Interpretation of Batch Reactor Data

Interpretation of Batch Reactor Data

Reactor Design and Rate Equations

• The primary goal of reactor design is to determine the reactor's size (volume) for a specific conversion.

• This determination requires the reactor's design equation, which includes a rate equation or rate law.

• A complete rate equation is essential for determining the required reactor size.

Determining Reaction Order and Rate Constant

• This chapter focuses on methods to determine the order of reaction ($n$) and the rate constant ($k$) using kinetic data.

• Kinetic data involves concentration as a function of time (e.g., $C<em>{A0}, C</em>{A1}, C<em>{A2}$ at $t = 0, t</em>1, t_2$).

• The determined values of $n$ and $k$ are used in the rate equation: $-r<em>A = kC</em>A^n$.

• Example: If $n = 1$ and $k = 0.015 \text{ min}^{-1}$, the rate equation is -rA = 0.015 CA \,\text{mol/(L\cdot min)}.

Two-Step Process for Rate Equation Determination

1. Concentration Dependency: Find the reaction order at a fixed temperature.

2. Temperature Dependency: Find the rate constant ($k$) and its variation with temperature ($T$).

Rate Equation for Irreversible Reactions

• For an irreversible reaction like $aA + bB \rightarrow \text{Products}$, the rate equation is:
  $-r<em>A = k C</em>A^\alpha C_B^\beta$

• To use this equation in reactor design, the constants $k$, $\alpha$, and $\beta$ must be known.

• These constants are determined experimentally by carrying out reaction runs, collecting kinetic data, and analyzing it.

Experimental Batch Reactor

• This chapter uses an experimental batch reactor to obtain rate and kinetic data.

• A batch reactor is a container that holds the reactants while they react.

• Experimental batch reactors are typically operated isothermally and are primarily used for collecting homogeneous kinetic data.

Collecting Kinetic Data

• Kinetic data for a chemical reaction is collected in an experimental batch reactor.

• Measurements of concentration, pressure, or volume are taken at various times during the reaction.

Methods for Collecting Kinetic Data

• Several methods exist for collecting kinetic data:

  • (i) Concentration Measurement: Measure the concentration of a reaction participant (usually the limiting component) at various times.

  • (ii) Total Pressure Measurement: Measure the total pressure of a constant-volume reaction system at different times.

  • (iii) Volume Measurement: Measure the volume of a constant-pressure reaction system at different times.

  • (iv) Physical Property Measurement: Measure a physical property of the fluid (e.g., refractive index, electrical conductivity) at different times.

Phase-Specific Measurements

• Liquid Phase Reactions: Concentration-time measurements are commonly used.

• Gas Phase Reactions: Pressure-time or volume-time measurements are used, especially when the number of moles changes.

Constant-Volume Batch Reactor for Gas Phase Reactions

• In a constant-volume batch reactor, the total pressure changes as the reaction proceeds (e.g., $2A \rightarrow R$).

• Measurements of total pressure are taken at different times.

• The reaction system may increase (e.g., $A \rightarrow 2R$) or decrease (e.g., $2A \rightarrow R$) in pressure.

Constant-Pressure Batch Reactor for Gas Phase Reactions

• In a constant-pressure batch reactor, the volume of the reaction system changes as the reaction proceeds.

• The volume may increase (e.g., $A \rightarrow 2R$) or decrease (e.g., $2A \rightarrow R$).

• Measurements of the reaction system's volume are taken at different times.

Kinetic Runs and Data Tabulation

• Batch reactor experiments performed this way are called kinetic runs.

• The data obtained are called kinetic data.

• The collected data are usually tabulated as:

  • (i) $C<em>{A1}, C</em>{A2}, C<em>{A3}, \dots$ at $t</em>1, t<em>2, t</em>3, \dots$

  • (ii) $P<em>1, P</em>2, P<em>3, P</em>4, P<em>5, \dots$ at $t</em>1, t<em>2, t</em>3, t<em>4, t</em>5, \dots$

  • (iii) $V<em>1, V</em>2, V<em>3, V</em>4, V<em>5, \dots$ at $t</em>1, t<em>2, t</em>3, t<em>4, t</em>5, \dots$

Methods for Analyzing Rate Data

• Data are used to find a rate equation that fits the data satisfactorily.

• Various methods for interpreting kinetic data/rate data to determine reaction order:

  • (i) Integral Method

  • (ii) Differential Method

  • (iii) Initial Rate Method

  • (iv) Half-Life Method

  • (v) Ostwald's Isolation Method

Comparison of Methods

• Integral and differential methods require only one experiment to find the specific reaction rate (rate constant) and reaction order.

• Half-life and initial rate methods require multiple experiments at different initial conditions.

Integral vs. Differential Method

• Integral Method:

  • Assume a rate equation, integrate, and check if the plot of a concentration function vs. time yields a straight line.

  • If a reasonably good straight line is obtained, the assumed rate equation fits the data.

• Differential Method:

  • Assume a rate equation and test its fit directly without integration.

  • First, find $\frac{dN}{dt}$ from the data before fitting.

Ease of Use and Complexity

1. Integral method is easier to use for testing data fit.

2. Differential method is more complicated and time-consuming.

Utility

1. Integral method is useful for testing relatively simple rate equations.

2. Differential method is useful for testing more complicated rate expressions.

Data Availability

1. Integral method can be used with scattered data.

2. Differential method requires accurate and large amounts of data.

Method Capability

1. Integral method can only test specific rate equation forms.

2. Differential method can evolve the rate equation to fit the data.

General Recommendation

• Start with the integral method.

• If testing different simple rate equations fails, use the differential method.

Constant-Volume Batch Reactor Defined

• Constant-volume batch reactor refers to the volume of the reaction system (reaction mixture), not the volume of the reactor itself.

• It means the reaction system undergoes no change in volume as the reaction progresses.

• These systems are also called constant density systems.

Density and Mass Conservation

• Density is defined as mass/volume. $\rho = \frac{m}{V}$

• For a given reaction system, its mass remains constant (law of conservation of mass).

• If volume is constant, density is also constant.

Examples of Constant-Volume Systems

1. Gas Phase Reactions in Sealed Vessels: Carried out in a closed vessel to measure pressure and temperature.

   • The volume of the reaction mixture will not change, resulting in gas mixture compression and pressure increase.

   • The laboratory batch reactor is a typical example.

2. Gas Phase Reactions with No Change in Number of Moles: Where the number of moles of product equals the number of moles of reactant.
   *Consider the reaction:$CO + H<em>2O \rightleftharpoons CO</em>2 + H_2$
   *Two moles of reactants form two moles of products; thus the volume does not change at a given temperature and pressure

3. Almost All Liquid Phase Reaction Systems: There is no change in density during the course of reaction.

Rate of Reaction

• The rate of reaction of any component $i$ is given by: $r<em>i = \frac{1}{V} \frac{dN</em>i}{dt}$. (2.1)

• For constant-volume systems, V is taken inside the differential: $r<em>i = \frac{d(N</em>i/V)}{dt} = \frac{dC<em>i}{dt}$, as $C</em>i = N_i/V$. (2.3)

• For reactant A disappearing, $-r<em>A = (-\frac{dC</em>A}{dt})$.

Alternative Formulation

We have: $r<em>i = \frac{1}{V} \frac{dN</em>i}{dt}$
Also $N<em>i = C</em>i V$
So: $r<em>i = \frac{1}{V} \frac{d(C</em>i V)}{dt}$
$r<em>i = \frac{1}{V} [V \frac{dC</em>i}{dt} + C<em>i \frac{dV}{dt}]$ For constant-volume systems, $\frac{dV}{dt} = 0$ as $V = V</em>0$, and thus $dV = 0$
$r<em>i = \frac{1}{V} [V \frac{dC</em>i}{dt}] = \frac{dC<em>i}{dt}$ For reactant A: $-r</em>A = (+\frac{dC_A}{dt})$

Ideal Gas Law Application

• If A is a gaseous material and the ideal gas law applies:

  • $C<em>A = \frac{N</em>A}{V} = \frac{P_A}{RT}$ (2.5)

  • $-r<em>A = -\frac{dC</em>A}{dt} = -\frac{d(P<em>A/RT)}{dt} = \frac{-1}{RT} \frac{dP</em>A}{dt}$

• For any component $i$ in gaseous form: $r<em>i = \frac{1}{RT} \frac{dP</em>i}{dt}$ (2.6)

Concentration and Conversion for Constant-Volume Systems

*To determine the rate of reaction as a function of conversion, know how the concentration of the reactant varies with its conversion
*The concentration of A is the number of moles of A per unit volume: $C<em>A = \frac{N</em>A}{V}$ (2.7)
*Where $C<em>A$ is the concentration and $N</em>A$ is the number of moles of A remaining (unreacted) at any time $t$ during the course of reaction.
Let $X<em>A = \frac{\text{moles of A reacted}}{\text{initial moles of A}} = \frac{N</em>{A0} - N<em>A}{N</em>{A0}}$
So $N<em>{A0}X</em>A$ is the moles of A reacted
Material balance of A: $N<em>A$ unreacted = A Initially - A Reacted $N</em>A = N<em>{A0} - N</em>{A0}X<em>A$ $N</em>A = N<em>{A0}(1-X</em>A)$
Substituting $N<em>A$ into equation 2.7: $C</em>A = \frac{N<em>A}{V} = \frac{N</em>{A0}(1-X_A)}{V}$

For a constant volume system:
$V = V<em>0$ $C</em>A = \frac{N<em>{A0}(1-X</em>A)}{V<em>0}$ Since $\frac{N</em>{A0}}{V<em>0} = C</em>{A0}$
$C<em>A = C</em>{A0}(1-X_A)$
$This equation relates the concentration of A at any time t to the initial concentration of A through the fractional conversion of A, for any constant-volume system$

Finding Concentrations of Other Components

Consider the reaction: $aA + bB \rightarrow cC + dD$
Let $N<em>{A0}$ moles of A and $N</em>{B0}$ moles of B be in the reactor at $t = 0$.
Let $X<em>A$ be the fractional conversion of A at any time $t$. Moles of A reacted/consumed = $N</em>{A0}X<em>A$ Moles of A remaining : $N</em>A = N<em>{A0}(1-X</em>A)$
Rewrite the reaction as: $A + \frac{b}{a}B \rightarrow \frac{c}{a}C + \frac{d}{a}D$
For every mole of A reacted $b/a$ moles of B must react.
Moles of B Reacted = $\frac{b}{a} \times \text{moles of A reacted} = \frac{b}{a}N<em>{A0}X</em>A$
Moles of B remaining at any time t are:
$N<em>B = N</em>{B0} - \frac{b}{a}N<em>{A0}X</em>A$
The concentration of B at any time t is given by
$C<em>B = \frac{N</em>B}{V} = \frac{N<em>{B0} - (b/a)N</em>{A0}X_A}{V}$

For constant volume batch systems
$C<em>B = \frac{N</em>{B0} - (b/a)N<em>{A0}X</em>A}{V<em>0} = C</em>{B0} - \frac{b}{a}C<em>{A0}X</em>A$ (2.8)
If $b=a$:
$C<em>B = C</em>{B0} - C<em>{A0}X</em>A$ or
$C<em>B = C</em>{B0}(1-X_B)$

Because $X<em>B = \frac{\text{moles of B reacted}}{\text{moles of B initially present}}$ Moles of B reacted $= N</em>{B0}X<em>B = (b/a)N</em>{A0}X<em>A$ $N</em>{B0}X<em>B = (b/a)N</em>{A0}X<em>A$ Dividing both sides by $V</em>0$
$C<em>{B0}X</em>B = \frac{b}{a}C<em>{A0}X</em>A$
If $b=a$:
$C<em>{A0}X</em>A = C<em>{B0}X</em>B$

Similarly:
$C<em>C = \frac{N</em>C}{V} = \frac{N<em>{C0} + (c/a)N</em>{A0}X<em>A}{V}$ For $V = V</em>0$
$C<em>C = C</em>{C0} + (c/a)C<em>{A0}X</em>A$ (2.9)
And
$C<em>D = C</em>{D0} + (d/a)C<em>{A0}X</em>A$ (2.10)

For flow systems the concentration of A at a given point in the system is given by
$C<em>A = F</em>A/v$ (2.11)
Where v is the volumetric flow rate at that time. If $F<em>{A0}$ is the entering flow rate to the reactor in moles/time and $X</em>A$ is the conversion of A, then:
Change in moles of A within the reactor = $-F<em>{A0}X</em>A$
Molal flow rate of A from the reactor = $F<em>A = F</em>{A0} (1-X_A)$

For liquids, the change in volume with reaction is negligible, so
$v = v<em>0$ The concentration of A is given by $C</em>A = \frac{F<em>A}{v</em>0} = \frac{F<em>{A0}(1-X</em>A)}{v<em>0}$ and $C</em>A = C<em>{A0}(1-X</em>A)$(for constant-volume systems) (2.12)

However, for gas phase reactions, the volumetric flow rate most often changes during the course of the reaction due to the changing number of moles or temperature or pressure.

Similarly for component B
$C<em>B = \frac{F</em>B - (b/a)F<em>{A0}X</em>A}{v}$
For $v = v<em>0$ $C</em>B = \frac{F<em>B - (b/a)F</em>{A0}X<em>A}{v</em>0}$
$C<em>B = C</em>{B0} - (b/a)C<em>{A0}X</em>A$ (2.15)

Where $C<em>{A0} = \frac{F</em>{A0}}{v_0} = \frac{\text{mol}}{l}$ or $\frac{\text{mol}}{dm^3}$ or $\frac{\text{kmol}}{m^3}$

Integral Method of Analysis of Rate Data

For determining the order of reaction by the integral method, assume the reaction order and integrate the differential rate equation. For our guess (regarding the order) to be correct, a plot of concentration function vs time has to be linear (should give a straight line). If the plot is non-linear, then the assumed value for the reaction order is incorrect (our guess is not correct) and a new value of it should be tried. The integral method uses a trial-and-error procedure to determine the reaction order. This method is often used to evaluate the rate constants at different temperatures to determine the activation energy when the reaction order is known.

The general stepwise procedure of analyzing the kinetic data by the integral method is given below:

(i) Assume/Guess a rate form (rate expression). For a constant-volume system, the equation for the disappearance of A is of the form:
$-r<em>A = -\frac{dC</em>A}{dt} = k f(C)$ (2.21)
For example, for $A \rightarrow R$, if we assume the disappearance of A to be second order, then the rate form is: $-r<em>A = -\frac{dC</em>A}{dt} = kC_A^2$

(ii) Rearrange the rate equation to get it in the following form:
$-\frac{dC<em>A}{f(C)} = k dt$ (2.22) Here $f(C)$ only involves concentrations of the reacting materials, which may be expressed in terms of $C</em>A$. Hence, integrate equation (2.22) either analytically or graphically to give
$-\int<em>{C</em>{A0}}^{C<em>A} \frac{dC</em>A}{f(C)} = \int_{0}^{t} k dt = kt$ (2.23)

(iii) From the experimental concentration versus time data, determine the numerical values of the integral of equation (2.23) for various t and construct a plot of the concentration function, i.e., the integral of equation (2.23) against time. As the concentration function (so obtained) is proportional to time, the plot should yield a straight line with a slope equal to k

(iv) If the above plot yields a straight line passing through the origin then it may be said that the assumed rate equation (being tested) satisfactorily fits the data, i.e., our guess regarding the order of reaction is correct. The slope of such plot gives us the value of the rate constant. If we do not get a straight line then the assumed rate equation (i.e., order of reaction) is rejected as it being incorrect and another rate equation is to be tried.

In the integral method of analysis of rate data, we must know the appropriate concentration function corresponding to a particular rate equation that is proportional to the time (i.e., linear with time). One must be thoroughly familiar with the methods of obtaining linear plots for first, second, etc. order reactions.

Integrated Rate Equations for Different Order Reactions

Irreversible Unimolecular-Type First-Order Reactions

When the rate of reaction is directly proportional to the concentration of only one reactant, the reaction is said to be of the first order.
Consider the reaction:
$A \rightarrow \text{Products}$ (2.24 a)
The rate of disappearance of A is given by
$-r<em>A = -\frac{dC</em>A}{dt} = kC<em>A$ (2.24 b) Rearranging, we get $\frac{dC</em>A}{C<em>A} = k dt$ Integrating the above equation within the limits: At time 0, $C</em>A = C<em>{A0}$ At time t, $C</em>A = C<em>A$ $\int</em>{C<em>{A0}}^{C</em>A} \frac{dC<em>A}{C</em>A} = \int<em>{0}^{t} k dt$ $-[ln C</em>A]<em>{C</em>{A0}}^{C<em>A} = k [t]</em>0^t$
$-ln \frac{C<em>A}{C</em>{A0}} = kt$
$ln \frac{C<em>{A0}}{C</em>A} = kt$ (2.25)
$k = \frac{1}{t} ln \frac{C<em>{A0}}{C</em>A}$ (2.26)
Equation (2.25) is the integrated rate equation for the first-order reaction in terms of concentration.

Characteristics

It follows that k depends only on the ratio $C<em>A/C</em>{A0}$, and is measured in units of reciprocal time, (t-1).
Hence, the rate constant of the first order reaction will be expressed in sec-1, min-1, etc. and is independent of the units of concentration.
A plot of $-ln \frac{C<em>A}{C</em>{A0}}$ versus t would give a straight line passing through the origin showing that the reaction is first order. The value of first order rate constant is equal to the slope of the line.

Half Life

An important characteristic of the first order kinetics is that the half life is independent of the initial concentration of the reactant $t_{1/2} = \frac{0.693}{k}$

$t = \frac{ln \frac{C<em>{A0}}{C</em>A}}{k} = \frac{2.303 log \frac{C<em>{A0}}{C</em>A}}{k}$
$t<em>{1/2} = \frac{2.303}{k}log2$ $t</em>{1/2} = \frac{0.693}{k}$ (2.27)
Equation (2.27) gives a direct relation between half-life and rate constant.

Decomposition of nitrogen pentaoxide in the gas phase as well as in organic solvent like CCl4, hydrolysis of methyl acetate and inversion of cane sugar are some examples of first order kinetics.

Now we will develop the integrated rate expression for first order reaction in terms of conversion.

Fractional conversion (or simply conversion) is a convenient variable often used in place of concentration in engineering work. Therefore, most of the results which follow will be presented in terms of both CA and XA.

The fractional conversion $X_A$ of reactant A is defined as the fraction of reactant converted to product.

Let $N<em>{A0}$ and $N</em>A$ be the moles of A at t=0 and t=t respectively, then
$X<em>A = \frac{\text{moles A reacted}}{\text{initial moles of A}} = \frac{N</em>{A0} - N<em>A}{N</em>{A0}}$
$X<em>A = 1 - \frac{N</em>A}{N<em>{A0}}$ $1 - X</em>A = \frac{N<em>A}{N</em>{A0}}$
$N<em>A = N</em>{A0}(1 - X_A)$

For a first order reaction of the type $A \rightarrow \text{Products}$
The rate is given by: $-r<em>A = -\frac{dC</em>A}{dt} = kC<em>A$ (2.28) We have: $C</em>A = \frac{N<em>A}{V} = \frac{N</em>{A0}(1-X<em>A)}{V} = C</em>{A0}(1-X<em>A)$ On differentiation it gives: $dC</em>A = -C{A0}dX<em>A$ Putting values of $dC</em>A$ and $C<em>A$ in equation (2.28), we get $C</em>{A0}\frac{dX<em>A}{dt} = k C</em>{A0}(1 - X<em>A)$ Rearranging and integrating gives $\int</em>{0}^{X<em>A} \frac{dX</em>A}{1 - X<em>A} = k\int</em>{0}^{t} dt$
$-ln(1 - X<em>A) = kt$ (2.29) A plot of $-ln(1 - X</em>A)$ versus t yields a straight line passing through the origin with a slope equal to k. Equation (2.29) is the integrated rate equation for first order reaction in terms of conversion.

Irreversible Bimolecular-Type Second Order Reactions

When the rate of reaction is directly proportional to the concentrations raised to power unity of two different reactants or to the square of the concentration of one reactant, the reaction is said to be of second order.

1. For the reaction $2A \rightarrow Products$

2. For the reaction $A + B \rightarrow Products$, with equal concentrations of A and B, the expression can be given as:
   $-r<em>A = -\frac{dC</em>A}{dt} = k C_A^2$ (2.30)

For $A + B \rightarrow Products, -r<em>A = kC</em>A C<em>B = kC</em>A^2$ since $C<em>A = C</em>B$ as the stoichiometric coefficient of both A and B is one (the same).
$\frac{dC<em>A}{C</em>A^2} = k dt$
$\int<em>{A0}^{C</em>A} \frac{dC<em>A}{C</em>A^2} = k \int<em>{0}^{t} dt$ $-[\frac{1}{C</em>A}]<em>{C</em>{A0}}^{C<em>A} = kt$ $\frac{1}{C</em>A} - \frac{1}{C<em>{A0}} = kt$ $\frac{C</em>A - C<em>{A0}}{C</em>A C<em>{A0}} = kt$ (2.31) $C</em>A = \frac{C<em>{A0}}{1+C</em>{A0}}$ (2.32)
Equation (2.31) is the integrated rate expression for second order reaction in terms of concentration.

Now we will develop the integrated rate expression in terms of conversion.

For the reaction $2A \rightarrow \text{Products}$
$-r<em>A = -\frac{dC</em>A}{dt} = k C<em>A^2$ $C</em>A = C<em>{A0}(1-X</em>A)$
$dC<em>A = C</em>{A0}dX<em>A$ so $-dC</em>A = C<em>{A0}dX</em>A$
$\frac{dX<em>A}{(1-X</em>A)^2} = kC<em>{A0}dt$ $Y = 1-X</em>A$
$-dX<em>A = dY$ $\int \frac{dX</em>A}{(1-X<em>A)^2} = kC</em>{A0}\int dt$
$-\int\frac{dY}{Y^2} = kC<em>{A0}t$ $[1/Y]</em>1^{1-X<em>A} = kC</em>{A0}t$
$(\frac{1}{1-X<em>A}-1) = kC</em>{A0}t$
$X<em>A = kC</em>{A0}t(1-X<em>A)$ $\frac{X</em>A}{1-X<em>A} = kC</em>{A0}t$ (2.33)
Equation (2.33) is the integrated rate equation for second order reaction in terms of conversion.
Characteristics:
A plot of 1/CA v/s t should yield a straight line with a slope equal to k and an intercept equal to 1/CA, a plot of X/(1-X) v/s t should yield a straight line with a slope equal to CAk.

It is evident from equations (2.31) and (2.33) that the dimensions of a second order rate constant of the reaction are (molar concentration x time)-1. It is customary to express the second order rate constant in the units L/(mol.s).

The integrated rate equation for : 2A → product is
CA0 CA = k t
The half life of a reaction, i.e., the time required to reduce the concentration of the reactant to half of its original value can be found by employing the condition CA = CA/2 at the half life time t=t1/2 in the above equation.
$t<em>{1/2} = \frac{1}{kC</em>{A0}}$ (2.35)

Half Life Second Order

Thus, the half life for a second order reaction is inversely proportional to the first power of initial concentration. The half life method cannot be used with reactions where initial concentrations of A and B are different, since A and B will have different times for half reaction.

Consider a reaction of the type $A+B \rightarrow Products$ (2.36)

The corresponding rate equation is:
$-r<em>A = -\frac{dC</em>A}{dt} = k \cdot C<em>A \cdot C</em>B$ (2.36 a)

Let CA and CB be the initial concentrations of A and B respectively.
Let $M = \frac{C<em>B}{C</em>A}$ be the initial molar ratio of reactants.
Let CA and CB be the concentrations of A and B at any time t.
Let X and X, be the fractional conversions of A and B respectively.

Amount of A reacted = XACA.
$X<em>A = \frac{C</em>{A0} - CA}{C<em>{A0}}$ Amount of B reacted = XBCBo. $X</em>B = \frac{C<em>{B0} - CB}{C</em>{B0}}$
The amounts of A and B which have reacted at any time t are equal and given by
Xa CA = XB C Bo
$C<em>A = C</em>{A0}(1 - X<em>A)$ $-dC</em>A = C<em>A - dX</em>A$
$\frac{dX<em>A}{C</em>A^2(1-X<em>A^2)} = k dt$ $k = \frac{1}{t(C</em>B - C<em>A)} ln{\frac{C</em>A<em>C{B0}}{C{B}</em>C<em>{A0}}}$ We have: $-dA = C</em>A^2 dt = k C<em>A^2$ and then from integral solution we have $\frac{1}{C</em>A} - \frac{1}{C_{A0}} = k t$ or, CA CA0 = CA02 k t

Solving for Unequal Concentrations

$-r<em>A = C</em>A^2(1-X<em>A^2) = k C</em>A (1 - X<em>A) C</em>{B0} (1 - X<em>B)$ We have $C</em>A^2=\frac{CC<em>A^2}{\Delta X^2}$ Since for bimolecular reactions $C</em>A C<em>B$ are equal $-dC</em>A = C<em>A^2 dX</em>A$
$-dC<em>A = C</em>A^2 = dX<em>A(1-X</em>A)$
$C<em>A d(XA^-1)$ $-X</em>A^-1 - C<em>A^-1 = t$ where t is initial time and X equals to initial Rewrite our equation to have form to be solved where $dX = -a</em>1 a=2t$ and $X = (-a<em>1)/2$ where $a = -A$: $\int</em>{CA<em>}^{CA</em>^<em>}\frac{dCA d(XA^-1)}{C^{2+}(1)} X^{j+}$ However, it is more convenient to represent equations as the following: $-dXdt = kCA(1-CA)$ and hence we have solution of
$ln(C^</em>1A((t+C^<em>1CA))-ln(C^</em>1A((t))-ln(C<em>{CA*0})$ and using $\frac{k</em>c}{d}$ for the rate expression for $(t\rightarrow \infty)$

We have solution of:
XA 1 M − XA ln CAO = (M − 1))kCAT ; M ̸= 1
and from that solution $C_B$ after rewriting, is defined as:
ln C A CB CA CB0 = (CB0 − CA0 )kt + ln CAO CB0
where ln CB CA = (CBO − CAO )kt + ln CB0 CA0

Zero-order Reactions:
When the rate of reaction is independent of the concentration of the reactant, the reaction is said to be zero order.
A zero order reaction is represented as
$A \rightarrow product$
$-r<em>A = -\frac{dC</em>A}{dt} = kC<em>A^0 \rightarrow -\frac{dC</em>A}{dt} = k$ (2.40 b)
The integrated rate equation for zero order reaction in terms of concentration is
CA0 − CA = kt (2.41)
$C<em>A = C</em>{A0}(1 - X<em>A)$ CACAO XA = k t + CACAO \rightarrow CA0XA = kt (2.42) The overall dimentions equation using the form $ln(t)\approx ln(\frac{-E</em>a}{RTk<em>c}) + A$ is A plot of CA CAO against t yields a straight line passing through origin with a slope equal to k. half life which is proportional to the initial concentration of the reactant: Substituting the contition $t = t</em>{1/2}$ when $C<em>A = \frac{CA}{2}$ we get: $t</em>{1/2} = CA0 \frac{2 kt}{( 2 k)} = k \frac{t}{0}$
The half life is given as
C10\int^t0 (k)s dt\int^{\frac{A}{0}} {A+I} C220 −CA+I1 \vert A = CA0,
$C00−CA = K(t−0) where in final rewritten form is time t2 to 12$ where 12 α CA
and half reaction is directly proportional to concentration of reactant in reactant.
Some heterogeneous reactions of