Which Of The Following Represents The Integrated Rate Law For A First-Order Reaction?
- Marvin Harvey
Want to join the conversation? –
why does the natural log of pressure have no unit?
It still has the original unit of pressure because all we’ve done to it by taking the ln(P) is a mathematical operation. We haven’t added a new unit or eliminated the original unit. So technically the unit of ln(P) should still be Torr and they should have indicated the unit on graph. Hope that helps.
Could anyone please show the derivation of the function for the first-order reactions?
This is grade-12/college-level but if you’re curious I will show you below. So for a first order reaction – we have the reaction equals the rate constant times the concentration of the (only) reactant -> R = k 1. Then we choose to re-write R as -Δ/Δt and we get -Δ/Δt = k 2. Then we bring -Δt to the right side Δ = -kΔt 3. Then we bring to the left side Δ/ = -kΔt 4. Then we integrate (the left side with respect to A and the right side with respect to t) ∫Δ 1/ = -k∫Δt Ln = -kt 5. Then we evaluate both integrals from 0 to t Lnₜ – Ln₀ = -kt-(-k0) 6. Then we bring Ln₀ to the right Lnₜ = -kt -0 + Ln₀ 7. Finally, we have our answer: Lnₜ = -kt + Ln₀ 8.Notes: i. For the connection to y=mx+b The natural log of the concentration of A at a given time t -> Lnₜ is basically Y, and is equal to the natural log of the initial concentration of A -> Ln₀ which is basically b, minus the rate constant ->k (basically m, aka the slope of the line) multiplied by time (basically x). So we get a linear graph of the form Y=mx+b ii. The reason the it is negative at the beginning -Δ/Δt and at -kt is because the rate is positive, but the change in reactant is negative because it is decreasing, so we build in a negative sign to cancel it and make the rate positive.
What does he mean by the “natural log” at 0:43 ? hi, at 2:25, Jay said if the coefficient of A is 2, then -kt will become -2kt, so why don’t we generalize it as -akt, where a is the coefficient of reactant A? then according to this new convention, won’t the half life equation be = e^(-akt) and t1/2 = ln2÷ak? thank you!
Yeah, the calculus works out to that.
I’m having trouble finding how you got -2.08 x 10 ^-4. When I add up the y and x numbers to do m= change in y divided by change in x, I get -16.189 divided by – 39800 = 4.067×10^-4
The slope of a line is defined as the change in the y-direction divided by the change in the x-direction (rise over run). As a formula looks like m = Δy/Δx; where m is the slope, Δy is the change in y, and Δx is the change in x. Change here being a difference (subtraction) between two point’s x and y coordinates. So we can also write the slope formula as m = (y2-y1)/(x2-x1); where x1 and y1 are the coordinates for the first point and x2 and y2 are the coordinates for the second point. Now assuming it is a perfectly straight line, the slope should be constant at all points on the line and so we can pick any two points to calculate the slope. I’ll choose the first and last points; (0,6.219) and (15000,3.109). So x1 is 0, y1 is 6.219, x2 is 15000, and y2 is 3.109. Substituting these into the previous formula yields: m = (3.109 – 6.219)/(15000 – 0) = -2.07 x 10^(-4). Which is reasonably close to what Jay got in the video. The discrepancy between my answer and Jay’s is due to him using a graphing computer of some sort which takes into account all the coordinate points and also shows that the line is not perfectly straight. At any case even doing it by hand our answers should agree for the most part. With your calculation I’m not sure why or what you added together, but having a change in y of -16.189 and a change in x of -39800 is wildly wrong. Additionally if you divide those two numbers you get a positive slope (dividing a negative by a negative) and judging solely off the graph of the line it should have a negative slope. Hope that helps.
What is integrated rate law for first order reaction?
First-Order Integrated Rate Law: This equation can be used for any first-order reactions of the form rate=k r a t e = k where k is the rate constant in units of s−1, is the concentration of reactant A, and the rate is in units of concentration per time.
What is a 1st order reaction?
A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.
What are the integrated rate laws for first and second-order reactions?
The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.
What does integrated rate law represent?
On the other hand, integrated rate laws express the reaction rate as a function of the initial concentration and a measured (actual) concentration of one or more reactants after a specific amount of time (t) has passed ; they are used to determine the rate constant and the reaction order from experimental data.
What is first order rate equation?
The unit of rate constant for a first-order reaction is sec − 1. For first-order reactions, equation ln = -kt + ln 0 is the same as the equation of a straight line (y = mx + c) with slope -k.
What is first order reaction with example?
Examples of First-Order Reaction Aspirin hydrolysis and the combination of t-butyl bromide with water to produce t-butanol are two examples of first-order reactions. The hydrolysis of the anticancer medication cisplatin is another process that displays apparent first-order kinetics.
What is order of reaction Mcq?
Order of reaction is equal to the number of molecules whose concentration is changing with time. It can be zero or in fractions or an integer.
Why is it called first-order reaction?
Example of First-Order Reaction – Here is an example to help you understand the concept more clearly. Consider the hydrolysis of, during the hydrolysis, the concentration of ethyl acetate is 0.02 mol/L whereas the amount of water is 20 mol/L as the process of hydrolysis involves a large amount of water. Let us say, the process of hydrolysis attains completion in time t.
- The reaction can be represented as
- CH 3 COOC 2 H5 + H 2 O —–> CH 3 COOH + C 2 H 5 OH
- (ethyl acetate) (Water) (Acetic acid) (ethyl alcohol)
|Components||CH 3 COOC 2 H 5||H 2 O||CH 3 COOH||C 2 H 5 OH|
For the above-mentioned reaction, the rate equation can be given as, Rate = k Here we see that the concentration of water is very high and thus does not change much during the course of the reaction. Thus the rate of the reaction can be said to be independent of the change in the concentration of H 2 O.
- Rate = k
- Here, the term kt takes into account the value of the constant concentration of water.
- where K = K’
We see that the reaction behaves as a first-order reaction. Such reactions are termed pseudo-first-order reactions.
- Another example of pseudo-first-order reaction is the inversion of cane sugar, given by the following reaction,
- C 12 H 22 O 11 + H 2 O —–> C 6 H 12 O 6 + C 6 H 12 O 6
- (Cane sugar) (Water) (Glucose) (Fructose)
- The rate equation can be given as,
- Rate = k
A first-order reaction can be defined as a chemical reaction for which the reaction rate is entirely dependent on the concentration of only one reactant. In such reactions, if the concentration of the first-order reactant is doubled, then the reaction rate is also doubled. Similarly, if the first-order reactant concentration is increased five-fold, it will be accompanied by a 500% increase in the reaction rate. Pseudo first order reaction: The reaction which is bimolecular but order is one is called pseudo first order reaction. This happens when one of the reactants is in large excess.E.g., acidic hydrolysis of ester (ethyl acetate). CH 3 COOC 2 H 5 + H 2 O ——> CH 3 COOH + C 2 H 5 OH The unit of the rate of the reaction (k) is (mol L -1 ) 1-n s -1, where n is the order of the reaction. For a pseudo first-order reaction n=1. So, (mol L -1 ) 1-n s -1 = (mol L -1 ) 1-1 s -1 = s -1 Examples: 1. Acidic hydrolysis of ester (ethyl acetate). CH 3 COOC 2 H 5 + H 2 O ——> CH 3 COOH + C 2 H 5 OH 2. The inversion of cane sugar, given by the following reaction, C 12 H 22 O 11 + H 2 O —–> C 6 H 12 O 6 + C 6 H 12 O 6 To learn more about the pseudo-first-order reactions and other related topics, you can download BYJU’S – The Learning App. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz Visit BYJU’S for all Chemistry related queries and study materials
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View Quiz Answers and Analysis : Pseudo First Order Reaction – Rate Law, Order of Reaction & Examples
What is a first-order rate constant?
Elimination Rate Constant ( k ) – The elimination rate constant (usually a first-order rate constant) represents the fraction of xenobiotics that is eliminated from the body during a given period of time. For instance, when the value of the elimination rate constant of a xenobiotic is 0.25 per hour, this means that ∼25% of the amount remaining in the body is excreted each hour. Figure 2, Time course of blood concentration of a xenobiotic following intravenous administration ((a) log-linear scale; (b) linear scale); C p 0, blood or plasma concentration at time zero. (c) Schematic representation of a one-compartment open model: i.v., intravenous administration; V, volume of the compartment; k, elimination rate constant (see text).
What is the rate law of 2A B → C?
The rate equation for the reaction 2A + B → C is found to be a : r = k. The correct statement in relation of this reaction is that the Rate of formation of C is twice the rate of disappearance of A No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Suggest Corrections 2 : The rate equation for the reaction 2A + B → C is found to be a : r = k. The correct statement in relation of this reaction is that the
Which is integrated rate equation?
The integrated rate equation is Rt = log C0 – log Ct.
What are the units for a first-order rate law?
In a first-order reaction, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows: \ \label \] If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth.
Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s −1 ). The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows: \ = _0e^ \label \] where 0 is the initial concentration of reactant A at t = 0; k is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places.
Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation \(\ref \) predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation \(\ref \) and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t : \ = \ln_0 − kt \label \] Because Equation \(\ref \) has the form of the algebraic equation for a straight line, y = mx + b, with y = \ln and b = \ln 0, a plot of \ln versus t for a first-order reaction should give a straight line with a slope of − k and an intercept of \ln 0, Figure \(\PageIndex \): Graphs of a first-order reaction. The expected shapes of the curves for plots of reactant concentration versus time (top) and the natural logarithm of reactant concentration versus time (bottom) for a first-order reaction. First-order reactions are very common.
- We have already encountered two examples of first-order reactions: the hydrolysis of aspirin and the reaction of t -butyl bromide with water to give t -butanol.
- Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
- Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults.
The structures of cisplatin and its hydrolysis product are as follows: Figure \(\PageIndex \) Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure \(\PageIndex \) is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active.
Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure \(\PageIndex \) have been studied extensively to find ways of maximizing the concentration of the active species.
Note If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order. The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Ta ble \(\PageIndex \).
The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table \(\PageIndex \) shows that the reaction rate doubles when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M).
Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k 1,
|Experiment||0 (M)||Initial Rate (M/min)|
|1||0.0060||9.0 × 10 −6|
|2||0.012||1.8 × 10 −5|
|3||0.024||3.6 × 10 −5|
|4||0.030||4.5 × 10 −5|
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
|Experiment||0 (M)||Initial Rate (M/s)|
|1||0.010||1.6 × 10 −8|
|2||0.015||2.4 × 10 −8|
|3||0.030||4.8 × 10 −8|
|4||0.040||6.4 × 10 −8|
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction Asked for: reaction order and rate constant Strategy:
Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species. Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant. Solution The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to, Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to,
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k, C We can calculate the rate constant ( k ) using any row in the table. Selecting Experiment 1 gives the following: 1.60 × 10 −8 M/s = k (0.010 M) 1.6 × 10 −6 s −1 = k Exercise \(\PageIndex \) Sulfuryl chloride (SO 2 Cl 2 ) decomposes to SO 2 and Cl 2 by the following reaction: SO 2 Cl 2 (g) → SO 2 (g) + Cl 2 (g) Data for the reaction at 320°C are listed in the following table.
|Experiment||0 (M)||Initial Rate (M/s)|
|1||0.0050||1.10 × 10 −7|
|2||0.0075||1.65 × 10 −7|
|3||0.0100||2.20 × 10 −7|
|4||0.0125||2.75 × 10 −7|
Answer first order; k = 2.2 × 10 −5 s −1 We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Figure \(\PageIndex \): The Hydrolysis of Cisplatin, a First-Order Reaction. These plots show hydrolysis of cisplatin at pH 7.0 and 25°C as (a) the experimentally determined concentrations of cisplatin and chloride ions versus time and (b) the natural logarithm of the cisplatin concentration versus time.
- The straight line in (b) is expected for a first-order reaction.
- The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way.
- When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure \(\PageIndex \),
The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure \(\PageIndex \) for t = 100 min ( = 0.0086 M) and t = 1000 min ( = 0.0022 M), \(\begin \textrm &=\dfrac ]_ -\ln _ } } \\-k&=\dfrac }=\dfrac }=-1.51\times10^ \;\mathrm } \\k&=1.5\times10^ \;\mathrm }\end \) The slope is negative because we are calculating the rate of disappearance of cisplatin.
Also, the rate constant has units of min −1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure \(\PageIndex \) are in minutes rather than seconds. The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law.
This must be true if the experiments were carried out under the same conditions. The First-Order Integrated Rate Law Equation: https://youtu.be/_JskhfxBAMI Example \(\PageIndex \) If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M ( k = 1.6 × 10 −6 s −1 ) ? Given: initial concentration, rate constant, and time interval Asked for: concentration at specified time and time required to obtain particular concentration Strategy:
Substitute values for the initial concentration ( 0 ) and the calculated rate constant for the reaction ( k ) into the integrated rate law for a first-order reaction. Calculate the concentration () at the given time t, Given a concentration, solve the integrated rate law for time t,
Solution The exponential form of the integrated rate law for a first-order reaction ( Equation \(\ref \) ) is = 0 e − kt, A Having been given the initial concentration of ethyl chloride ( 0 ) and having the rate constant of k = 1.6 × 10 −6 s −1, we can use the rate law to calculate the concentration of the reactant at a given time t,
Substituting the known values into the integrated rate law, \(\begin _ }&=_0e^ \\&=\textrm (e^ \textrm ^ )}}) \\&=0.0189\textrm \end \) We could also have used the logarithmic form of the integrated rate law ( Equation \(\ref \) ): \(\begin \ln_ }&=\ln _0-kt \\ &=\ln 0.0200-(1.6\times10^ \textrm ^ ) \\ &=-3.912-0.0576=-3.970 \\ _ }&=e^ \textrm \\ &=0.0189\textrm \end \) B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for t,
Equation \(\ref \) gives the following: \(\begin \ln_t &=\ln_0-kt \\kt &=\ln_0-\ln_t=\ln\dfrac ]_0} ]_t} \\ t &=\dfrac \left (\ln\dfrac ]_0} ]_t} \right )=\dfrac \textrm ^ }\left(\ln \dfrac } }\right) \\ &=\dfrac \textrm ^ }=8.7\times10^5\textrm =240\textrm =2.4\times10^2\textrm \end \) Exercise \(\PageIndex \) In the exercise above, you found that the decomposition of sulfuryl chloride (SO 2 Cl 2 ) is first order, and you calculated the rate constant at 320°C.
Which plot represents a first-order reaction?
For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
How to find the rate constant of a first-order reaction from a graph?
How is the rate constant k found from a graph? Once the graph shows a straight line, the rate constant is found from the slope of the line. For zero- and first-order reactions, the rate constant is the negative slope, and for second-order reactions, the rate constant is the slope.
Which is the correct expression for integrated rate law of nth order reaction?
K=1t(n−1) Where C and C0are the concentration of reactant at time t and initially respectively.
Which of the following is not a first order reaction?
The hydrolysis of ethyl acetate (CH3COOC2H5) by sodium hydroxide yields ethanol (C2H5OH) and sodium acetate (NaO2CCH3) by the reaction showed follows second order kinetics.
What is 1st and 2nd order reaction?
Key Takeaways: Reaction Orders in Chemistry –
Chemical reactions may be assigned reaction orders that describe their kinetics.The types of orders are zero-order, first-order, second-order, or mixed-order.A zero-order reaction proceeds at a constant rate. A first-order reaction rate depends on the concentration of one of the reactants. A second-order reaction rate is proportional to the square of the concentration of a reactant or the product of the concentration of two reactants.
Which of the following is true for order of reaction?
(d) The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression.
Is sn1 a first order reaction?
The S N 1 mechanism – A second model for a nucleophilic substitution reaction is called the ‘ dissociative’, or ‘ S N 1’ mechanism: in this picture, the C-X bond breaks first, before the nucleophile approaches: This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as sp 2 hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three sp 2 hybrid orbitals is an empty, unhybridized p orbital. In the second step of this two-step reaction, the nucleophile attacks the empty, ‘electron hungry’ p orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry. We saw that S N 2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of S N 1 reactions? In the model S N 1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack.
Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, sp 2 -hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted.
In other words, racemization has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of R and S alcohols after an S N 1 reaction with water as the incoming nucleophile. Exercise 11.4.1 Draw the structure of the intermediate in the two-step nucleophilic substitution reaction above. The S N 1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. Exercise 11.4.2 Draw structures representing TS1 and TS2 in the reaction above. Use the solid/dash wedge convention to show three dimensions. Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining.
- This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step.
- Also recall that an S N 1 reaction has first order kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding.
- Exercise 11.4.3 Consider two nucleophilic substitutions that occur uncatalyzed in solution.
Assume that reaction A is S N 2, and reaction B is S N 1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant.
What is integrated rate equation class 12?
Integrated Rate Equation | Using Integrated Rate Laws for zero order An equation representing the dependence of the rate of reaction on the concentration of reacting species is known as a differential rate equation. The instantaneous rate of reaction is expressed as the slope of the tangent at any instant of time in the concentration-time graph.
Hence, it is very difficult to determine the rate of reaction from the concentration-time graph. Therefore, we integrate the differential rate equation to obtain a relation between the concentration at different points and rate constant. This equation is known as the integrated rate equation. For reactions of different order, we observe different integrated rate equations.
Zero order reaction: In zero order reaction, the rate of reaction depends upon the zeroth power of concentration of reactants. Zero order reactions are very rarely observed. Some examples of zero order reactions are: thermal decomposition of HI on gold surface, decomposition of gaseous ammonia on a etc.
- A → B
- \(\begin Rate = – \frac =k^\circ\end \)
- \(\begin -\frac = k\end \)
- => d = -k dt
- Integrating both sides:
⇒ = -kt + c.(1)
- Where, c= constant of integration,
- At time, t=0,
- \(\begin = _0\end \)
- Putting the limits in equation (1) we get the value of c,
- \(\begin _0 = c\end \)
- Using the value of c in equation (1) we get,
- \(\begin = -kt + _0\end \)
The above equation is known as the integrated rate equation for zero order reactions. We can observe the above equation as an equation of straight line with concentration of reactant on y-axis and time on x-axis. The slope of the straight line signifies the value of rate constant, k. First order reaction: In first order reaction, the rate of reaction depends upon the first power of concentration of reactants. Natural and artificial radioactive decay of unstable nuclei are examples of first order reactions. A general equation for a first order reaction with rate constant k is derived below:
- A → B
- \(\begin Rate = – \frac = k\end \)
- \(\begin = \frac = -k dt\end \)
- Integrating both sides:
- => ln = -kt + c.(2)
- Where, c= constant of integration,
- At time, t=0,
- \(\begin = _0\end \)
- Putting the limits in equation (2) we get the value of c
- \(\begin _0 = c \end \)
- Using the value of c in the above equation, we get,
- ln = -kt + ln A 0
We observe that the equation above can be plotted as a straight line with ln on y-axis and time (t) on x-axis. The negative of the slope of this straight line gives us the value of the rate constant, k.
- We can also determine the value of the rate constant, k from the equation as:
- \(\begin ln ~\frac = -kt\end \)
- \(\begin k = – \frac } \end \)
- Concentration at any moment of time can be given as,
- \(\begin = _0 e^ \end \)
- Thus, we can determine the concentration and rate of reaction at any moment with the help of an integrated rate equation for zero and first order reaction.
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View Quiz Answers and Analysis : Integrated Rate Equation | Using Integrated Rate Laws for zero order
What is the second-order integrated rate law?
Video transcript – – Let’s say we have a hypothetical reaction where reactant A turns into products. And let’s say the reaction is second order with respect to A. If the reaction is second order with respect to A, then we can write the rate of the reaction is equal to the rate constant k times the concentration of A to the second power since this is a second order reaction.
- We can also write that the rate of the reaction is equal to the negative of the change in the concentration of A over the change in time.
- If we set these two ways of writing the rate of reaction equal to each other, and use of calculus, including the concept of integration, we will arrive at the integrated rate law for a second order reaction.
The integrated rate law for a second order reaction says that one over the concentration of reactant A at some time t, is equal to the rate constant k times the time plus one over the initial concentration of A. Notice how the integrated rate law has the form of y is equal to mx plus b, which is the equation for a straight line.
- So if we graph one over the concentration of A on the y axis, and time on the x axis, so let’s go ahead and put that in here, one over the concentration of A on the y axis, and time on the x axis, we will get a straight line and the slope of that line is equal to the rate constant k.
- So slope is equal to K and the y intercept is equal to one over the initial concentration of A.
So the point where our line intersects the y axis is equal to one over the initial concentration of A. Let’s look at an example of a second order reaction. C5H6 is cyclopentadiene, and two molecules of cyclopentadiene will react with each other to form dicyclopentadiene.
Our goal is to use the data from this data table to prove that this reaction is second order. However, we have to be careful because in our balanced equation, we have a two as a coefficient in front of cyclopentadiene. Going back to our hypothetical reaction where reactant A turned into products, there’s also one as a coefficient in front of the A.
And if there’s a one as a coefficient in front of the A, we can use this form of the integrated rate law for a second order reaction. However, for our problem, we have a two as a coefficient in front of cyclopentadiene. And that means we need to have a stoichiometric coefficient of 1/2 in here, and that changes the math.
Now when we set these two rates of reactions equal to each other, and we use calculus and we integrate to get our integrated rate law, because of this one half, we end up with a two in front of the K. So thinking about y is equal to mx plus b, now the slope of the line is equal to two K. So now for our reaction, we can write our integrated rate law as one over the concentration of cyclopentadiene at some time t, is equal to two kt plus one over the initial concentration of cyclopentadiene.
So if we look at our data table, we have time in seconds and we have the concentration of cyclopentadiene, but we need to have one over the concentration of cyclopentadiene so we need a new column. So we’re going to calculate one over the concentration of cyclopentadiene.
So if the concentration of cyclopentadiene when time is equal to zero seconds is 0.0400 molar, if we take one divided by 0.0400, we would get 25.0. To save some time I’ve gone ahead and filled in the rest of this column. Notice as time increases so as we go from zero seconds to 50 to 100 to 150 to 200, the concentration of cyclopentadiene decreases because cyclopentadiene is turning into dicyclopentadiene.
Next, we need to graph our data. So we’re gonna have one over the concentration of cyclopentadiene on the y axis and time on the x axis. Our first point, so when time is equal to zero, one over the concentration of cyclopentadiene is 25.0. So if we go down to our graph, we can see that when time is equal to zero, our first point here is 25.0.
- And plotting the other points gives us a straight line.
- Next, we need to find the slope of this straight line.
- And there are many ways to do that.
- One way to do it is to use a graphing calculator.
- And when I used a graphing calculator to find the slope of this line, I found that the slope is equal to 0.1634.
Thinking about y is equal to mx plus b, our slope should be equal to two k. So to find the rate constant k, we need to divide the slope by two, which gives us 0.0817. To find the units for K, remember that slope is equal to change in y over change in x, and on our Y axis, our units are one over molar, and the x axis the units are seconds.
So therefore, we can write the rate constant k is equal to 0.0817. It’s to be one over molar divided by seconds which is the same thing as one over molar times seconds. It’s important to point out that most textbooks don’t cover how the two as a coefficient changes the integrated rate law. And so a lot of textbooks will simply say that the slope of the line for the second order integrated rate law, is equal to K.
So a lot of books would just say the final answer for the rate constant is 0.163. So you’ll see you’ll see a lot of textbooks say that the rate constant would be 0.163, one over molar times seconds. However, since the coefficient in front of cyclopentadiene is a two, technically this rate constant is the correct one.
What is intermediate rate law?
A reaction intermediate is a chemical species that is formed in one elementary step and consumed in a subsequent step. The slowest step in a reaction mechanism is known as the rate-determining step. The rate-determining step limits the overall rate and therefore determines the rate law for the overall reaction.