Pharmaceutical Analytical Chemistry I Lecture Notes PDF

# Pharmaceutical Analytical Chemistry I (PC_101) ## Lecture No. (09) ### **Assistant Prof. Ashraf Mohamed Taha** ### **Pharm. D** #### 6. Effect of Concentration on Reaction Rates: The Rate Law - One of the goals in a chemical kinetics study is to derive an equation that can be used to predict t...

# Pharmaceutical Analytical Chemistry I (PC_101) ## Lecture No. (09) ### **Assistant Prof. Ashraf Mohamed Taha** ### **Pharm. D** #### 6. Effect of Concentration on Reaction Rates: The Rate Law - One of the goals in a chemical kinetics study is to derive an equation that can be used to predict the relationship between the rate of reaction and the concentrations of reactants. - Such an experimentally determined equation is called a rate law, or rate equation. #### 6.1. The Rate law - So far, we have learned that the reaction rate is proportional to the concentration of reactants and that the proportionality constant *k* is called the rate constant. - The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. For the general reaction: $aA + bB \longrightarrow cC + dD$ - The rate law takes the form of Equation 6: $rate = k[A]^x[B]^y$ - Where *x* and *y* are numbers that must be determined experimentally. - Note that, in general, *x* and *y* are not equal to the stoichiometric coefficients *a* and *b*. - When we know the values of *x, y*, and *k*, we can use Equation (6) to calculate the reaction rate, given the concentrations of *A* and *B*. - The exponents *x* and *y* specify the relationships between the concentrations of reactants *A* and *B* and the reaction rate. Together, they give us the overall reaction order, defined as the sum of the powers to which all reactant concentrations appearing in the rate law are raised. - For Equation (6) the overall reaction order is *x + y*. - Alternatively, we can say that the reaction is *x*th order in *A*, *y*th order in *B*, and *(x+y)*th order overall. #### Example 4 - To see how to determine the rate law of a reaction, let us consider the reaction between fluorine and chlorine dioxide: $F_2(g) + 2CIO_2(g) \longrightarrow 2FCIO_2(g)$ - One way to study the effect of reactant concentration on reaction rate is to determine how the initial rate depends on the starting concentrations. - It is preferable to measure the initial rates because: - As the reaction proceeds, the concentrations of the reactants decrease, and it may become difficult to measure the changes accurately. - Also, there may be a reverse reaction of the type which would introduce an error in the rate measurement. - $products \longrightarrow reactants$ - Both complications are virtually absent during the early stages of the reaction. - Table 2 shows three rate measurements for the formation of FCIO₂. | [F₂] (M) | [CIO₂] (M) | Initial Rate (M/s) | |---|---|---| | 0.10 | 0.010 | 1.2 x 10⁻³ | | 0.10 | 0.040 | 4.8 x 10⁻³ | | 0.20 | 0.010 | 2.4 x 10⁻³ | - Looking at entries 1 and 3, we see that as we double [F₂] while holding [CIO₂] constant, the reaction rate doubles. Thus, the rate is directly proportional to [F₂]. - Similarly, the data in entries 1 and 2 show that as we quadruple [CIO₂] at constant [F₂], the rate increases by four times, so that the rate is also directly proportional to [CIO₂]. - We can summarize our observations by writing the rate law as: $rate = k[F_2][CIO_2]$ #### 7. The Relation between the Reactant Concentration and Time - The rate laws we have examined so far enable us to calculate the rate of a reaction from the rate constant and reactant concentrations. - In this section, we will show that rate laws can also be converted into equations that show the relationship between concentrations of reactants or products and time. - The mathematics required to accomplish this conversion involves calculus. #### 7.1. Zero-Order Reactions - First- and second-order reactions are the most common reaction types. - Reactions whose order is zero are rare. - For a zero-order reaction: - Zero-Order equation: $A \longrightarrow product$ - Where: - $rate = \frac{\Delta[A]}{\Delta t}$ - This can be expressed as: $rate = k[A]^0$ - The rate of a zero-order reaction is a constant, independent of reactant concentration. - Combining this with the previous equation: - $\frac{\Delta[A]}{\Delta t} = k$ - Using calculus, we can show that: - $\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k \int_{0}^{t} dt$ - $ln[A] = -kt + ln[A]_0$ - This equation has the form of a Linear Equation. As Figure 10 shows, a plot of [A] versus t gives a straight line with slope = -k and y-intercept = [A]₀. #### Reaction Half-life - To calculate the half-life of a zero-order reaction, we set [A] = [A]₀/2 in Equation (9) and obtain Equations 10, 11 - $kt = [A]_0-[A]_t$ - $t = \frac{[A]_0-[A]_t}{k} $ - $t_{1/2} = \frac{[A]_0}{2k} $ - By rearrangement Equation 10 we obtained Equation 11 - $t_{1/2} = \frac{[A]_0}{2k} $ - Zero order does not mean that the rate is zero. It just means that the rate is independent of the concentration of A present. #### Example 5 - Many of the known zero-order reactions take place on a metal surface. - An example is the decomposition of nitrous oxide (NO) to nitrogen and oxygen in the presence of platinum (Pt): - $2N_2O(g) \longrightarrow 2N_2(g) + O_2(g)$ - When all the binding sites on Pt are occupied, the rate becomes constant regardless of the amount of N₂O present in the gas phase. #### 7.2. First-Order Reactions - First-Order Equation: - A first-order reaction is a reaction whose rate depends on the reactant concentration raised to the first power. - In a first-order reaction of the type: $A \longrightarrow product$ - The rate is: - $rate = \frac{\Delta[A]}{\Delta t}$ - From the rate law (in this case), we can express the relation between rate and concentration in Equation 12: - $rate = k[A]$ - Combining the last two equations (Eq. 1 and Eq. 12) for the rate we get Equation 13 - $\frac{\Delta[A]}{\Delta t} = k[A]$ - Using calculus: - $\int \frac{d[A]}{[A]} = -k \int dt$ - $ln[A] = -kt + ln[A]_0$ - We can show from Equation (13) that: - $ln(\frac{[A]}{[A]_0}) = -kt$ - Where *In* is the natural logarithm, and *[A]* and *[A]₀* are the concentrations of *A* at times *t=0* and *t=t*, respectively. - It should be understood that *t=0* need not correspond to the beginning of the experiment; it can be any time when we choose to start monitoring the change in the concentration of A. - Equation (14) can be rearranged as follows: - $ln[A] = -kt + ln[A]_0$ - Equation (14) has the form of the Linear Equation *y = mx + b*, in which *m* is the slope of the line that is the graph of the Linear equation: - $ln[A] = (-k)(t) + ln[A]_0$ - Consider Figure 11. As we would expect during a reaction, the concentration of reactant A decreases with time [Figure 11(a)]. - For a first-order reaction, if we plot *ln[A]* versus time (*y* versus *x*), we obtain a straight line with a slope equal to *k* and a y-intercept equal to *ln[A]₀* [Figure 11(b)]. Thus, we can calculate the rate constant from the slope of this plot. #### Reaction Half-life - We can obtain an expression for *t₁/₂* for a first-order reaction as follows: - $t = \frac{1}{k} ln(\frac{[A]_0}{[A]})$ - By the definition of half-life, when *t=t₁/₂*, *[A]t = [A]₀/2* and applying in Equation 15 we obtain Equations 16 and 17 - $t_{1/2} = \frac{1}{k} ln(\frac{[A]_0}{[A]_0/2})$ - $t_{1/2} = \frac{1}{k} ln2$ - Equation (17) tells us that the half-life of a first-order reaction is independent of the initial concentration of the reactant. - Thus, it takes the same time for the concentration of the reactant to decrease from 1.0 M to 0.50 M, say, as it does for a decrease in concentration from 0.10 M to 0.050 M (Figure 12). - Measuring the half-life of a reaction is one way to determine the rate constant of a first-order reaction. - The half-life, *t₁/₂*, is that it gives us an estimate of the magnitude of the rate constant--the shorter the half-life, the larger the *k*. #### Example 6: - The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 x 10⁻⁴ s⁻¹ at 500°C - $CH₂CH₂CH₂ \longrightarrow CH₃CH=CH₂$ - a. If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? - $ln(\frac{[A]}{[A]_0}) = -kt$ - $ln([A]) = -(6.7 * 10^{-4} s^{-1}) * (8.8 min * \frac{60 s}{1min}) + ln(0.25)$ - $[A] = 0.18M$ - b. How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? - $ln(\frac{[A]}{[A]_0}) = -kt$ - $ln(\frac{0.15 M}{0.25 M}) = -(6.7 * 10^{-4} s^{-1} )t$ - $t = 13 min$ - c. How long (in minutes) will it take to convert 74 percent of the starting material? - $ln(\frac{[A]}{[A]_0}) = -kt$ - $ln(\frac{0.26 M}{1.00 M}) = -(6.7 * 10^{-4} s^{-1} )t$ - $t = 33 min$ #### Reaction Half-life - As a reaction proceeds, the concentration of the reactant(s) decreases. - Another measure of the rate of a reaction, relating concentration to time, is the half-life, *t₁/₂*, which is the time required for the concentration of a reactant to decrease to half of its initial concentration. #### Example 7: - The decomposition of ethane (C₂H₆) to methyl radicals is a first-order reaction with a rate constant of 5.36 x 10⁻⁴ s⁻¹ at 700°C - $C_2H_6 \longrightarrow 2CH_3$ - $t_{1/2} = \frac{ln2}{k}$ - $t_{1/2} = \frac{ln2}{( 5.36 * 10^{-4} s^{-1})} * \frac{1min}{60s}$ - $t_{1/2} = 21.5 min$ #### 7.3. Second-Order Reactions - Second-Order Equation: - A second-order reaction is a reaction whose rate depends on the concentration of one reactant raised to the second power or on the concentrations of two different reactants, each raised to the first power. - The simpler type involves only one kind of reactant molecule: - $A \longrightarrow product$ - The rate is: - $rate = \frac{\Delta[A]}{\Delta t}$ - From the rate law (in this case), we can express the relation between rate and concentration in Equation 18: - $rate = k[A]^2$ - Combining the last two Equations (Eq. 1 and Eq. 18) for the rate we get: - $\frac{\Delta[A]}{\Delta t} = k[A]^2$ - Using calculus: - $\int \frac {d[A]}{[A]^2} = -k \int dt$ - $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$ - Equation (20) has the form of a Linear Equation. As Figure 13 shows, a plot of 1/[A] versus t gives a straight line with slope = k and y-intercept = 1/[A]₀. #### Reaction Half-life - We can obtain an equation (Eq. 21) for the half-life of a second-order reaction by setting [A]t = [A]₀/2 in Equation (20) - $\frac{1}{[A]_0/2} = kt_{1/2} + \frac{1}{[A]_0}$ - $t_{1/2} = \frac{1}{k[A]_0}$ - Note that the half-life of a second-order reaction is inversely proportional to the initial reactant concentration. - This result makes sense because the half-life should be shorter in the early stage of the reaction when more reactant molecules are present to collide with each other. - Measuring the half-lives at different initial concentrations is one way to distinguish between a first-order and a second-order reaction. #### Example 8: - Iodine atoms combine to form molecular iodine in the gas phase. - $I(g) + I(g) \longrightarrow I_2(g)$ - This reaction follows second-order kinetics and has a high rate constant of 7.0 x 10⁹ M⁻¹s⁻¹ at 23°C. - a. If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min. - $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$ - $\frac{1}{[A]} = (7.0 * 10^{9} M^{-1}s^{-1})(2.0 min * \frac{60s}{1 min})+ \frac{1}{0.086 M}$ - $[A] = 1.2 * 10^{-10} M$ - b. Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M. - $t_{1/2} = \frac{1}{k[A]_0}$ - $t_{1/2} = \frac{1}{(7.0 * 10^{9} M^{-1}s^{-1})(0.60 M)}$ - $t_{1/2} = 2.4 * 10^{-10} s$ - $t_{1/2} = \frac{1}{(7.0 * 10^{9} M^{-1}s^{-1})(0.42 M)}$ - $t_{1/2} = 3.4 * 10^{-10} s$ #### Table 3: Rate Law Summary Table | Order | Rate Law | Units of *k* | Integrated Rate Law | Straight-Line Plot | Half-Life Expression | |---|---|---|---|---|---| | 0 | $Rate = k[A]^0$ | M s⁻¹ | $[A] = -kt + [A]_0$ | $\frac{1}{[A]}$ vs. t <br/>y-intercept = [A]₀ <br/> Slope = k | $t_{1/2} = \frac{[A]_0}{2k}$ | | 1 | $Rate = k[A]^1$ | s⁻¹ | $ln([A]) = -kt + ln([A]_0)$ | $ln([A])$ vs. t <br/>y-intercept = *ln[A]₀* <br/>Slope = k | $t_{1/2} = \frac{ln2}{k}$ | | 2 | $Rate = k[A]^2$ | M⁻¹s⁻¹ | $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$ | $\frac{1}{[A]}$ vs. t <b/>y-intercept = $\frac{1}{[A]_0}$ <br/>Slope = k | $t_{1/2} = \frac{1}{K[A]_0}$ |