Do you mean "what happens to the rate of a second order reaction as the concentration doubles?" If so, you would have to look at the order of each reactant: Show
Rate = k[A][B] and Rate = k[A]2 are both second order reaction. In the first case, A and B are both first order and in the second, A is second order. If a reactant is first order, then when its concentration is doubled, the rate of the reaction doubles; if the concentration is tripled, the rate
triples, etc. If a reactant is 0 order, then changing its concentration does not affect the rate of reaction. IntroductionA first-order reaction is one in which the rate of reaction is proportional to the concentration of the reactant. To put it another way, doubling the concentration doubles the reaction rate. A first-order reaction can have one or two reactants, as in the case of the decomposition reaction. Table of Content
What is a First-Order Reaction?A first-order reaction can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant. In other words, a first-order reaction is a chemical reaction in which the rate varies based on the changes in the concentration of only one of the reactants. Thus, the order of these reactions is equal to 1. Examples of First-Order Reactions
Differential Rate Law for a First-Order ReactionA differential rate law can be employed to describe a chemical reaction at a molecular level. The differential rate expression for a first-order reaction can be written as: Rate = -d[A]/dt = k[A]1 = k[A] Where,
Integrated Rate Law for a First-Order ReactionIntegrated rate expressions can be used to experimentally calculate the value of the rate constant of a reaction. To obtain the integral form of the rate expression for a first-order reaction, the differential rate law for the first-order reaction must be rearranged as follows. \(\begin{array}{l}\frac{-d[A]}{dt} = k[A]\end{array} \) \(\begin{array}{l}\Rightarrow \frac{d[A]}{[A]} = -kdt\end{array} \) Integrating both sides of the equation, the following expression is obtained. \(\begin{array}{l}\int_{[A]_0}^{[A]}\frac{d[A]}{[A]} = -\int_{t_0}^{t}kdt\end{array} \) Which can be rewritten as: \(\begin{array}{l}\int_{[A]_0}^{[A]}\frac{1}{[A]}d[A] = -\int_{t_0}^{t}kdt\end{array} \) Since \(\begin{array}{l}\int\frac{1}{x} = ln(x) \end{array} \) , the equation can be rewritten as follows:ln[A] – ln[A]0 = -kt ln[A] = -kt + ln[A]0 (or) ln[A] = ln[A]0 – kt Raising each side of the equation to the exponent ‘e’ (since eln(x) = x), the equation is transformed as follows: \(\begin{array}{l}e^{ln[A]} = e^{ln[A]_0 – kt}\end{array} \) Therefore, \(\begin{array}{l}[A] = [A]_0 e^{-kt}\end{array} \) This expression is the integrated form of the first-order rate law. Graphical Representation of a First-Order ReactionThe concentration v/s time graph for a first-order reaction is provided below.  For first-order reactions, the equation ln[A] = -kt + ln[A]0 is similar to that of a straight line (y = mx + c) with slope -k. This line can be graphically plotted as follows. Thus, the graph for ln[A] v/s t for a first-order reaction is a straight line with slope -k. Half-Life of a First-Order ReactionThe half-life of a chemical reaction (denoted by ‘t1/2’) is the time taken for the initial concentration of the reactant(s) to reach half of its original value. Therefore, At t = t1/2 , [A] = [A]0/2 Where [A] denotes the concentration of the reactant and [A]0 denotes the initial concentration of the reactant. Substituting the value of A = [A]0/2 and t = t1/2 in the equation [A] = [A]0 e-kt: \(\begin{array}{l}\frac{[A]_0}{2} = [A]_0 e^{-kt_{1/2}}\end{array} \) \(\begin{array}{l}\Rightarrow \frac{1}{2} = e^{-kt_{1/2}}\end{array} \) Taking the natural logarithm of both sides of the equation to eliminate ‘e’, the following equation is obtained. \(\begin{array}{l}ln(\frac{1}{2}) = -kt_{1/2}\end{array} \) \(\begin{array}{l}\Rightarrow t_{1/2} = \frac{0.693}{k}\end{array} \) Thus, the half-life of a first-order reaction is equal to 0.693/k (where ‘k’ denotes the rate constant, whose units are s-1). Frequently Asked Questions – FAQsA 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. The differential rate law for a first-order reaction can be expressed as follows: Rate = -d[A]/dt = k[A] The integrated rate equation for a first-order reaction is: [A] = [A]0e-kt Where,
What is the relationship between the half-life and the rate constant for a first-order reaction?The half-life of a chemical reaction is the time required for the concentration of the reactants to reach half of their initial value. For first-order reactions, the relationship between the reaction half-life and the reaction rate constant is given by the expression: t1/2 = 0.693/k Where ‘t1/2’ denotes the half-life of the reaction and ‘k’ denotes the rate constant. What are the units of the rate constant for a first-order reaction?For first-order reactions, the rate constant is expressed in s1 (reciprocal seconds). The units of the rate constant can be determined using the following expression: Units of k = M(1-n).s-1 (where ‘n’ is the order of the reaction) Since the reaction order of a first-order reaction is equal to 1, the equation is transformed as follows: Units of k = M(1-1).s-1 = s-1 For a first-order reaction, if a graph is plotted with ln[A] on the Y-axis and time on the X-axis, what will it look like?The graph will be a straight line with a slope of -k. To learn more about first-order reactions and other important concepts in chemical kinetics, register with BYJU’S and download the mobile application on your smartphone. What happens if you double the concentration of a first order reaction?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.
What happens to the initial rate if you double the concentration of a second order reaction?Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1).
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