{"id":12525,"date":"2026-05-09T15:23:21","date_gmt":"2026-05-09T15:23:21","guid":{"rendered":"https:\/\/www.vedprep.com\/exams\/?p=12525"},"modified":"2026-05-09T15:23:21","modified_gmt":"2026-05-09T15:23:21","slug":"integrated-rate-equations","status":"publish","type":"post","link":"https:\/\/www.vedprep.com\/exams\/iit-jam\/integrated-rate-equations\/","title":{"rendered":"Master Integrated rate equations (Zero, First, Second order) For JAM 2027"},"content":{"rendered":"<p>Found within chemistry&#8217;s core tools, <strong>integrated rate equations<\/strong> link how much reactant remains over time. As time passes, concentration changes follow patterns unique to each reaction type. Because of this, exam success often depends on recognizing these forms clearly. One way they prove useful is exposing whether a process moves in first, second, or zero order. Their structure also allows extraction of the rate constant without ambiguity. Understanding them means seeing how fast transformations unfold under set conditions.<\/p>\n<p>A straight line appears when you plot how fast a zero-order reaction goes in <strong>integrated rate equations<\/strong>. At any moment t, what&#8217;s left of substance A follows this math: start with how much was there originally, subtract the chunk lost over time. That chunk depends on a steady number called k, which never changes during the process.<\/p>\n<p>The graph slopes downward like a hill sliding into dusk.<br \/>\nInstead of curving, the math for a one-step reaction shows ln([A]\/[A]\u2080) equals -kt, otherwise written as ln([A]) equals -kt plus ln([A]\u2080). When you map ln([A]) against time, it forms a downward-slanting line. With two-part reactions, the formula shifts to 1\/[A] equaling kt plus 1\/[A]\u2080. Plotting 1\/[A] over time gives an upward-tilted straight path.<\/p>\n<h2><strong>Integrated Rate Equations (Zero, First, Second Order) for IIT JAM<\/strong><\/h2>\n<p>Among topics listed in the <a href=\"https:\/\/jam2026.iitb.ac.in\/files\/syllabus_CY.pdf\" rel=\"nofollow noopener\" target=\"_blank\"><strong>IIT JAM chemistry syllabus<\/strong><\/a>, Integrated rate equations &#8211; covering zero, first, and second order &#8211; appear within Physical and Chemical Changes. Within this section, focus shifts toward core ideas in chemical kinetics. Understanding how fast reactions proceed becomes a central point of study. Though broad in scope, the material centers on measuring reaction speed through mathematical expressions.<\/p>\n<p>You will find this subject in common textbooks &#8211; take Atkins&#8217; Physical Chemistry by Peter Atkins together with Julio de Paula &#8211; for instance, where <strong>integrated rate equations<\/strong> get a full breakdown. Look also at the official IIT JAM Chemistry Syllabus; it spells out exactly what areas matter for the test.<\/p>\n<p>The key concepts that are typically covered under this topic include:<\/p>\n<ul>\n<li>Zero-order reactions<\/li>\n<li>First-order reactions<\/li>\n<li>Second-order reactions<\/li>\n<\/ul>\n<p>These concepts are critical for understanding the kinetics of chemical reactions and are frequently asked in competitive exams such as CSIR NET, IIT JAM, and GATE.<\/p>\n<h2><strong>Integrated Rate Equations for Chemical Kinetics<\/strong><\/h2>\n<p>Starting from the rate law, one finds that <strong>integrated rate equations<\/strong> connecting reactant levels with elapsed time. Essential for probing reaction speed, this step transforms raw rates into usable forms. From such math emerges a clear view of how substances change over time. These derived expressions follow strictly from initial rate assumptions, shaping predictions without guesswork.<\/p>\n<p>What drives the speed of a reaction ties back to its order, pulled straight from the powers in the rate equation. Depending on that number, scientists sort reactions into one of three groups &#8211; zero, first, or second. These aren\u2019t just labels; each comes with its own math rule showing how much reactant fades moment by moment. From start to finish, it\u2019s about matching behavior to form.<\/p>\n<ul>\n<li>A straight line drops steadily when zero-order reactions unfold &#8211; concentration fades without regard to how much substance remains. Over time, what you see follows a simple form: left-over amount equals starting point minus a steady drain marked by k times t. This path stays predictable, shaped only by time and that unchanging pace called k. Each moment chips away just the same, no matter the crowd present at first.<\/li>\n<li>First-order reaction: The rate of reaction depends on the concentration of one reactant, and the integrated rate equation is ln([A]\/[A]\u2080) = -kt.<\/li>\n<li>When two substances influence speed together, or a single substance squared affects pace, it follows second-order kinetics. At such times, the math unfolds as 1 divided by concentration equals time multiplied by rate constant plus initial reciprocal amount. This pattern emerges clearly when tracking change over duration under fixed conditions.<\/li>\n<\/ul>\n<p>Grasping how these rate laws connect matters because they form the backbone of chemical kinetics questions in exams like IIT JAM. Though each order behaves differently, working through them helps learners tackle time-concentration patterns with ease. Once applied, such formulas reveal trends in decay speed, duration of half change, and shifts across time. Because real problem solving demands precision, using the correct expression makes all the difference.<\/p>\n<h2><strong>Worked Example: Zero-Order Reaction<\/strong><\/h2>\n<p>A zero-order reaction shows a rate that does not shift when reactant levels change. Take the process A turning into B &#8211; here, substance A follows zero-order behavior. Found within such cases, the formula unfolds as [A] equals minus kt plus [A]\u2080. This expression holds [A], representing how much A remains at moment t. Standing firm throughout, k acts as the steady rate factor. Begin with what was first present &#8211; that value is labeled [A]\u2080.<\/p>\n<p>The following data is given for the reaction: [A]\u2080 = 0.5 M, k = 0.1 M s\u207b\u00b9. Calculate the concentration of A at t = 2 s, 5 s, and 10 s.<\/p>\n<table>\n<tbody>\n<tr>\n<th>Time (s)<\/th>\n<th>Concentration of A (M)<\/th>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>0.5<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>[A] = -0.1 \u00d7 2 + 0.5 = 0.3<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>[A] = -0.1 \u00d7 5 + 0.5 = 0.2<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>[A] = -0.1 \u00d7 10 + 0.5 = 0.1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Over time, the amount of A drops in a straight line. That kind of pattern shows up only when the reaction moves at a steady pace, no matter how much A is left.<br \/>\nFive seconds in, A sits at 0.2 M. Because it follows a zero-order pattern, you can track how much reactant remains using math meant for that behavior.<\/p>\n<h2><strong>Misconception: Confusing Rate Laws and Integrated Rate Equations<\/strong><\/h2>\n<p>Although they look similar, rate laws differ from integrated versions in function. How fast a reaction proceeds relates directly to substance amounts present. One form shows instantaneous behavior through an expression: speed equals k times A raised to n, B to m. Here, k stands fixed, while exponents reflect dependence on each component. These powers reveal individual influence without assuming mechanism. Such relationships emerge from observation rather than derivation.<\/p>\n<p>Still, a built-up speed formula shows how much of a substance remains as minutes pass. Take one-step reactions: here, the math rule runs ln([A]\u209c\/[A]\u2080) = -kt, with [A]\u209c standing for amount left at moment t, [A]\u2080 the starting level, while k marks the steady pace number.<\/p>\n<h2><strong>Exam Strategy: Mastering Integrated Rate Equations for IIT JAM<\/strong><\/h2>\n<p>Mastering<strong> integrated rate equations<\/strong> matters greatly when preparing for IIT JAM. Though rooted in chemical kinetics, this concept tends to trouble many learners. From concentrations evolving over time, these expressions emerge &#8211; revealing both reaction order and rate constant values.<\/p>\n<p>Beginning with problem-solving sharpens understanding of rate constants alongside time-based concentration shifts. Study resources from <a href=\"https:\/\/www.vedprep.com\/online-courses\"><strong>VedPrep<\/strong> <\/a>&#8211; detailed videos, targeted exercises &#8211; support steady progress through complex topics. Where focus meets structured support, clarity around <strong>integrated rate equations<\/strong> grows. Preparation shaped by consistent effort, combined with clear explanations, leads toward stronger performance in IIT JAM settings.<\/p>\n<p>Key concepts to review include:<\/p>\n<ul>\n<li>Derivation of <strong>integrated rate equations<\/strong> for zero, first, and second-order reactions<\/li>\n<li>Graphical analysis of concentration-time data<\/li>\n<li>Calculating rate constants and half-life<\/li>\n<\/ul>\n<p><a href=\"https:\/\/www.vedprep.com\/online-courses\/iit-jam\"><strong>VedPrep&#8217;s<\/strong> <\/a>resources can help students develop a thorough understanding of these concepts and improve their problem-solving skills.<\/p>\n<h2><strong>Final Thoughts<\/strong><\/h2>\n<p>Understanding <strong>Integrated rate equations<\/strong> &#8211; whether zero, first, or second order &#8211; involves more than recalling expressions. It means forming a sense of how reactions change mathematically with time. To someone preparing for IIT JAM 2027, shifting quickly between differential rate laws and their integrated versions matters greatly. Recognizing these relationships using graph gradients becomes useful under timed conditions. Such awareness grows sharper with practice, yet clarity often arrives unexpectedly.<\/p>\n<p>To know more in detail from our faculty, watch our YouTube video:<\/p>\n<p class=\"responsive-video-wrap clr\"><iframe title=\"Chemical Kinetics | Integrated Rate Equation |CSIR NET | GATE | IIT JAM |Lec-2| VedPrep Chem Academy\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/hS_2lZu0SNE?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/p>\n<h2><strong>Frequently Asked Questions (FAQs)<\/strong><\/h2>\n<style>#sp-ea-15336 .spcollapsing { height: 0; overflow: hidden; transition-property: height;transition-duration: 300ms;}#sp-ea-15336.sp-easy-accordion>.sp-ea-single {margin-bottom: 10px; border: 1px solid #e2e2e2; }#sp-ea-15336.sp-easy-accordion>.sp-ea-single>.ea-header a {color: #444;}#sp-ea-15336.sp-easy-accordion>.sp-ea-single>.sp-collapse>.ea-body {background: #fff; color: #444;}#sp-ea-15336.sp-easy-accordion>.sp-ea-single {background: #eee;}#sp-ea-15336.sp-easy-accordion>.sp-ea-single>.ea-header a .ea-expand-icon { float: left; color: #444;font-size: 16px;}<\/style><div id=\"sp_easy_accordion-1778320715\">\n<div id=\"sp-ea-15336\" class=\"sp-ea-one sp-easy-accordion\" data-ea-active=\"ea-click\" data-ea-mode=\"vertical\" data-preloader=\"\" data-scroll-active-item=\"\" data-offset-to-scroll=\"0\">\n\n<!-- Start accordion card div. -->\n<div class=\"ea-card ea-expand sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153360\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153360\" aria-controls=\"collapse153360\" href=\"#\"  aria-expanded=\"true\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-minus\"><\/i> What is an integrated rate equation?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse collapsed show\" id=\"collapse153360\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153360\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>An integrated rate equation is a mathematical expression derived from the differential rate law that shows the relationship between the concentration of reactants and the time elapsed during a chemical reaction.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153361\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153361\" aria-controls=\"collapse153361\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Why are integrated rate equations important for IIT JAM 2027?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153361\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153361\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>These equations allow students to calculate the concentration of a reactant at any specific time, determine the rate constant (<span class=\"math-inline\" data-math=\"k\" data-index-in-node=\"193\">k<\/span>), and find the half-life, which are core components of the Physical Chemistry syllabus.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153362\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153362\" aria-controls=\"collapse153362\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is a zero-order reaction?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153362\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153362\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>A zero-order reaction is one where the rate of the reaction is independent of the concentration of the reactants. The rate remains constant throughout the process.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153363\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153363\" aria-controls=\"collapse153363\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is the integrated rate equation for a zero-order reaction?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153363\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153363\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The equation is <span class=\"math-inline\" data-math=\"[A] = -kt + [A]_0\" data-index-in-node=\"84\">[A] = -kt + [A]<sub>0<\/sub><\/span>, where <span class=\"math-inline\" data-math=\"[A]\" data-index-in-node=\"109\">[A]<\/span>\u00a0is the final concentration, <span class=\"math-inline\" data-math=\"[A]_0\" data-index-in-node=\"141\">[A]<sub>0<\/sub><\/span> is the initial concentration, <span class=\"math-inline\" data-math=\"k\" data-index-in-node=\"177\">k<\/span>\u00a0is the rate constant, and <span class=\"math-inline\" data-math=\"t\" data-index-in-node=\"205\">t<\/span>\u00a0is time.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153364\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153364\" aria-controls=\"collapse153364\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What does the slope represent in a zero-order concentration vs. time graph?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153364\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153364\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>When plotting <span class=\"math-inline\" data-math=\"[A]\" data-index-in-node=\"94\">[A]<\/span> versus <span class=\"math-inline\" data-math=\"t\" data-index-in-node=\"105\">t<\/span>, the slope of the straight line is equal to <span class=\"math-inline\" data-math=\"-k\" data-index-in-node=\"151\">-k<\/span>.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153365\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153365\" aria-controls=\"collapse153365\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is a first-order reaction?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153365\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153365\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>A first-order reaction is a reaction where the rate depends on the concentration of only one reactant raised to the power of one.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153366\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153366\" aria-controls=\"collapse153366\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> How do you identify a first-order reaction graphically?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153366\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153366\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>A plot of <span class=\"math-inline\" data-math=\"\\ln[A]\" data-index-in-node=\"70\">ln[A]<\/span>\u00a0versus time (<span class=\"math-inline\" data-math=\"t\" data-index-in-node=\"90\">t<\/span>) will yield a straight line with a slope of <span class=\"math-inline\" data-math=\"-k\" data-index-in-node=\"136\">-k<\/span>\u00a0and a y-intercept of <span class=\"math-inline\" data-math=\"\\ln[A]_0\" data-index-in-node=\"160\">ln[A]<sub>0<\/sub><\/span>.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153367\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153367\" aria-controls=\"collapse153367\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What defines a second-order reaction?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153367\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153367\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>A second-order reaction has a rate proportional to either the square of the concentration of one reactant or the product of the concentrations of two different reactants.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153368\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153368\" aria-controls=\"collapse153368\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What are the units of the rate constant (k) for a zero-order reaction?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153368\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153368\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The units are <span class=\"math-inline\" data-math=\"M \\cdot s^{-1}\" data-index-in-node=\"90\">M \u00b7 s<sup>-1<\/sup><\/span>\u00a0(moles per liter per second).<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-153369\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse153369\" aria-controls=\"collapse153369\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What are the units of $k$ for a first-order reaction?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse153369\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-153369\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The units are <span class=\"math-inline\" data-math=\"s^{-1}\" data-index-in-node=\"71\">s<sup>-1<\/sup><\/span>\u00a0(per second), making it independent of concentration units.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1533610\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1533610\" aria-controls=\"collapse1533610\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Can a reaction have a fractional order?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1533610\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-1533610\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Yes, while we primarily focus on zero, first, and second orders, many complex reactions involve fractional orders which also have specific integrated forms.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1533611\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1533611\" aria-controls=\"collapse1533611\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is the difference between a rate law and an integrated rate law?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1533611\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-1533611\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The rate law relates rate to concentration, while the integrated rate law relates concentration to time.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1533612\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1533612\" aria-controls=\"collapse1533612\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Which textbook is best for Integrated Rate Equations for IIT JAM?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1533612\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-1533612\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p><i data-path-to-node=\"20\" data-index-in-node=\"71\">Atkins' Physical Chemistry<\/i> is the gold standard, though <i data-path-to-node=\"20\" data-index-in-node=\"127\">Puri, Sharma &amp; Pathania<\/i> is also highly recommended for Indian competitive exams.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1533613\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1533613\" aria-controls=\"collapse1533613\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Are integrated rate equations used in the \"Physical and Chemical Changes\" unit?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1533613\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-1533613\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Yes, they form the mathematical backbone of Chemical Kinetics within that unit in the IIT JAM syllabus.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1533614\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1533614\" aria-controls=\"collapse1533614\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is a pseudo-first-order reaction?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1533614\" data-parent=\"#sp-ea-15336\" role=\"region\" aria-labelledby=\"ea-header-1533614\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>It is a second-order reaction that behaves like a first-order reaction because one of the reactants is present in such great excess that its concentration remains effectively constant.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Integrated rate equations (Zero, First, Second order) For IIT JAM are mathematical expressions that relate reactant concentrations to time for chemical reactions. These equations help in determining the order of a reaction and the rate constant, which are critical in understanding the kinetics of a reaction. For a zero-order reaction, the integrated rate equation is [A] = -kt + [A]\u2080, where [A] is the concentration of the reactant at time t, [A]\u2080 is the initial concentration, and k is the rate constant. In contrast, the integrated rate equation for a first-order reaction is ln([A]\/[A]\u2080) = -kt, which can also be expressed as ln([A]) = -kt + ln([A]\u2080).<\/p>\n","protected":false},"author":12,"featured_media":12524,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":"","rank_math_seo_score":85},"categories":[23],"tags":[7375,2923,11644,11643,11645,11646,11647,2922],"class_list":["post-12525","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-iit-jam","tag-chemical-kinetics-for-iit-jam","tag-competitive-exams","tag-first","tag-integrated-rate-equations-zero","tag-second-order-for-iit-jam","tag-second-order-for-iit-jam-notes","tag-second-order-for-iit-jam-questions","tag-vedprep","entry","has-media"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12525","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/users\/12"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/comments?post=12525"}],"version-history":[{"count":8,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12525\/revisions"}],"predecessor-version":[{"id":15393,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12525\/revisions\/15393"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/media\/12524"}],"wp:attachment":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/media?parent=12525"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/categories?post=12525"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/tags?post=12525"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}