A first order reaction takes 80 minutes for 60% decomposition calculate t1/2
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A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration. The Differential RepresentationDifferential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. The differential equation describing first-order kinetics is given below: \[ Rate = - \dfrac{d[A]}{dt} = k[A]^1 = k[A] \label{1}\] The "rate" is the reaction rate (in units of molar/time) and \(k\) is the reaction rate coefficient (in units of 1/time). However, the units of \(k\) vary for non-first-order reactions. These differential equations are separable, which simplifies the solutions as demonstrated below. The Integral RepresentationFirst, write the differential form of the rate law. \[ Rate = - \dfrac{d[A]}{dt} = k[A] \nonumber\] Rearrange to give: \[ \dfrac{d[A]}{[A]} = - k\,dt \nonumber\] Second, integrate both sides of the equation. \[ \begin{align*} \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]} &= -\int_{t_o}^{t} k\, dt \label{4a} \\[4pt] \int_{[A]_{o}}^{[A]} \dfrac{1}{[A]} d[A] &= -\int_{t_o}^{t} k\, dt \label{4b} \end{align*}\] Recall from calculus that: \[ \int \dfrac{1}{x} = \ln(x) \nonumber\] Upon integration, \[ \ln[A] - \ln[A]_o = -kt \nonumber\] Rearrange to solve for [A] to obtain one form of the rate law: \[ \ln[A] = \ln[A]_o - kt \nonumber\] This can be rearranged to: \[ \ln [A] = -kt + \ln [A]_o \nonumber\] This can further be arranged into y=mx +b form: \[ \ln [A] = -kt + \ln [A]_o \nonumber\] The equation is a straight line with slope m: \[mx=-kt \nonumber\] and y-intercept b: \[b=\ln [A]_o \nonumber\] Now, recall from the laws of logarithms that \[ \ln {\left(\dfrac{[A]_t}{ [A]_o}\right)}= -kt \nonumber\] where [A] is the concentration at time \(t\) and \([A]_o\) is the concentration at time 0, and \(k\) is the first-order rate constant. Because the logarithms of numbers do not have any units, the product \(-kt\) also lacks units. This concludes that unit of \(k\) in a first order of reaction must be time-1. Examples of time-1 include s-1 or min-1. Thus, the equation of a straight line is applicable: \[ \ln [A] = -kt + \ln [A]_o.\label{15}\] To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction. To create another form of the rate law, raise each side of the previous equation to the exponent, \(e\): \[ \large e^{\ln[A]} = e^{\ln[A]_o - kt} \label{16}\] Simplifying gives the second form of the rate law: \[ [A] = [A]_{o}e^{- kt}\label{17}\] The integrated forms of the rate law can be used to find the population of reactant at any time after the start of the reaction. Plotting \(\ln[A]\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\). More information can be found in the article on rate laws. This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law. Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function \(y=e^x\) so efficiently describes such changes is that dy/dx = ex; that is, ex is its own derivative, making the rate of change of \(y\) identical to its value at any point. Graphing First-order ReactionsThe following graphs represents concentration of reactants versus time for a first-order reaction. Plotting \(\ln[A]\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\). Half-lives of first order reactionsThe half-life (\(t_{1/2}\)) is a timescale on which the initial population is decreased by half of its original value, represented by the following equation. \[ [A] = \dfrac{1}{2} [A]_o \] After a period of one half-life, \(t = t_{1/2}\) and we can write \[\dfrac{[A]_{1/2}}{[A]_o} = \dfrac{1}{2}=e^{-k\,t_{1/2}} \label{18}\] Taking logarithms of both sides (remember that \(\ln e^x = x\)) yields \[ \ln 0.5 = -kt\label{19}\] Solving for the half-life, we obtain the simple relation \[ t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{k}\label{20}\] This indicates that the half-life of a first-order reaction is a constant.  Example 1: Estimated Rate Constants The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant? SolutionUse Equation 20 that relates half life to rate constant for first order reactions: \[k = \dfrac{0.693}{600 \;s} = 0.00115 \;s^{-1} \nonumber\] As a check, dimensional analysis can be used to confirm that this calculation generates the correct units of inverse time. Notice that, for first-order reactions, the half-life is independent of the initial concentration of reactant, which is a unique aspect to first-order reactions. The practical implication of this is that it takes as much time for [A] to decrease from 1 M to 0.5 M as it takes for [A] to decrease from 0.1 M to 0.05 M. In addition, the rate constant and the half life of a first-order process are inversely related. Example \(\PageIndex{1}\): Determining Half life If 3.0 g of substance \(A\) decomposes for 36 minutes the mass of unreacted A remaining is found to be 0.375 g. What is the half life of this reaction if it follows first-order kinetics? Solution There are two ways to approach this problem: The "simple inspection approach" and the "brute force approach" Approach #1: "The simple Inspection Approach" This approach is used when one can recognize that the final concentration of \(A\) is \(\frac{1}{8}\) of the initial concentration and hence three half lives \(\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)\) have elapsed during this reaction. \[t_{1/2} = \dfrac{36\, \text{min}}{3}= 12 \; \text{min} \nonumber\] This approach works only when the final concentration is \(\left(\frac{1}{2}\right)^n\) that of the initial concentration, then \(n\) is the number of half lives that have elapsed. If this is not the case, then approach #2 can be used. Approach #2: "The brute force approach" This approach involves solving for \(k\) from the integral rate law (Equation \ref{17}) and then relating \(k\) to the \(t_{1/2}\) via Equation \ref{20}. \[\begin{align*} \dfrac{[A]_t}{[A]_o} &= e^{-k\,t} \\[4pt] k &= -\dfrac{\ln \dfrac{[A]_t}{[A]_o}}{t} \\[4pt] &= -\dfrac{\ln \dfrac{0.375\, g}{3\, g}}{36\, \text{min}} \\[4pt] &= 0.0578 \, \text{min}^{-1} \end{align*}\] Therefore, via Equation \ref{20} \[t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{0.0578 \, \text{min}^{-1}} \approx 12\, \text{min} \nonumber\] The first approach is considerably faster (if the number of half lives evolved is apparent). Exercise \(\PageIndex{2a}\) Calculate the half-life of the reactions below:
Use the half life reaction that contains initial concentration and final concentration. Plug in the appropriate variables and solve to obtain:
Exercise \(\PageIndex{2b}\) Determine the percent \(\ce{H2O2}\) that decomposes in the time using \(k=6.40 \times 10^{-5} s^{-1}\)
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Contributors and Attributions
2.3: First-Order Reactions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. What is the value of t1 2 in first order reaction?The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions.
What is ratio of t1 2 and t90% of a first order reaction?Answer. =3.3 times that of half-life.
Is 60% of a first order reaction was completed in 60 minutes 50% of the same reaction would be completed in?If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately (log 4 = 0.60 log 5 = 0.69) UPLOAD PHOTO AND GET THE ANSWER NOW! Solution : `K = (2.303)/(60) "log" (100)/(40) implies 0.152`
`k = (0.693)/(T_(1//2)) T_(1//2) = (0.693)/(0.152) = 45.37` min . What is halfThe half-life of a reaction (t1/2), is the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. Its application is used in chemistry and medicine to predict the concentration of a substance over time.
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