Tag: calculating rate of reaction graphically

Quizzes (0) Questions (19)

Questions Related to calculating rate of reaction graphically

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

$H _2 + l _2 \rightarrow 2 Hl$ (An elementary reaction)
If the volume of the container containing the gaseous mixture is increased to two times, then final rate of the reaction

  1. Become four time
  2. Become $\dfrac{1}{4} th$ of the original rate
  3. Become $2$ times
  4. Become $\dfrac{1}{2}$ of the original rate
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

Assuming an element reaction $H _2O _2+ 3I^-+ 2H^+\to 2H _2O+ I _3^-.$ The effect on the rate of this reaction brought about by doubling the concentration of $I^-$ without changing the order?

  1. The rate would increases by a factor of $3$
  2. The rate would increase by a factor of $8$
  3. The rate would decrease by a factor of $1/3$
  4. The rate would increase by a factor of $9$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

Statement 1: The temperature of a substance always increases as heat energy is added to the system.
Statement 2: The average kinetic energy of the particles in the system increases with an increase in temperature.

  1. Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1

  2. Both the Statement 1 and Statement 2 are correct and Statement 2 is NOT the correct explanation of Statement 1.

  3. Statement 1 is correct but Statement 2 is not correct.

  4. Statement 1 is not correct but Statement 2 is correct.

  5. Both the Statement 1 and Statement 2 are not correct.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Energy can neither be created nor be destroyed. It is only converted in one form or the other. Addition of heat energy increases the kinetic energy which is dependent on absolute temperature.

$\therefore $ Temperature also increases when heat energy is added.
Hence, Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

Instanteneous rate of reaction can be found be :

  1. slope of a rate of reaction vs time

  2. slope of a concentration vs time graph

  3. taking any two points on the graph
  4. both $B$ and $C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We determine an instantaneous rate at time t:

  1. by calculating the negative of the slope of the curve of concentration of a reactant versus time at time t.
  2. by calculating the slope of the curve of concentration of a product versus time at time t.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

For the reaction, $2{ N } _{ 2 }{ O } _{ 5 }\left( g \right) \longrightarrow 4N{ O } _{ 2 }\left( g \right) +{ O } _{ 2 }\left( g \right) $, if the concentration of $N{ O } _{ 2 }$ increases by $5.2\times { 10 }^{ -3 }M$ in $100$ sec, then the rate of reaction is:

  1. $1.3\times { 10 }^{ -5 }M{ s }^{ -1 }$
  2. $5\times { 10 }^{ -4 }M{ s }^{ -1 }$
  3. $7.6\times { 10 }^{ -4 }M{ s }^{ -1 }$
  4. $2\times { 10 }^{ -3 }M{ s }^{ -1 }$
  5. $2.5\times { 10 }^{ -5 }M{ s }^{ -1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\vartheta =\cfrac { 1 }{ 4 } \cfrac { d }{ dt } [{ NO } _{ 2 }]=\cfrac { 1 }{ 4 } (5.2\times { 10 }^{ -5 }M{ s }^{ -1 })\ =1.3\times { 10 }^{ -5 }{ Ms }^{ -1 }$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

A reaction is represented as $2A + B \mapsto  2C + 3D$. The concentration of C at 10 s is 4 moles $l^{-1}$. The concentration of C at 20 seconds is 5.2 moles $l^{-1}$. The rate of reaction of B in the same time interval could be :

  1. $-0.12$ mole $l^{-1} S^{-1}$
  2. $-0.6$ mole $l^{-1} S^{-1}$
  3. $-0.06$ mole $l^{-1} S^{-1}$
  4. $-1.2$ mole $l^{-1} S^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$r _{B} = \dfrac{1}{2}\dfrac{\Delta C}{\Delta t}= \dfrac{-1}{2}\dfrac{(1.2)}{10}$


$(\Delta C= 5.2 -4 = 1.2, \Delta t= 10 sec.)$

$= -0.06$ mole l$^{-1}$ sec$^{-1}$
Hence the answer is $C$.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

For $S{O _2}C{l _{2\left( g \right)}} \to S{O _{2\left( g \right)}} + C{l _{2\left( g \right)}},$ Pressures of $S{O _2}C{l _2}$ at $t = 0$ and $t = 20$ minutes respectively are $700mm$ and $350mm.$ When $\log \left( {{P _0}/p} \right)$ is plotted against time ($t$), slope equals to:

  1. $1.505 \times {10^{ - 2}}{s^{ - 1}}$
  2. $1.202 \times {10^{ - 3}}{\min ^{ - 1}}$
  3. $1.505 \times {10^{ - 2}}{\min ^{ - 1}}$
  4. $0.3465\ {\min ^{ - 1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$k=\dfrac{2.303}{t}log\dfrac{P _0}{P}$ [1st order reaction]

Plot of $log\dfrac{P _0}{P}$ against $t$ will give slope $=  \dfrac{k}{2.303}$
Using given data, $k=\dfrac{2.303}{20}log \dfrac{700}{350}$
$\dfrac{k}{2.303}=\dfrac{log2}{20}=1.5\times 10^{-2}s^{-1}$
Slope $=1.5\times 10^{-2}s^{-1}$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

From the concentrations of R at different times given below. Determine  the average rate of the reaction range: R $\rightarrow$ P in given intervals of time.

t (s) 0 5 10 20 30
$10^{-3}\, \times\, [R] \,(mol\, L^{-1})$ 160 80 40 10 2.5
  1. $3.5\times\,10^{2}$ to $0.42 \, \times\, 10^{2}$ $mol.L^{-1}\, s^{1}$
  2. $7\times\,10^{2}$ to $0.84 \, \times\, 10^{2}$ $mol.L^{-1}\, s^{1}$
  3. $8\times\,10^{3}$ to $0.37 \, \times\, 10^{3}$ $mol.L^{-1}\, s^{1}$
  4. $16\times\,10^{3}$ to $0.75 \, \times\, 10^{3}$ $mol.L^{-1}\, s^{1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average Rate $=\cfrac{ change \ in \ concentration }{ change \ in \ time }$


$\therefore$ here we wanted to find reaction range

Initially avergae rate$=\cfrac{ (160-80) }{ (5-0) }=$$\cfrac{ 80 }{ 5 }$
$=16$

Here, ${ 10 }^{ 3 }$ already given.

$\therefore$ Average rate $=16 \times {10}^{3}\ mol {L}^{-1} {s}^{-1}$

Final average rate $=\cfrac{ (10-2.5) }{(30-20) }=\cfrac{ 7.5 }{ 10 }=0.75$

$\therefore$ Average Rate $=0.75 \times {10}^{3}\ mol {L}^{-1} {s}^{-1}$