Decomposition of A follows first order kinetics by the following equation.
\[4A(g)\xrightarrow{{}}B(g)+2C(g)\]
If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is half-life of A? (Assume only A is present initially).
If \[{{k}_{1}}\] is much smaller than \[{{k}_{2}},\] the most suitable qualitative plot of Potential Energy (PE) versus reaction coordinate (R.Co) for the above reaction.
For the reaction \[A+2B\to \] products(started with concentrations taken in stoichiometric proportion), the experimentally determined rate law is \[-\frac{d[B]}{dt}=k\sqrt{[A]}\sqrt{[B]}\]
Two consecutive irreversible first order reaction can be represented by \[A\xrightarrow[{}]{{{k}_{1}}}B\xrightarrow[{}]{{{k}_{2}}}C\].
The rate equation for A is integrated to obtain
\[{{[A]}_{t}}={{[A]}_{0}}{{e}^{-{{k}_{1}}t}}\] and \[{{[B]}_{t}}=\frac{{{k}_{1}}[{{A}_{0}}]}{{{k}_{2}}-{{k}_{1}}}[{{e}^{-{{k}_{1}}t}}-{{e}^{-{{k}_{2}}t}}]\].
At what time will B be present in the greatest concentration?
Figure shows a graph in \[{{\log }_{10}}\,k\,vs\,\frac{1}{T}\] where, k is rate constant and T is temperature. The straight line BC has slope, \[\tan \,\theta \,=-\frac{1}{2.303}\] and an intercept of 5 on y-axis. Thus, \[{{E}_{a}},\] the energy of activation, is
Two substances A and B are present such that\[[{{A}_{0}}]=4[{{B}_{0}}]\] and half-life of A is 5 minute and that of B is 15 min. If they start decaying at the same time following 1st order kinetics, how much time later will the concentration of both of them would be same?
For the\[2A(g)\xrightarrow{\,}\,3B(g),\,{{t}_{1/2}}=12\,\min \]. Initial pressure exerted by A is 640 mm of Hg. The pressure of the reaction mixture after the time period of 36 min will be
\[(aq)\xrightarrow{{}}B(aq)+C(aq)\] is a first order reaction.
Time
t
\[\infty \]
Moles of reagent
\[{{x}_{1}}\]
\[{{x}_{2}}\]
Reaction progress is measured with the help of titration of reagent P, if all A, B and C reacted with reagent have n factors \[\left[ \text{n factor}:n=\frac{mol.wt.}{eq.wt.} \right]\] in the ratio 1 : 2 : 3 with the reagent. The k in terms of t, \[{{x}_{1}}\] and \[{{x}_{2}}\] is
The reaction cis\[-X\text{trans}-X\] is first order in both directions. At \[\text{25}{}^\circ \text{C,}\] the equilibrium constant is 0.10 and the rate constant \[{{k}_{f}}=3\times {{10}^{-4}}{{s}^{-1}}\]. In an experiment starting with the pure cis-form, how long would it take for half of the equilibrium amount of the trans-isomer to be formed?
The acid catalyzed hydrolysis of an organic, compound 'A' at \[30{}^\circ C\] has a time for half change of 100 minute when carried out in a buffer solution at \[(pH=5)\] and 10 minute when carried out at \[(pH=4)\]. Both times of half change are independent of the initial concentration of A. If rate constant K is defined by \[\frac{-d[A]}{dt}=K{{[A]}^{a}}{{[{{H}^{+}}]}^{b}},\] the values of a and b respectively are
A reaction that is of the first order with respect to reactant A has a rate constant \[6{{\min }^{-1}}\]. If we start with [A]\[0.5\,mol\,{{l}^{-1}},\] when would [A] reach the value \[0.05\,mol\,{{l}^{-1}}\]
\[6{{I}^{-}}(aq)+Br{{O}_{3}}^{-}(aq)+6{{H}^{+}}(aq)\xrightarrow{{}}3{{I}_{2}}(aq)+\] \[B{{r}^{-}}(aq)+3{{H}_{2}}O(\ell )\] these data were obtained when this reaction was studies.
\[[{{I}^{-}}],M\]
\[[Br{{O}_{3}}^{-}],M\]
\[[{{H}^{+}}],M\]
Reaction rate \[(mol\,{{L}^{-1}}{{s}^{-1}})\]
0.0010
0.0020
0.010
\[8.0\times {{10}^{-5}}\]
0.0020
0.0020
0.010
\[1.6\times {{10}^{-4}}\]
0.0020
0.0040
0.010
\[1.6\times {{10}^{-4}}\]
0.0010
0.0040
0.020
\[1.6\times {{10}^{-4}}\]
What is the units of the rate constant for this reaction?
For a first-order homogeneous gaseous reaction \[A\to 2B+C,\] if the total pressure after time t was \[{{P}_{t}}\] and after long time \[(t\to \infty )\] was \[{{P}_{\infty }}\] then k in terms of \[{{P}_{t}},\,{{P}_{\infty }}\] and t is:
A gaseous compound A reacts by three independent first-order processes (as shown in the figure) with rate constants \[2\times {{10}^{-3}},\,3\times {{10}^{-3}}\] and \[1.93\times {{10}^{-3}}{{s}^{-1}}\] products B, C and D respectively. If initially pure A was taken in a closed container with P = 8 atm, then the partial pressure of B (in atm) after 100 s from start of experiment will be:
In a first-order reaction, the reacting substance has half-life period of 10 min. What fraction of the substance will be left after an hour the reaction has occurred?
A graph of volume of hydrogen released vs time for the reaction between zinc and dil. HCl is given in figure. On the basis of this mark the correct option.
A)
Average rate upto 40s is \[\frac{{{V}_{3}}-{{V}_{2}}}{40}\]
doneclear
B)
Average rate upto 40 seconds is \[\frac{{{V}_{3}}-{{V}_{2}}}{40-30}\]
doneclear
C)
Average rate upto 40 seconds is \[\frac{{{V}_{3}}}{40}\]
doneclear
D)
Average rate upto 40 seconds is \[\frac{{{V}_{3}}-{{V}_{1}}}{40-20}\]
If a first-order reaction is completed to the extent of 60% and 20% in time intervals, \[{{t}_{1}}\] and \[{{t}_{2}},\] what is the ratio \[{{t}_{1}}:{{t}_{2}}\]
Consider the following case of completing 1st order reactions.
After the start of the reaction at t = 0 with only A, the [C] is equal to the [D] at all times. The time in which all three concentrations will be equal is given by -
The forward rate constant for the reversible gaseous reaction \[{{C}_{2}}{{H}_{6}}2C{{H}_{3}}\] is\[3.14\times {{10}^{+2}}{{s}^{-1}}\] at 200 K. What is the rate constant for the backward reaction at this temperature, if \[{{10}^{-5}}\] moles of \[C{{H}_{3}}\] and 100 mol of \[{{C}_{2}}{{H}_{6}}\] are present in 10 L vessel at equilibrium?
In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as \[f=\left[ 1-\left( C/{{C}_{0}} \right) \right],\] where \[{{C}_{0}}\] and C are the concentrations of the reactant at the start and after time t. For a first-order reaction:
The inversion of cane sugar proceeds with half-life of 500 min at pH = 5 for any concentration of sugar. However, if pH = 6, the half-life changes to 50 min. The rate law expression for sugar inversion can be written as: