doubled, the rate of reaction is also doubled. This means that, Rate µ [B]    i.e., order with respect to B is 1

(iii)    Keeping the concentrations of A and B constant, if concentration of C is doubled, the rate of reaction remains unaffected. This means that rate is independent of the concentration of C i.e., order with respect to C is zero. Hence the overall rate law expression will be, Rate = k[A]2 [B] [C]0

            ∴  Overall order of reaction = 2 + 1 + 0 = 3.

 

Molecularity of Reaction

“It is the sum of the number of molecules of reactants involved in the balanced equation”.

Or

“It is the minimum number of reacting particles (Molecules, atoms or ions) that collide in a rate determining step to form product or products”.

Molecularity of a complete reaction has no significance and overall kinetics of the reaction depends upon the rate determining step. Slowest step is the rate-determining step. This was proposed by Van’t Hoff.

Example :               NH4NO2 → N2 + 2H2O                  (Unimolecular)

                               NO + O3 → NO2 + O2                            (Bimolecular)

                               2NO + O2 → 2NO2                         (Trimolecular)

  • Molecularity of a reaction can’t be Zero, negative or fractional.
  • Molecularity of a reaction is derived from the mechanism of the given reaction.
  • Molecularity cannot be greater than three because more than three molecules may not mutually collide with each other.

A2 +  \frac { 3 }{ 2 }   B2 → A2B3 

Mechanism

            Step I         :         2A → A2                                 (Bimolecular)

            Step II       :           A2 +  \frac { 1 }{ 2 }   B2 → A2B             (Trimolecular)

            Step III     :         A2B + B2 → A2B3                  (Bimolecular)

Example : Decomposition of H2O2

                     2H2O2 → 2H2O + O2            (Overall reaction Mechanism)

                     H2O2 → H2O + O                  (Slow)

                     H2O + O → H2O + O2            (Fast)

                     Rate = K[H2O2] ;  The reaction is unimolecular   

(1)     Pseudo Unimolecular Reaction : Reaction whose actual order is different from that expected using rate law expression are called pseudo-order reaction. For example, RCl + H2O → ROH + HCl

Expected rate law :  Rate = k[RCl][H2O] ; Expected order = 1 + 1 =2

Actual rate law :  Rate = k[RCl] ; Actual order = 1

Because of water is taken in excess amount; therefore, its concentration may be taken constant. The reaction is therefore, pseudo first order. Similarly the acid catalysed hydrolysis of ester, viz.,

RCOOR’ + H2O ⇌ RCOOH + R’OH

(follow first order kinetic) : Rate = k[RCOOR’]

Those reactions which may have order of reaction as one while molecularity of reaction 2 or more than two are as follows :

Examples :

(i)      2N2O5 → 4NO2 + O2 ; Order = 1 ; molecularity = 2

(ii)     CH3COOC2H5 + H2O  \underrightarrow { \quad H^{ + }\quad }   CH3COOH + C2H5OH ;

          r = k[CH3COOC2H5]            Order = 1, Molecularity = 2

  (iii)    inversion of cane sugar :

            C12H22O11 + H2O \underrightarrow { \quad H^{ + }\quad }   C6H12O6 + C6H12O6

            Sucrose               glucose                fructose

                     Order = 1, Molecularity = 2

  (iv)    (CH3)3CCl + OH– → (CH3)3COH + Cl–

                             Order = 1, Molecularity = 2

  (v)     2H2O2 \underrightarrow { \quad Pt\quad }   2H2O + O2

                      Order = 1, Molecularity = 2

 

Difference between Molecularity and Order of reaction

Molecularity Order of Reaction
It is the number of molecules of reactants terms taking part in elementary step of a reaction. It is sum of the power of the concentration terms of reactants in the rate law expression.
Molcularity is a theoretical value Order of a reaction is an experimental value
Molecularity can neither be zero nor fractional. Order of a reaction can be zero, fractional for integer.
Moleculaity has whole number values only i.e., 1, 2, 3, etc. Order of a reaction may have negative value.
It is assigned for each step of mechanism separately. It is assigned for overall reaction.
It is independent of pressure and temperature. It depends upon pressure and temperature.