Question
How will you represent first-order reactions graphically.
✓
Answer
(1) A graph of rate of a reaction and concentration : The differential rate law for first-order reaction, $A \rightarrow $ Products is represented as, Rate = $
latex-\frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]$
$\therefore $ Rate $= k x [A]_t$ (y = mx). When the rate of a first order reaction is plotted against concentration,$[A]_t$, a straight line graph is obtained.
With the increase in the concentration $[A]_t$, rate R, increases. The slope of the line gives the value of rate constant k.
(2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration $[A]$, of the reactant decreases exponentially with time. The variation in the concentration can be represented as,$[ A ]_t=[ A ]_0\ e ^{-k t}$
where $[A]_0$ and $[A]_t$ are initial and final concentrations the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.
(3) A graph of $log_{10} (a – x)$ against time t :
$k=\frac{2.303}{t} \log _{10}\left(\frac{a}{a-x}\right)$
$\therefore k=\frac{2.303}{t}\left[\log _{10} a-\log _{10}(a-x)\right]$
$\therefore \frac{k}{2.303} \times t=\log _{10} a-\log _{10}(a-x)$
$\therefore \log _{10}(a-x)$
$=-\frac{k}{2.303} \times t+\log _{10} a(y=-m x+c)$
When $log_{10}(a – x)$ is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.
(4) A graph of half-life period and concentration : The half-life period, $t_{1/2}$ of a first order reaction is given by, where k is the rate constant.
For the given reaction at a constant temperature, $t_{1/2}$ is constant and independent of the concentration of the reactant.
Hence when a graph of $t_{1/2}$ is plotted against concentration, a straight line parallel to the concentration axis (slope = zero) is obtained.
(5) A graph of $log_{10} [latex]\left(\frac{a}{a-x}\right)$ against time : The rate constant, for a first order reaction is represented as,
$k=\frac{2.303}{t} \log _{10}\left(\frac{a}{a-x}\right)=\frac{2.303}{t} \log _{10} \frac{\left[ A _0\right]}{[ A ]_t}$ where $\left[ A _0\right]$ and $[ A ]_{ t }$ are the respective initial and final concentrations of the reactant after time $t$.
$\therefore \log \left(\frac{a}{a-x}\right)=\frac{k}{2.303} \times t(y=m x)$
When $\log _{10}\left(\frac{a}{a-x}\right)$ is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of $k / 2.303$. From this slope, the rate constant can be calculated.
latex-\frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]$
$\therefore $ Rate $= k x [A]_t$ (y = mx). When the rate of a first order reaction is plotted against concentration,$[A]_t$, a straight line graph is obtained.
With the increase in the concentration $[A]_t$, rate R, increases. The slope of the line gives the value of rate constant k.
(2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration $[A]$, of the reactant decreases exponentially with time. The variation in the concentration can be represented as,$[ A ]_t=[ A ]_0\ e ^{-k t}$
where $[A]_0$ and $[A]_t$ are initial and final concentrations the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.
(3) A graph of $log_{10} (a – x)$ against time t :
$k=\frac{2.303}{t} \log _{10}\left(\frac{a}{a-x}\right)$
$\therefore k=\frac{2.303}{t}\left[\log _{10} a-\log _{10}(a-x)\right]$
$\therefore \frac{k}{2.303} \times t=\log _{10} a-\log _{10}(a-x)$
$\therefore \log _{10}(a-x)$
$=-\frac{k}{2.303} \times t+\log _{10} a(y=-m x+c)$
When $log_{10}(a – x)$ is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.
(4) A graph of half-life period and concentration : The half-life period, $t_{1/2}$ of a first order reaction is given by, where k is the rate constant.
For the given reaction at a constant temperature, $t_{1/2}$ is constant and independent of the concentration of the reactant.
Hence when a graph of $t_{1/2}$ is plotted against concentration, a straight line parallel to the concentration axis (slope = zero) is obtained.
(5) A graph of $log_{10} [latex]\left(\frac{a}{a-x}\right)$ against time : The rate constant, for a first order reaction is represented as,
$k=\frac{2.303}{t} \log _{10}\left(\frac{a}{a-x}\right)=\frac{2.303}{t} \log _{10} \frac{\left[ A _0\right]}{[ A ]_t}$ where $\left[ A _0\right]$ and $[ A ]_{ t }$ are the respective initial and final concentrations of the reactant after time $t$.
$\therefore \log \left(\frac{a}{a-x}\right)=\frac{k}{2.303} \times t(y=m x)$
When $\log _{10}\left(\frac{a}{a-x}\right)$ is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of $k / 2.303$. From this slope, the rate constant can be calculated.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Observe the above figure carefully. The two types of circles in this figure represent two types of constituent particles of a solid.
Question 1.
Will you call the arrangement of particles in this solid regular or irregular ?
Question 2.
Is the arrangement of constituent particles in directions $\overrightarrow{ A B }, \overrightarrow{ C D }$ $\overrightarrow{ E F }$ same or different?
From the rate expressions for the following reactions, determine their order :
(a) $2 N _2 O _{5( g )} \rightarrow 4 NO _{2( g )}+ O _{2( g )}:$ Rate $= k \left[ N _2 O _5\right]$
(b) $CHCl _{3( g )}+ Cl _{2( g )} \rightarrow CCl _{4( g )}+ HCl _{( g )}:$ Rate $= k \left[ CHL _3\right]\left[ Cl _2\right]^{1 / 2}$
(c) $C _2 H _5 Cl _{( g )} \rightarrow C _2 H _{4( g )}+ HCl _{( g )}$ : Rate $= k \left[ C _2 H _5 Cl \right]$
(d) $2 NO _{2( g )}+ F _{2( g )} \rightarrow 2 NO _2 F _{( g )} \rightarrow$ : Rate $= k \left( NO _2\right]\left[ F _2\right]$
→(a) $2 N _2 O _{5( g )} \rightarrow 4 NO _{2( g )}+ O _{2( g )}:$ Rate $= k \left[ N _2 O _5\right]$
(b) $CHCl _{3( g )}+ Cl _{2( g )} \rightarrow CCl _{4( g )}+ HCl _{( g )}:$ Rate $= k \left[ CHL _3\right]\left[ Cl _2\right]^{1 / 2}$
(c) $C _2 H _5 Cl _{( g )} \rightarrow C _2 H _{4( g )}+ HCl _{( g )}$ : Rate $= k \left[ C _2 H _5 Cl \right]$
(d) $2 NO _{2( g )}+ F _{2( g )} \rightarrow 2 NO _2 F _{( g )} \rightarrow$ : Rate $= k \left( NO _2\right]\left[ F _2\right]$
What is abnormal colligative property? Explain the reasons.
How is dioxygen prepared in laboratory from $KClO_3?$
→Optical isomerism in $2$-chlorobutane. Explain
→Give the structures of the following compounds :
Sulfides of cation of group II are precipitated in acidic solution $(H_2S + HCl)$ whereas sulfides of cations of group IIIB are precipitated in ammoniacal solution of $H_2S$. Comment on the relative values of solubility product of sulfides of these.
→Derive the relationship between degree of dissociation and dissociation constant of a weak base.
###
Derive the expression of Ostwald’s dilution law in case of a weak base.
###
Derive the expression of Ostwald’s dilution law in case of a weak base.
What are monohydric alcohols? How are they classified?
How instantaneous rate of reaction is determined?