Chemical Kinetics Test

Result

Chemical Kinetics Test - 20

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$A_{(g)} \rightarrow B_{(g)}$$ is a first order reaction. The initial concentration of A is 0.2mol $$l^{-1}$$ . After 10 minutes, the concentration of B is found to be 0.18mol $$l^{-1}$$ . Calculate rate constant of the reaction?

A
0.2303

B
2.303

C
0.693

D
0.01

Solution

At t=0, $$[A]=0.2$$ and $$[B]=0.$$
At t=10 min, $$[A]=0.2-0.18=0.02$$, $$[B]=0.18.$$
$$k=\frac{1}{t} ln \frac{[C_o]}{[C_o - C_t]}=\frac{1}{10} ln\frac{[0.2]}{[0.02]} =0.2302.$$
Question 2
1 / -0

The concentration of reaction decreases from 0.2M to 0.05M in 5 minutes. The rate of reaction in $$mol. lit^{-1}.s^{-1}$$ is:

A
$$8.3 \times 10^{-4}$$

B
$$0.05$$

C
$$0.0005$$

D
$$0.15$$

Solution

The expression for the rate of the reaction is $$Rate=\displaystyle \frac{change\ in\ concentration}{time\ interval}.$$

Substitute values in the above expression.

$$Rate=\displaystyle \frac{0.2 - 0.05}{{5 \times 60}}=0.0005 \ mol.lit^{-1}.s^{-1}.$$
Question 3
1 / -0

The rate constant of a first order reaction is $$0.0693$$ $$min^{-1}$$.
Time (in minutes) required for reducing an initial concentration of $$20mol$$ $$lit^{-1}$$ to $$2.5mol$$ $$lit^{-1}$$ is :

A
$$40$$

B
$$30$$

C
$$20$$

D
$$10$$

Solution

For a $$1^{st}$$ order reaction,

$$t=\dfrac{2.303}{k}\ log\left [ \dfrac{R_o}{R} \right ]$$

$$t=\dfrac{2.303}{0.0693}\ log\left [ \dfrac{20}{2.5} \right ]$$

t $$= 30.01 min$$

Option B is correct.
Question 4
1 / -0

From the graph, pick out the correct one :

A
$$E*$$ for the forward reaction is $$E_{A}-E_{B} $$

B
$$E*$$ for the backward reaction is $$E_{C}-E_{A} $$

C
$$E*$$ for reverse reaction is $$\,> \,E*$$ for forward reaction

D
$$E*$$ for forward reaction $$\,>\,E*$$ for backward reaction

Solution

$$E^*\ for\ reverse = E^*\ for\ forward\ + \Delta H$$
$$so\ E^*\ reverse\ >\ E^*\ forward$$
Question 5
1 / -0

A first order reaction was commenced with 0.2 M solution of the reactants. If the molarity of the solution falls to $$0.02M$$ after 100 minutes the rate constant of the reaction is:

A
$$2\times 10^{-2}min^{-1}$$

B
$$2.3\times 10^{-2}min^{-1}$$

C
$$4.6\times 10^{-2}min^{-1}$$

D
$$2.3\times 10^{-1}min^{-1}$$

Solution

$$k=\frac{1}{t}ln\left [ \frac{Co}{Ct} \right ]$$
Where, $$Co$$=initial concentration and $$Ct$$=change in concentration
$$k=\frac{1}{100}ln\left [ \frac{0.2}{0.02} \right ]$$
$$=\frac{ln\ 10}{100}$$
$$=2.3\times 10^{-2}\ \ min^{-1}$$
Question 6
1 / -0

The amount left after completion of average life period in a first order reaction is :

A
$$\dfrac{a(e-1)}{e+1}$$

B
$$\dfrac{a}{e-1}$$

C
$$\dfrac{a(e-1)}{e}$$

D
$$\dfrac{a}{e}$$

Solution

The average life period of a first order reaction is equal to the reciprocal of its rate constant.
Thus $$\lambda = \frac {1} {k}.$$
The amount left after completion of average life period in a first order reaction is$$[A]=[A]_0e^{-kt}=[A]_0e^{-k \times \frac {1} {k}}=\frac{a}{e}.$$
Question 7
1 / -0

With respect to the figure which of the following statement is correct?

A
$$E_{A}$$ for forward reaction is $$E_{A}-E_{B} $$

B
$$E_{A}$$ for forward reaction is $$E_{C}-E_{A} $$

C
$$E_{A}$$ for reverse reaction is greater than $$E_{A}$$for forward reaction

D
$$E_{A}$$ for forward reaction is greater than $$E_{A}$$for backward reaction

Solution

The activation energy for the backward reaction is equal to the sum of the activation energy of the forward reaction and the enthalpy change of the reaction.
Thus the activation energy for the backward reaction is greater than the activation energy for forward reaction.
Question 8
1 / -0

Statement - I : The reactions $$2NO+O_{2}\rightarrow 2NO_{2}$$ and $$2CO+O_{2}\rightarrow 2CO_{2}$$ proceed at the same rate because they are similar.
Statement - II Above two reactions will have different activation energies.

A
Both statements are true Statement - II is correct explantion of Statement - I

B
Both statements are true but Statement -II is not correct explantion of Statement - I

C
Statement - I is true but Statement - II is false.

D
Statement - I is false but Statement - II is true.

Solution

they will not proceed at the same time because they have different activation energy
So [D]
Question 9
1 / -0

A first-order reaction is carried out with an initial concentration of $$10$$ mole per litre and $$80$$% of the reactant changes into the product. Now if the same reaction is carried out with an initial concentration of $$5$$ mol per litre, the percentage of the reactant changing to the product is:

A
$$40$$

B
$$80$$

C
$$160$$

D
$$\text{cannot be calculated}$$

Solution

$$a_{0}=10\ M$$
$$80\%$$ changed into product
$$\therefore a_{t}=10\dfrac{10\times 80}{100}=2\ M$$
If $$a_{0}=5\ M$$, In the same time duration same percentage of initial concentration changes into products because in first order reaction is independent from initial concentration.
Question 10
1 / -0

A first order reaction A $$\rightarrow$$ products has k$$=$$6.93 $$min^{-1}$$ . If 90% of first order reaction B $$\rightarrow$$ products is completed in the time taken by 50% of the first reaction, what is the rate constant for the second reaction?

A
0.76 min

B
23.03min

C
5.6 min

D
6.7 min

Solution

$$A\rightarrow prod.$$
$$K=6.93$$
$$t_{^1/_2} = \frac{ln\ 2}{6.93}$$
$$= 0.100$$
$$Now\ for\ B\rightarrow prod.\ K= \frac{1}{0.100}ln \left [ \frac{a}{a-.9} \right ]$$
$$K= \frac{ln\ 10}{0.1}$$
$$= 23.02\ min$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 20

Result

Chemical Kinetics Test - 20

Score

-

Rank

-

SHARING IS CARING

Question 1

0.2303

2.303

0.693

0.01

Solution

Question 2

$$8.3 \times 10^{-4}$$

$$0.05$$

$$0.0005$$

$$0.15$$

Solution

Question 3

$$40$$

$$30$$

$$20$$

$$10$$

Solution

Question 4

$$E*$$ for the forward reaction is $$E_{A}-E_{B} $$

$$E*$$ for the backward reaction is $$E_{C}-E_{A} $$

$$E*$$ for reverse reaction is $$\,> \,E*$$ for forward reaction

$$E*$$ for forward reaction $$\,>\,E*$$ for backward reaction

Solution

Question 5

$$2\times 10^{-2}min^{-1}$$

$$2.3\times 10^{-2}min^{-1}$$

$$4.6\times 10^{-2}min^{-1}$$

$$2.3\times 10^{-1}min^{-1}$$

Solution

Question 6

$$\dfrac{a(e-1)}{e+1}$$

$$\dfrac{a}{e-1}$$

$$\dfrac{a(e-1)}{e}$$

$$\dfrac{a}{e}$$

Solution

Question 7

$$E_{A}$$ for forward reaction is $$E_{A}-E_{B} $$

$$E_{A}$$ for forward reaction is $$E_{C}-E_{A} $$

$$E_{A}$$ for reverse reaction is greater than $$E_{A}$$for forward reaction

$$E_{A}$$ for forward reaction is greater than $$E_{A}$$for backward reaction

Solution

Question 8

Both statements are true Statement - II is correct explantion of Statement - I

Both statements are true but Statement -II is not correct explantion of Statement - I

Statement - I is true but Statement - II is false.

Statement - I is false but Statement - II is true.

Solution

Question 9

$$40$$

$$80$$

$$160$$

$$\text{cannot be calculated}$$

Solution

Question 10

0.76 min

23.03min

5.6 min

6.7 min

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$$E$$ for reverse reaction is $$\,> \,E$$ for forward reaction

$$E$$ for forward reaction $$\,>\,E$$ for backward reaction