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Determining rates graphically - class-XII

Description: determining rates graphically
Number of Questions: 19
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Tags: reaction kinetics chemistry electrochemistry and chemical kinetics chemical kinetics
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When a chemical reaction takes place, during the course of the reaction the rate of reaction?

  1. Keeps on increasing with time

  2. Remains constant with time

  3. Keeps on decreasing with time

  4. Shows irregular trend with time


Correct Option: C
Explanation:

Rate of reaction $\propto$ conc. of reactants
As the reaction proceeds, concentration of the reactants decreases hence the rate also keeps on decreasing with time.

In a reaction $2HI \rightarrow H _{2} + I _{2}$, the concentration of $HI$ decreases from $0.5\ mol\ L^{-1}$ to $0.4\ mol\ L^{-1}$ in $10$ minutes. What is the rate of reaction during this interval?

  1. $5\times 10^{-3} M\ min^{-1}$

  2. $2.5\times 10^{-3} M\ min^{-1}$

  3. $5\times 10^{-2} M\ min^{-1}$

  4. $2.5\times 10^{-2} M\ min^{-1}$


Correct Option: A
Explanation:

Average rate $= -\dfrac {1}{2} \dfrac {\triangle [R]}{\triangle t} = -\dfrac {1}{2}\times \dfrac {0.4 - 0.5}{10}$


$= \dfrac {1}{2}\times \dfrac {0.1}{10} = 5\times 10^{-3}M\ min^{-1}$.

Which of the following statements is correct?

  1. The rate of a reaction decreases with passage of time as the concentration of reactants decreases.

  2. The rate of a reaction is same at any time during the reaction.

  3. The rate of a reaction is independent of temperature change.

  4. The rate of a reaction decreases with increase in concentration of reaction(s).


Correct Option: A
Explanation:

According to rate law expression,
$Rate \propto [R]$
Thus, rate of a reaction decreases with passage of time as the concentration of reactants decreases.

The rate constant of a zero order reaction is 0.2 mol $d{ m }^{ -3 }{ h }^{ -3 }$. If the concentration minutes is 0.05 mol $d{ m }^{ -3 }$. Then its initial concentration would be:

  1. $6.05 mol d{ m }^{ -3 }$

  2. $0.15 mol d{ m }^{ -3 }$

  3. $0.25 mol d{ m }^{ -3 }$

  4. none of the above


Correct Option: B

Consider the chemical reaction:
$N _2(g)+3H _2(g)\rightarrow 2NH _3(g)$


The rate of this reaction can be expressed; in terms of time and of concentration of $N _2(g), H _2(g)$ $NH _3(g)$. Identify the correct relationship amongst the rate expressions.

  1. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=+\dfrac{1}{2}\dfrac{d[NH _3]}{dt}$

  2. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{3d[H _2]}{dt}=\dfrac{2d[NH _3]}{dt}$

  3. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$

  4. Rate $=-\dfrac{d[N _2]}{dt}=\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$


Correct Option: A
Explanation:

As the reaction proceed, the concentration of the reactant decrease and that of the product increases.
Thus, the rate of the reaction of the reactant and product can be given as follow:
$Rate =-\cfrac{d}{dt}[N _2]=-\cfrac{1}{3}\cfrac{d}{dt}[H _2]=+\cfrac{1}{2}\cfrac{d}{dt}[NH _3]$

Rate of formation of $SO _3$ in the following reaction $2SO _2+O _2\rightarrow 2SO _3$ is $100g$ $min^{-1}$. 


Then the rate of disappearance of $O _2$ is:

  1. $50g$ $min^{-1}$

  2. $40g$ $min^{-1}$

  3. $200g$ $min^{-1}$

  4. $20g$ $min^{-1}$


Correct Option: A
Explanation:

$2{ SO } _{ 2 }+{ O } _{ 2 }\rightarrow 2{ SO } _{ 3 }$


rate $=\dfrac { 1 }{ 2 } \dfrac { \left[ { \triangle SO } _{ 2 } \right]  }{ \triangle t } =\dfrac { \triangle \left[ { O } _{ 2 } \right]  }{ \triangle t } =\dfrac { 1 }{ 2 } \dfrac { \triangle \left[ { SO } _{ 3 } \right]  }{ \triangle t } $

rate of formation of ${ SO } _{ 3 }=$ rate of disappearance of ${ O } _{ 2 }$

$\dfrac { 1 }{ 2 } \dfrac { \triangle \left[ { SO } _{ 3 } \right]  }{ \triangle t } =-\dfrac { \triangle \left[ { O } _{ 2 } \right]  }{ \triangle t } $

$-\dfrac { \triangle w _{{ O } _{ 2 }}   }{ \triangle t }=\dfrac { 1 }{ 2 } \times \dfrac{100}{80}\times 32g{ min }^{ -1 }=20g{ min }^{ -1 }$

So, the correct option is $D$

If concentration of reactants is increased by a factor x then the rate constant k becomes:

  1. $\ln{\frac{k}{x}}$

  2. $\frac{k}{x}$

  3. $k+x$

  4. $k$


Correct Option: D
Explanation:

Rate Constant doesn't depend on the concentration of the reactants.

So, If the conc of reactants is increased by a factor x, then the rate constant k becomes k.

An aqueous solution of $CH _3COOH$ has a pH = $3$ and acid dissociation constant of $CH _3COOH$ is $10^{-5}$. What will be the concentration of acid taken initially?

  1. $0.1$M

  2. $0.11$M

  3. $0.09$M

  4. $0.101$M


Correct Option: A
Explanation:

${H}^{+} = \sqrt{{K} _{a} \times {C}}$


${10}^{-6} = {10}^{-5} \times {C}$

${C} = 0.1\space M$

For the non-equilibrium process, $A + B \rightarrow Products$, the rate is first order with respect to $A$ and second-order with respect to $B$. If $1.0$ mole each of $A$ and $B$ are introduced into a 1-litre vessel and the initial rate was $1.0 \times 10^{-2}$ mol/litre-sec. The rate (in mol $litre^{-1} sec^{-1}$) when half of the reactants have been used:

  1. $1.2 \times 10^{-3}$

  2. $1.2 \times 10^{-2}$

  3. $2.5 \times 10^{-4}$

  4. none of these


Correct Option: A
Explanation:

$rate = k[A]{[B]}^{2}$


Initially 1 mol each of A and B are present and the rate was $1 \times {10}^{-2}\space mol/lit-sec$

When Half of the Reactants are used, the rate becomes $\dfrac{1}{2} \times {[\dfrac{1}{2}]}^{2} \times {10}^{-2}$ = $1.2 \times {10}^{-3}$

 Time  0  5min  10min  15min
 [A]  20mol  18mol  16mol  16 mol

For the reaction $A\longrightarrow Products$; $\frac { -d[A] }{ dt } =k$ and at different time interval, IAI values are given. At $20$ minute, rate will be :

  1. $12 mol /min$

  2. $10 mol/min$

  3. $8 mol/min$

  4. $0.4 mol/min$


Correct Option: D
Explanation:

The concentration decreases linearly with time

Hence moles of A at $20 min$
$=12$
Therefore, $k=\dfrac{-d[A]}{dt}$
$=-\dfrac{12-14}{20-15}$
$=0.4\ mol/min$

$H _2 + l _2 \rightarrow 2 Hl$ (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


Correct Option: A

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 $9$


Correct Option: B

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.


Correct Option: A
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.

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 $C$


Correct Option: B
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.

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. $2.5\times { 10 }^{ -5 }M{ s }^{ -1 }$


Correct Option: A
Explanation:

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

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 $l^{-1} S^{-1}$


Correct Option: C
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 $C$.

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. $0.3465\ {\min ^{ - 1}}$


Correct Option: A
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 $=1.5\times 10^{-2}s^{-1}$

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 $0.75 \, \times\, 10^{3}$ $mol.L^{-1}\, s^{1}$


Correct Option: D
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 $=0.75 \times {10}^{3}\ mol {L}^{-1} {s}^{-1}$

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