Chemical Kinetics Test

Result

Chemical Kinetics Test - 51

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Question 1
1 / -0

What is the half life of a radioactive substance if $$75%$$ of its given amount disintegrate in $$60\ min$$?

A
$$30\ min$$

B
$$45\ min$$

C
$$75\ min$$

D
$$90\ min$$

Solution

$$t=\cfrac {2.303}{k}\log \left(\cfrac {N_o}{N}\right)$$
Case- (i) $$60= \cfrac {2.303}{k} \log \left(\cfrac {4}{3}\right)$$
Case- (ii) $$t= \cfrac {2.303}{k} \log (2)$$
Dividing we get $$\cfrac {60}{t}= \cfrac {\log 4-\log 3}{\log 2}$$
$$\Rightarrow \cfrac {60}{t}= \cfrac {1}{0.415}$$
$$\Rightarrow t= 60 \times 0.415= 24.9 min= 30 min$$
Question 2
1 / -0

The decomposition of $$N_2O_5$$ is a first order reaction as with a rate constant of $$5\times 10^{-4}s^{-1}$$. If the initial concentration of $$N_2O_5$$ is 0.25 M, then calculate its concentration after 2 minutes.
$$N_2O_5(g) \rightarrow 2NO_2(g) + \dfrac{1}{2}O_2(g)$$

A
0.08 M

B
0.235 M

C
0.5M

D
1 M
Question 3
1 / -0

When ln k is plotted against $${ 1 }/{ T }$$, the slope was found to be -10.7 $$\times $$ $${ 10 }^{ 3 }$$ K, activation energy for the reaction would be :

A
-78.9 kJ $${ mol }^{ -1 }$$

B
2.26 kJ $${ mol }^{ -1 }$$

C
88.9 kJ $${ mol }^{ -1 }$$

D
10.7 kJ $${ mol }^{ -1 }$$

Solution

From the Arrhenius equation, we get
$$lnk=lnA-\dfrac{E_a}{RT}$$
Given,
$$-\dfrac{E_a}{R}=-10.7\times10^3$$

$$\Rightarrow E_a=8.314\times10.7\times10^3\ Jmol^{-1}$$
$$\Rightarrow E_a=88.9\ kJmol^{-1}$$
Question 4
1 / -0

The slope in the activation energy curve is 5.42 $$\times$$ $$10^{3}$$ . The value of the activation energy is approximately

A
104 J $$mol^{-1}$$

B
104 MJ $$mol^{-1}$$

C
104 KJ $$mol^{-1}$$

D
104 J $$mol^{-1}K^{-1}$$
Question 5
1 / -0

For zero order reaction, A$$ \to $$,a graph of rate vs time has slope equal to:( where k$$=$$rate of reaction)

A
k

B
-k

C
zero

D
-2.303 k

Solution
Question 6
1 / -0

$$3/4\ th$$ of first order reaction was completed in $$32\ min, 15/16$$ the part will be completed in ?

A
$$24\ min$$

B
$$64\ min$$

C
$$16\ min$$

D
$$32\ min$$

Solution
Question 7
1 / -0

The rate expression for a reaction for a reaction is $$ \dfrac{-dC}{dt}=\dfrac {\alpha C_0}{1+\beta C_0}$$ where $$\alpha,\beta$$ are constants and C is the concentration of reactant at time 't'. The half -life for this reaction is ?

A
$$\cfrac 1 \alpha ln 2+\cfrac {\beta C_0}{2\alpha}$$

B
$$\cfrac 1 \beta ln 2+\cfrac {\beta C_0}{2\alpha}$$

C
$$\cfrac {\beta ln 2}\alpha$$

D
$$\dfrac \alpha \beta$$

Solution
Question 8
1 / -0

For the first order reaction $$(C=C_{0}e^{-k})$$ and $$T_{av}=K_{1}^{-1}$$. After two average lives, concentration of the reactant is reduced to?

A
$$25\%$$

B
$$75\%$$

C
$$\dfrac{100}{e}\%$$

D
$$\dfrac{100}{e^{2}}\%$$

Solution
Question 9
1 / -0

The decomposition of $$NH_3$$ on platinum surface is zero order reaction. What are the rates of production of $$N_2$$ and $$H_2$$ if $$K= 2.5\times 10^{-4} mol^{-1} L s^{-1}$$?

A
$$2.5 \times 10^{-4}$$ and $$7.5 \times 10^{-4}$$ respectively

B
$$1.25 \times 10^{-4}$$ and $$3.75 \times 10^{-4}$$ respectively

C
$$3.75 \times 10^{-3}$$ and $$1.25 \times 10^{-3}$$ respectively

D
$$1.25 \times 10^{-3}$$ and $$3.25 \times 10^{-3}$$ respectively

Solution

The reaction is zero order. Hence, Rate of the reaction is equal to rate constant.
The reaction is,
$$2NH_3 \rightarrow N_2 + 3H_2$$
Therefore, $$\dfrac {d[{N_2}]}{dt}=2.5\times 10^{-4}$$
$$\dfrac{d[H_2]}{dt}=2.5\times 10^{-4}\times 3=7.5\times 10^{-4}$$
Question 10
1 / -0

The rate constant of a reaction increased by $$5%$$ when temperature is raised from $$27^{o}$$ to $$28^{o}\ C$$. The activation energy of the reaction is $$(\log\ 7=0.8450)$$

A
$$36.5\ KJ$$

B
$$16.5\ KJ$$

C
$$47.5\ KJ$$

D
$$27.5\ KJ$$

Solution