MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 4

The rate constant of a zero order reaction is

$0.2\:mol\:dm^{-3}h^{-1}$ . If the concentration of the reactant after 30 minutes is

$0.05\:mol\:dm^{-3}$ . Then its initial concentration would be:

0%
$0.15\:mol\:dm^{-3}$
0%
$1.05\:mol\:dm^{-3}$
0%
$0.25\:mol\:dm^{-3}$
0%
$4,00\:mol\:dm^{-3}$

Explanation

For a zero order reaction,

$t = \frac {[A_0] -[A] }{ k }$

$t = 30 min = \frac{1}{2} hr$

$k = 0.2 mol dm^{-3} h^{-1}$

$[A] = 0.05 mol dm^{-3}$

On putting these values in the equation above, we get

$A_o = 0.15 mol dm^{-3}$

90% of a first order reaction was completed in 100 min. How much time it will take for 80% completion of a reaction :

0%
90 min
0%
80 min
0%
70 min
0%
60 min

Explanation

We can use

$k_{1}=\frac{2.303}{t}log\frac{a}{a-x}$ formula first to calculate the K (reaction constant) for the reaction.
Now,putting given conditions in above equation we get

$k_{ 1 }=\frac { 2.303 }{ 100 } log\frac { 1 }{ 0.1 }$
this gives K=0.02303
now using same formula used above but for 80% yield we can calculate time required

$0.02303=\frac { 2.303 }{ x } log\frac { 1 }{ 0.2 }$
from this we get time =70 min

The thermal decomposition of a compound is of first order. If 50% of a sample of the compound is decomposed in 120 min how long will it take for 90% of the compound to decompose ?

0%
399 min
0%
410 min
0%
250 min
0%
120 min

Explanation

$\displaystyle k\, =\, \frac{0.6932}{120}$ (i)

$\displaystyle k\, =\, \frac{2.303}{t}\, log\, \frac{a}{0.10a}$

$\displaystyle =\, \frac{2.303}{t}\, log\, 10$ (ii)

Equating (i) and (ii).

$\displaystyle \frac{0.6932}{120}\, =\, \frac{2.303}{t}$
t = 399 min

In a zero order reaction half life is 100 sec. After how much time 78 fraction of reactant will be reacted?

0%
300 sec
0%
200 sec
0%
175 sec
0%
25 sec

The half-life period of a first order reaction is

$15\: minute.$ How much reaction will be completed in

$30\:minute$ ?

0%
$\;75\%$
0%
$\;80\%$
0%
$\;95\%$
0%
$\;100\%$

Explanation

$75\%$ completion time

$=T = 2t_{1/2} = 30$ min as

$t_{1/2} = 15$ min

Reaction will complete

$75\ %$ in

$30$ min only.

$70\%$ of a first order reaction was completed in

$70\ min$ . What is the half-life of the reaction?

0%
$4.2\ min$
0%
$42\ min$
0%
$4.2\ hr$
0%
$42\ s$

Explanation

As this is a first order reaction we can use

$k_{1}=\dfrac{2.303}{t}log\dfrac{a}{a-x}$ to get value of

$k_1$ .

So putting value is above equation we get

$k_{ 1 }=\dfrac { 2.303 }{ 70 } log\dfrac { 1 }{ 0.3 }$

Hence,

$k_!=$ 0.017

Now to calculate half-life we can use

$t_{1/2}=\dfrac { 0.693 }{ k }$

Putting values in above equation we get

Half-life

$=42$ min

A radioactive sample has a half-life period 1500 years. A sealed tube containing 1 g of the sample will contain _________ after 3000 year.

0%
1 g of the sample
0%
0.5 g of the sample
0%
0.25 g of the sample
0%
0.025g of the sample

The half-life period of a radioactive substance is 20 minutes. The time taken for 1 g of the substance to reduce to 0.25g will be:

0%
30 minutes
0%
40 minutes
0%
60 minutes
0%
10 minutes

For a reaction

$2A + B \rightarrow$ product, rate law is :

$-\displaystyle \frac{d[A]}{dt}=k[A]$
At a time when

$t =\displaystyle \frac{1}{k},$ concentration of the reactant is : (

$C_0 =$ initial concentration)

0%
$\displaystyle \frac{C_0}{e}$
0%
$C_0e$
0%
$\displaystyle \frac{C_0}{e^2}$
0%
$\displaystyle \frac{1}{C_0}$

Explanation

$2A+B\quad \rightarrow \quad products\quad \quad \quad \quad\displaystyle \frac { -d[A] }{ dt } =k[A]\\ A=A_{ 0 }e^{ kt }\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad when\quad t_{ 2 }l/k\\ A=A_{ 0 }e^{ 1 }\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad A=A_0/e$

80% of a first order reaction was completed in 70 min. How much time will it take for 90% completion of a reaction?

0%
114 min
0%
140 min
0%
100 min
0%
200 min

Explanation

Using

$k_{1}=\dfrac{2.303}{t}log\dfrac{a}{a-x}$ , we can find out the value of K. Using K time required for 90% reaction can be calculated.

By putting values in the above equation, we will get

$K=\dfrac { 2.303 }{ 70 } log\dfrac { 1 }{ 0.2 } = 0.023$

Now, again using the same equation as above, but for 90% completion, we can calculate the time.

$t=\dfrac { 2.303 }{ 0.023} log\dfrac { 1 }{ 0.1 }$

So, we get time

$= 100\ min$ .

A reaction which is of first order w.r.t reactant

$A$ , has a rate constant

$6$

${min}^{-1}$ . If we start with

$[A]=0.5mol$

${L}^{-1}$ , when would

$[A]$ reach the value of

$0.05 mol$

${L}^{-1}$ ?

0%
$0.384min$
0%
$0.15min$
0%
$3min$
0%
$3.84min$

Explanation

As we know, for a first order reaction,

$t=\cfrac{2.303}{k}\log{\cfrac{a}{a-x}}$

By putting values, we get

$t=\cfrac{2.303}{6}\log{\cfrac{0.5}{0.05}}=0.384$ min

For the reaction:

${H}_{2}+{Cl}_{2}\overset { sunlight }{ \longrightarrow}2HCl$ the order of reaction is :

0%
$0$
0%
$2$
0%
$1$
0%
$3$

At 500 K, the half-life period of a gaseous reaction at an initial pressure of 80 kPa is 350 s. When the pressure is 40 kPa, the half-life period is 175 s. The order of the reaction is:

0%
zero
0%
one
0%
two
0%
three

Explanation

$P_1= 80kPa, (t_{1/2})_1 = 350 s$

$P_2= 40kPa, (t_{1/2})_2 = 175 s$

$\dfrac{p_1}{p_2} = \dfrac{(t_{1/2})_1}{(t_{1/2})_2 }=\dfrac{a_1}{a_2}$

$\Rightarrow t_{1/2}\propto a (zero \ order)$

A mixture of 2 gases -A (reactive) and X (inert) is kept in a closed flask when A decomposes as per the following roaction at 500 k

$\displaystyle A(g)\rightarrow 2B(g)\rightarrow 2B(g)+3C(\varphi )+4D(s)$

$\displaystyle P^{\circ}=210mm Hg$

$\displaystyle P_{t}(10\, mins)=330mm\, Hg$

$\displaystyle P_{\infty t}=430mm\, Hg$
Vapour pressure of

$\displaystyle C(\varphi )$ at 500 K is 20 mm Hg
Calculate the half-life of A ?

0%
5 mins
0%
10 mins
0%
12 mins
0%
Given data is not sufficient to answer

When sugar is stirred with a spoon in a glass of water, more sugar is dissolved and faster. Why?

0%
Spoon acts as a catalyst
0%
On stirring, temperature increases
0%
Stirring increases the rate of interaction
0%
Spoon increases the attraction between the molecules of water and sugar

The time required for

$100\%$ completion of a zero order reaction is:

0%
$ak$
0%
$\dfrac{a}{2k}$
0%
$\dfrac{a}{k}$
0%
$\dfrac{2k}{a}$

Explanation

Hint: The rate of zero reactions is always equal to the rate constant of the reactions.

Correct Option:

$C$

Explanation for correct option:

Let's write the kinetic equation of zero-order reaction as:

$x = [A] - [A]_0 =k t$

Where

$x$ is the amount of reactant converted into a product,

$[A] =$ Final concentration

$[A]_0 -$ Initial concentration

$k$ is the rate constant and

$t$ is time.

If the initial concentration of reactants given is completely converted into a product, then we can say that it is

$100 \%$ completion reaction.

So, let us consider the initial concentration of reactant as a that is

$[A]_{\circ}=a$

$[A] = [A_0] - kt$

$0 = a - kt_{100}$

Time required for

$100\%$ completion is;

$a=k \times t_{100}$

Then,

$t_{100}=\dfrac{a}{k}$

Which among the following reactions is an example of instantaneous reaction under normal condition?

0%
$2{H}_{2}+{O}_{2}\rightarrow 2{H}_{2}O$
0%
${N}_{2}+{O}_{2}\rightarrow 2NO$
0%
$NaOH+HCl\rightarrow NaCl+{H}_{2}O$
0%
${C}_{12}{H}_{22}{O}_{11}+{H}_{2}O\rightarrow {C}_{6}{H}_{12}{O}_{6}+{C}_{6}{H}_{12}{O}_{6}$

Explanation

Instantaneous reactions are those which proceed with a very fast speed

$NaOH+HCl\rightarrow NaCl+{H}_{2}O$ is an strong acid strong base reaction and proceeds instantaneously.

Which of the following expression is correct for first order reactions (

$c_0$ refers to initial concentration of reactant) ?

0%
$t_{1/2}\, \propto\, c_0$
0%
$t_{1/2}\, \propto\, c_0^{-1}$
0%
$t_{1/2}\, \propto\, c_0^{-2}$
0%
$t_{1/2}\, \propto\, c_0^0$

Explanation

As we know,

$t_(1/2) \propto c_0^{1-n}$
so for first order

$t_{1/2}\, \propto\, c_0^0$

$99\%$ at a first order reaction was completed in

$32\:min.$ when will

$99.9\%$ of the reaction complete.(in min) :

0%
48
0%
46
0%
50
0%
45

Explanation

$k\times32\,=\,ln\begin{pmatrix}\displaystyle\frac{100}{1}\end{pmatrix}\;\;\;\;\;\;...(i)$

$\;\;\;\;\;k\times t\,=\,ln\begin{pmatrix}\displaystyle\frac{100}{0.1}\end{pmatrix}\;\;\;\;\;\;...(ii)$

$\;\;\;\;\;eq.\;(ii)/(i)$

$\;\;\;\;\;\displaystyle\frac{t}{32}\,=\,\displaystyle\frac{3\,ln\,10}{2\,ln\,10}$

$\;\;\;\;\;t\,=\,48\:min$

The rate equation for a reaction,

$\displaystyle A\longrightarrow B$ is

$\displaystyle r={ k\left[ A \right] }^{ 0 }$ . If the initial concentration of the reactant is

$a$

$mol.{ dm }^{ -3 }$ , the half-life period of the reaction is:

0%
$\displaystyle \frac { a }{ 2k }$
0%
$\displaystyle \frac { k }{ a }$
0%
$\displaystyle \frac { a }{ k }$
0%
$\displaystyle \frac { 2a }{ k }$

Explanation

Reaction is

$A\longrightarrow B$

Rate is given as

$r=k[A]^0=k$

Order of the reaction is 0.

Initial concentration of A is

$A_0=a$

Order of the reaction is zero.

Half-life,

$t_{1/2}=\dfrac{[A_0]}{2k}=\dfrac{a}{2k}$

The time taken for

$10$ % completion of a first order reaction is

$20$ min. Then, for

$19$ % completion, the reaction will take:

0%
$40$ min
0%
$60$ min
0%
$30$ min
0%
$50$ min

Explanation

When the reaction is 10% completed,

$k=\dfrac { 2.303 }{ 20 } \log { \dfrac { 100 }{ 100-10 } }$ ...(i)
When the reaction is 19% completed,

$k=\dfrac { 2.303 }{ t } \log { \dfrac { 100 }{ 100-19 } }$ ...(ii)
From equation (i) and (ii),

$\Rightarrow \dfrac { 2.303 }{ 20 } \log { \dfrac { 100 }{ 90 } } =\dfrac { 2.303 }{ t } \log { \dfrac { 100 }{ 81 } }$

$\Rightarrow \dfrac { 1 }{ 20 } \times 0.04575=\dfrac { 1 }{ t } \times 0.09151$

$\therefore t=40$ min.

For a zero order reaction :

0%
${ t }_{ 1/2 }\propto { R }_{ o }$
0%
${ t }_{ 1/2 }\propto \dfrac{1}{{ R }_{ o }}$
0%
${ t }_{ 1/2 }\propto { R }_{ o }^{ 2 }$
0%
${ t }_{ 1/2 }\propto \dfrac{1}{{ R }_{ o }^{2 }}$

Explanation

For a zero-order reaction ,

$[R]=-kt+[R_0]$

When

$t=t_{1/2}$ ,

$[R]=[R_0/2]$

$[R_0/2]=-kt_{1/2}+[R_0]$

$t_{1/2}=\cfrac{[R_0]}{2k}$

$t_{1/2}$ is proportional to

$[R_0]$

Hence, the correct option is

$\text{A}$

For a reaction

$X\longrightarrow\ Y$ , the graph of the product concentration (x) versus time (t) came out to be straight line passing through the origin. Hence the graph of

$\cfrac{-d[X]}{dt}$ and time would be:

0%
straight line with a negative slope and an intercept on y-axis
0%
straight line with positive slope and an intercept on y-axis
0%
a straight line parallel to x-axis
0%
a hyperbola

Explanation

If the product concentration is

$x$

For a zero-order reaction

$\cfrac{x}{t}=k$

Thus, the graph would be a straight line passing through the origin. So the given information is for zero order reaction. for zero order reaction, the rate of reaction is constant. Thus, the plot of rate vs time ie.,

$-\cfrac{d[X]}{dt}$ vs time will be a straight line parallel to the x-axis

A first order reaction is given as

$A\rightarrow$ products. Its integrated equation is:

0%
$k=\dfrac{2.303}{t}log\dfrac{a-x}{a}$
0%
$k=\dfrac{1}{t}log\dfrac{a}{a-x}$
0%
$k=\dfrac{2.303}{t}log\dfrac{a}{a-x}$
0%
$-k=\dfrac{1}{t}log\dfrac{a-x}{a}$

Explanation

A first-order reaction is given as

$A \longrightarrow \text{products}$ .

$\dfrac{d[A]}{dt} = -k[A]$

On solving differential equation,

$ln \dfrac{a-x}{a} = -kt$

$kt = ln \dfrac{a}{a-x}$

$\displaystyle k=\frac{1}{t}ln\frac{a} {a-x}$

$k$ is the rate constant,

$t$ is the time,

$a$ is the initial concentration of A and

$(a-x)$ is the concentration of A at time

$t$ . When natural logarithm is changed to logarithm at base 10, the equation becomes :

$\displaystyle k=\frac{2.303}{t}log _{10}\frac{a} {a-x}$

Hence, option

$C$ is correct.

The relationship between rate constant and half-life period of zero order reaction is given by:

0%
$\displaystyle t_{1/2} = [A_0]2k$
0%
$\displaystyle t_{1/2} = \frac{0.693}{k}$
0%
$\displaystyle t_{1/2} = \frac{[A_0]}{2k}$
0%
$\displaystyle t_{1/2} = \frac{2[A_0]}{2k}$

Explanation

The relationship between rate constant and half-life period of zero-order reaction is given by:

$t = \dfrac{[A_0] - [A]}{k}$

$t = t_{1/2}$ ,

$[A] = \dfrac 12 [A_0]$

$t_{1/2}=\dfrac { [{ A }_{ 0 }] }{ 2k }$

Hence, option

$C$ is correct.

In the first order reaction, 75% of the reactant gets disappeared in 1.386 h. The rate constant of the reaction is:

0%
$3.0 \times { 10 }^{ -3 }{ s }^{ -1 }$
0%
$2.8 { \times 10 }^{ -4}{ s }^{ -1 }$
0%
$17.2 { \times 10 }^{ -3 }{ s }^{ -1 }$
0%
$1.8 { \times 10 }^{ -3 }{ s }^{ -1 }$

Explanation

For a first-order reaction, the rate constant will be

$k=\dfrac { 2.303 }{ t } \log { \dfrac { a }{ a-x } }$

where,

K is rate constant.

t is time taken for x amount of reactant to convert to products.

a is initial concentration of reactant.

$t=1.386\times60\times60sec$

$x=0.75a$

$k=\dfrac { 2.303 }{ 1.386\times60\times60 } \log { \dfrac { a }{ a-0.75a } }$

$\Longrightarrow k= \dfrac { 2.303 }{ 1.386\times3600 } \times\log { 4 } =2.8\times10^{ -4 }$

The first order integrated rate equation is:

0%
$k\, =\, \displaystyle \frac {x}{t}$
0%
$k\, =\, -\displaystyle \frac {2.303}{t}\, log\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$
0%
$k\, =\, \displaystyle \frac {1}{t}\, ln\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$
0%
$k\, =\, \displaystyle \frac {1}{t}\, ln \displaystyle \frac {x}{\alpha(\alpha\, -\, x)}$

Explanation

For first order,

rate

$=\, \displaystyle \frac {d[R]}{dt}\, =\, k[R]$

$=\, \displaystyle \frac {d[R]}{[R]}\, =\, kdt[R]$ ....(i)

On integrating Eq. (i)

$\displaystyle \int \displaystyle \frac {d[R]}{[R]}\, =\, k\,\int dt$

In [R] = - kt + C ....(ii)

At t = 0, [R] = [R

$_0$ ]

[where, R = final concentration, ie,

$\alpha\, -\, x$ and

$R_0$ is the initial concentration, ir,

$\alpha$ ].

$[R_0]$ = C

On putting the value of C in Eq. (ii), we get

ln[R] = - kt + ln

$[R_0]$

- kt = ln [R] - ln

$[R_0]$

kt = ln

$[R_0]$ - ln [R]

$k\, =\, \displaystyle \frac {1}{t}\, ln\, \displaystyle \frac {[R_0]}{[R]}$

$k\, =\, \displaystyle \frac {1}{t}\, ln\, \displaystyle \frac {\alpha}{\alpha\, -\, x}$

Found the enthalpy change of the forward reaction ?

0%
B
0%
C
0%
D
0%
E

Assertion: Reactions happen faster at higher temperatures.
Reason: As temperatures increase, there is also an increase in the number of collisions with the required activation energy for a reaction to occur.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true
0%
Both Assertion and Reason are false

The minimum energy required for molecules to react and form compounds can be defined as :

0%
Reduction potential
0%
Ionization energy
0%
Electronegativity
0%
Heat of formation
0%
Activation energy

Explanation

Activation energy may also be defined as the minimum energy required to start a chemical reaction. The activation energy of a reaction is usually denoted by E_a and given in units of kilojoules per mole

$(kJ/mol)$ or kilocalories per mole

$(kcal/mol).$

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state

The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation.

The time required for

$100\%$ completion of zero order reaction is:

0%
$\displaystyle\frac{a}{k}$
0%
$\displaystyle\frac{a}{2k}$
0%
$\displaystyle\frac{2a}{k}$
0%
$\displaystyle\frac{k}{a}$

Explanation

$Explanation:$

For a zero-order reaction,

$x=kt$
Where,

$x =$ Amount of substance

$k =$ Rate constant

$t =$ Time

For 100% completion of the reaction, x = a

Therefore,

$a = kt$

$t=\cfrac ak$

Hence, the correct option is

$\text{A}$

$$x=kt

In a reaction, the potential energy of the reactants is 40 kJ/mol, the potential energy of the products is 10 kJ/mol and the potential energy of the activated complex is 55 kJ/mol. What is the activation energy for the reverse reaction?

0%
45 kJ/mol
0%
-30 kJ/mol
0%
15 kJ/mol
0%
35 kJ/mol
0%
-55 kJ/mol

Explanation

In a reaction the potential energy of the reactants is 40 kJ/mol, the potential energy of the products i 10 kJ/mol and the potential energy of the activated complex is 55 kJ/mol. What is the activation energy for the reverse reaction?
A
45 kJ/mol
The activation energy for the reverse reaction is the difference between the potential energy of the activated complex and the potential energy of the product of forward reaction. It is

$\displaystyle 55 - 10 = 45 \: kJ/mol$

Assertion: Increasing the concentration of reactants will cause a reaction to proceed faster.
Reason: More reactants lowers the activation energy of a reaction.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true
0%
Both Assertion and Reason are false

In a first order reaction, the concentration of the reactant decreases from 0.6 M to 0.3 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M in minutes is:

0%
1.2
0%
12
0%
30
0%
3

Explanation

In a first-order reaction, the concentration of the reactant decreases from

$0.6\ M$ to

$0.3\ M$ in

$15$ minutes. Since the concentration is reduced to one half, the half-life period is

$15$ minutes.

The concentration changes from

$0.1\ M$ to

$0.025\ M$ . Hence, the concentration reduces to one-fourth of the original concentration. The time taken for the concentration to change from

$0.1\ M$ to

$0.025\ M$ in minutes is equal to two half-life periods or

$\displaystyle 2 \times 15 = 30$ minutes.

Hence, option

$C$ is correct.

The addition of a catalyst to a reaction changes which of the following?

0%
Activation energy
0%
Entropy
0%
Enthalpy
0%
Nature of the products

How an increase in concentration is related to number of collisions?

0%
Directly
0%
Inversely
0%
Has no effect
0%
None

Based on the following reaction and rate data:

$\displaystyle A\left( g \right) +B\left( g \right) \rightarrow products$

Experiment	$[A]$ mol/L	$[B]$ mol/L	Initial Rate (M/s)
1	0.060	0.010	0.040
2	0.030	0.010	0.040
3	0.030	0.020	0.080

What is the order of the reaction with respect to

$A$ ?

0%
0
0%
1
0%
2
0%
3
0%
4

Explanation

Analyzing from row

$1$ and

$2$ ,change in concentration of

$A$ is not affecting initial rate of reaction so reaction is zero order with respect to

$A$ .

What does the exponential factor represent?

0%
The total number of reactants in a reaction.
0%
The amount of energy needed to start a chemical reaction.
0%
The fraction of reactants that have approached the activation energy hill and made it over per number of attempts.
0%
The fraction of reaction energy given off per unit of time.
0%
The fraction of products that have approached the activation energy hill and made it over per number of attempts.

Based on the following reaction and rate data:

$\displaystyle A\left( g \right) +B\left( g \right) \rightarrow products$

Experiment	$[A]$ mol/L	$[B]$ mol/L	Initial Rate (M/s)
1	0.060	0.010	0.040
2	0.030	0.010	0.040
3	0.030	0.020	0.080

What is the order of the reaction with respect to

$B$ ?

0%
0
0%
1
0%
2
0%
3
0%
4

Explanation

Analyzing from row

$2$ and

$3$ ,when change in concentration of

$B$ is doubled then initial rate of reaction also doubles so rate of reaction is varying in same order as concentration of B. Hence reaction is first order with respect to

$B$ .

Refer to the heating curve for

$H_{2}O$ above.

Which region indicates a solid?

0%
A
0%
B
0%
C
0%
D
0%
E

Refer to the heating curve for

$H_{2}O$ above.

Where is the temperature of

$H_{2}O$ changing at

$\dfrac {1^{\circ}C}{g.cal}$ ?

0%
A
0%
B
0%
C
0%
D
0%
E

Explanation

$1^oC$ /g.cal is specific heat for liquid water so

$C$ region represents this.

Assertion: The rate of the reaction of iron and hydorchloric acid will increase if the iron pieces are cut into smaller pieces.
Reason: Increasing surface area of a reactant helps to reduce activation energy.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true
0%
Both Assertion and Reason are false

Which letter shows the potential energy of the products?

0%
A
0%
B
0%
C
0%
D
0%
E

Explanation

The letter

$D$ shows the potential energy of the products.
The letter

$A$ shows the potential energy of the transition state.
The letter

$B$ shows the energy of activation.
The letter

$C$ shows the enthalpy change for the reaction.
The letter

$E$ shows the activation energy of the backward reaction.

In a multistep chemical process, the rate-limiting step is the step in the reaction with the :

0%
Highest activation energy & fastest reaction rate
0%
Highest activation energy & slowest reaction rate
0%
Lowest activation energy & fastest reaction rate
0%
Lowest activation energy & slowest reaction rate
0%
Greatest concentration of the reactants and products

Explanation

The rate determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate determining step can be compared to the neck of a funnel. In above above plot step $A$ is rate determining step.

Statement I : Catalysts decrease the rate of a chemical reaction
Because
Statement II : Catalysts decrease activation energy

0%
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1 .
0%
Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%
Statement 1 is correct but Statement 2 is not correct.
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Statement 1 is not correct but Statement 2 is correct.
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Both the Statement 1 and Statement 2 are not correct.

Which reaction in the diagram has the lowest activation energy?

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$A$
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$B$
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$C$
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$D$
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All reactions have equal activation energy

Explanation

In the given curves, activation energy will be lowest for the curve in which the peak is lowest of all as the initial energy of reactants is same for all curves.

$\Rightarrow \quad$

$D$ has the lowest activation energy.

How would the activation energy of the reaction diagram above be affected by an increase in temperature?

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The activation energy curve would grow taller.
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The activation energy curve would grow shorter.
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The activation energy curve would become inverted.
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The activation energy curve would not be affected by an increase in temperature.

In the following image, which reverse reaction has the highest activation energy?

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E to D
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G to E
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C to A
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This cannot be determined from the given information.

Two molecules collide and a reaction not occur. Which of the following is not a valid explanation for this?

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The molecules were not in the proper states of matter.
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The molecules did not have enough kinetic energy.
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The molecules were not oriented correctly when they struck each other.
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The temperature of the reaction mixture was not high enough.

Explanation

For a reaction to occur,

$(i)$ The two molecules must collide with proper orientation.

$(ii)$ They should have high kinetic energy.

$(iii)$ The temperature should be high.

$(iv)$ But the state of matter does not have any effect on the reaction.

The states of matter molecules do not affect the rate of reaction.

Activation energy, by definition, is the minimum amount of energy required for reactants to form products. Which of the following is not true of activation energy?

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The value of activation energy varies from reaction to reaction.
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If molecules are moving fast enough, the collisions will result in kinetic energy that can be used to bend, stretch, and eventually break the bonds of the reactants releasing the potential energy in a usable form.
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All reactions eventually form an intermediate that lowers the overall energy requirements of the reaction resulting in a relatively reasonable activation energy.
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Reactants only form products if the activation energy barrier has been exceeded with usable kinetic energy from the system and /or surroundings.

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 4 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 4 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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