Chemical Kinetics Test

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Chemical Kinetics Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

Following reaction are of first order
$$A\xrightarrow [ ]{ { k }_{ 1 }=2{ sec }^{ -1 } } 2B$$
$$A\xrightarrow [ ]{ { k }_{ 2 }=5{ sec }^{ -1 } } 3C$$
Which of the following statement $$(s)$$ is/are correct?

A
$$ \dfrac { { \left[ B \right] }_{ t } }{ { \left[ C \right] }_{ t } } =\dfrac { 2 }{ 5 }$$ at any time during reaction.

B
After completion of reaction $$\dfrac { { \left[ B \right] } }{ { \left[ C \right] } } =\dfrac { 2 }{ 15 } $$

C
$$ \dfrac { { \left[ B \right] }_{ t } }{ { \left[ C \right] }_{ t } } =\dfrac { 4 }{ 15 }$$ at any time during reaction

D
None of these

Solution

$$A \xrightarrow []{K_1} 2B$$ $$K_1= 2 sec^{-1}$$
For first order,
$$R_1=K_1(B)^1_t$$
$$ A \xrightarrow []{K_2} 3C$$ $$K_2=5sec^{-1}$$
$$R_2= K_2(C)^1_t$$
As the rate of the above reactions are not equal and depend upon concentration of (B) & (C) respectively.
Thus, $$\cfrac {(B)_t}{(C)_t}\neq \cfrac {2}{5}$$ at any time.
So, none of the statement is correct.
Question 2
1 / -0

For the reaction system:
$$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$$, volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to $$O_2$$ and second order with respect to $$NO$$, the rate of reaction will:

A
diminish to one-eighth of its initial value

B
increase to eight times of its initial value

C
increase to four times of its initial value

D
diminish to one-fourth of its initial value

Solution

Order of reaction with respect to $$O_2: first$$

Order of reaction with respect to $$NO: second$$

$$Rate = k [NO]^2[O]$$

Volume is decreased to half, then concentration of each reactants gets doubled.

$$Rate \ New= k[2NO]^2[2O]$$

$$\therefore$$ Rate of reaction will increase to eight times of it’s initial value.
Question 3
1 / -0

Which is correct about zero order reaction?

A
Rate of reaction depends on decay constant

B
Rate of reaction is independent of concentration

C
Unit of rate constant is concentration

D
Unit of rate onstant is concentration time

Solution

For a zero order reaction,
Rate=$$K{ \left[ A \right] }^{ 0 }$$
so, rate of reaction depends on K only. K is called rate constant or decay constant.
Question 4
1 / -0

The activation energy of a reaction is $$94.14$$ kj/mole, and the value of rate constant at $$313$$K is $$1.8\times 10^{-1}sec^{-1}$$. Calculate the frequency factor A.

A
$$[A]=8.25\times 10^{10}s^{-4}$$.

B
$$[A]=7.923\times 10^{10}s^{-4}$$.

C
$$[A]=9.194\times 10^{10}s^{-4}$$.

D
none of these

Solution

By using arrhenius equation,
$$ \implies \log k = \frac{-E_a}{2.303\space RT} + \log A$$
We get
$$ \implies \log A = \log(1.8 \times 10^{-5}) + \frac{94140}{2.303 \times 8.314 \times 313 }$$
$$ \implies (\log 1.8) - 5 + 15.7082)$$
$$\implies 0.2553 - 5 + 15.7082 = 10.9635$$
$$ \therefore \log A = (10.9634) = 9.914 \times 10^{10} $$
Question 5
1 / -0

Find the two third life $$(t_{2/3})$$ of a first order reaction in which $$K=5.48\times 10^{-14}sec^{-1}$$.

A
$$t_{2/3}=5\times 10^{13}$$ sec

B
$$t_{2/3}=2\times 10^{13}$$ sec

C
$$t_{2/3}=1.25\times10^{13}$$ sec

D
none of these

Solution

1st Order Reaction,

$$t = \dfrac{2.303}{k} log \dfrac{[100]}{[100 - x]}$$

Substituting the values, we get
$$t_{2/3} = \dfrac{2.303}{5.48 \times {10}^{-14}}\times log{[3]}$$

$$t_{2/3} = 2 \times {10}^{13}\space sec$$
Question 6
1 / -0

What is the activation energy of a reaction if its rate doubles when temperatures is raised from $$20^o$$C to $$35^o$$C?(R$$=8.314$$ J$$K^{-1}$$ $$mol^{-1}$$)

A
$$34.7$$ kJ $$mol^{-1}$$

B
$$15.1$$ kJ $$mol^{-1}$$

C
$$342$$ kJ $$mol^{-1}$$

D
$$269$$ kJ $$mol^{-1}$$

Solution

Let us consider the Arrhenius equation,

We have

$$\log\dfrac{K_2}{K_1}=-\dfrac{E_a}{2.303R}[\dfrac{T_1-T_2}{T_1T_2}]$$

Given that
Initial temperature $$T_1=20+273=293\;K$$
Final temperature $$T_2=35+273=308\;K$$

$$R=8.314\;J\;mol^{-1}K^{-1}$$

Since, the rate becomes double on raising the temperature,
$$ \therefore r_2=2r_1$$

$$\dfrac{r_2}{r_1}=2$$

Hence the rate constant, $$K \propto r$$

Since $$\frac{K_2}{K_1}=2$$

putting the values

$$\log2=-\dfrac{E_a}{2.303 \times 8.314}[\dfrac{293-308}{293 \times 308}]$$

$$E_a=34673.48\;J\;mol^{-1}$$

$$E_a=34.7\;KJ\;mol^{-1}$$

Hence, the correct option is $$A$$
Question 7
1 / -0

The rate constant for the decomposition of a hydrocarbon is $$2.418\times 10^{-5}S^{-1}$$ at $$546$$ K. If the energy of activation is $$179.9$$ kJ/mol. What will be the value of pre-exponential factor A?

A
$$[A]=9.245\times 10^{12}s^{-1}$$.

B
$$[A]=3.912\times 10^{12}s^{-1}$$.

C
$$[A]=7.83\times 10^{12}s^{-1}$$.

D
None of these

Solution

By using arrhenius equation,
$$ \implies \log k = \frac{-E_a}{2.303\space RT} + \log A$$
We get
$$ \implies \log A = \log(2.418 \times 10^{-5}) + \frac{179900}{2.303 \times 8.314 \times 546 }$$
$$ \implies (\log 2.418) - 5 + 15.7082)$$
$$\implies 0.3834 - 5 + 17.209 = 12.6$$
$$ \therefore \log A = (12.6) = 3.9 \times 10^{12} $$
Question 8
1 / -0

The activation energy for a certain reaction is $$334.4 kJ mol^{-1}$$. How many times larger is the rate constant at $$610$$ K than the rate constant at $$600$$ K?

A
Two times

B
Three times

C
Four times

D
None of these

Solution

Let $${ K }_{ 1 } and { K }_{ 2 }$$ be rate constants.
$$\log { \left( \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } \right) } =\cfrac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] $$

$$\log { \left( \cfrac { { K }_{ 2 } }{ { K }_{ 1 } } \right) } =\left( \cfrac { 334.4 \times { 10 }^{ 3 } }{ 2.303\left( 8.316 \right) } \right) \left[ \cfrac { 610-600 }{ 600 \times 610 } \right] $$

$$\cfrac { { K }_{ 2 } }{ { K }_{ 1 } } =3$$ $$\therefore \quad { K }_{ 2 }$$ is .greater by three times
Question 9
1 / -0

What is the activation energy for a reaction whose rate constant doubles when temperature changes from $$30^oC$$ to $$40^oC$$?

A
$$72.56\ KJmol^{-1}$$

B
$$45.45\ KJmol^{-1}$$

C
$$54.66\ kJ mol^{-1}$$

D
None of these

Solution

Let $${ K }_{ 1 }={ K }$$
then $${ K }_{ 2 }=2K$$
$${ T }_{ 1 }={ 30 }^{ 0 }C+273\\ =303K$$ $${ T }_{ 2 }={ 40 }^{ 0 }+273\\ =313K$$
$$\log { \left( \frac { { K }_{ 2 } }{ { K }_{ 1 } } \right) } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] $$
$$or,\quad \log { \left( \frac { { 2K } }{ { K } } \right) } =\frac { { E }_{ a } }{ 2.303X8.3 } \left[ \frac { 313-303 }{ 313X303 } \right] $$
$${ E }_{ a }=54571\quad KJ/mol\\ =54.5\quad KJ/mol\\ \approx \quad 54.66\quad KJ/mol$$
Question 10
1 / -0

Which one of is first order reaction ?

A
$$NH_4NO_2 \rightarrow N_2 + 2H_2O$$

B
$$2HI\rightleftharpoons H_2 + I_2$$

C
$$2NO_2 \rightarrow 2NO + O_2$$

D
$$2NO + O_2 \rightarrow 2NO_2$$

Solution

$${ NH }_{ 4 }{ NO }_{ 2 }\rightarrow { N }_{ 2 }+2{ H }_{ 2 }O$$
Rate $$=K{ \left[ { NH }_{ 4 }{ NO }_{ 2 } \right] }^{ 1 }$$
$$\therefore \\ $$ order of this reaction is 1.

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Chemical Kinetics Test - 42

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Chemical Kinetics Test - 42

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Question 1

$$ \dfrac { { \left[ B \right] }_{ t } }{ { \left[ C \right] }_{ t } } =\dfrac { 2 }{ 5 }$$ at any time during reaction.

After completion of reaction $$\dfrac { { \left[ B \right] } }{ { \left[ C \right] } } =\dfrac { 2 }{ 15 } $$

$$ \dfrac { { \left[ B \right] }_{ t } }{ { \left[ C \right] }_{ t } } =\dfrac { 4 }{ 15 }$$ at any time during reaction

None of these

Solution

Question 2

diminish to one-eighth of its initial value

increase to eight times of its initial value

increase to four times of its initial value

diminish to one-fourth of its initial value

Solution

Question 3

Rate of reaction depends on decay constant

Rate of reaction is independent of concentration

Unit of rate constant is concentration

Unit of rate onstant is concentration time

Solution

Question 4

$$[A]=8.25\times 10^{10}s^{-4}$$.

$$[A]=7.923\times 10^{10}s^{-4}$$.

$$[A]=9.194\times 10^{10}s^{-4}$$.

none of these

Solution

Question 5

$$t_{2/3}=5\times 10^{13}$$ sec

$$t_{2/3}=2\times 10^{13}$$ sec

$$t_{2/3}=1.25\times10^{13}$$ sec

none of these

Solution

Question 6

$$34.7$$ kJ $$mol^{-1}$$

$$15.1$$ kJ $$mol^{-1}$$

$$342$$ kJ $$mol^{-1}$$

$$269$$ kJ $$mol^{-1}$$

Solution

Question 7

$$[A]=9.245\times 10^{12}s^{-1}$$.

$$[A]=3.912\times 10^{12}s^{-1}$$.

$$[A]=7.83\times 10^{12}s^{-1}$$.

None of these

Solution

Question 8

Two times

Three times

Four times

None of these

Solution

Question 9

$$72.56\ KJmol^{-1}$$

$$45.45\ KJmol^{-1}$$

$$54.66\ kJ mol^{-1}$$

None of these

Solution

Question 10

$$NH_4NO_2 \rightarrow N_2 + 2H_2O$$

$$2HI\rightleftharpoons H_2 + I_2$$

$$2NO_2 \rightarrow 2NO + O_2$$

$$2NO + O_2 \rightarrow 2NO_2$$

Solution

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