Chemical Kinetics Test

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Chemical Kinetics Test - 37

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Question 1
1 / -0

When the initial concentration of the reaction is doubled, the half-life is also doubled. The order of the reaction will be:

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Half life $$\alpha $$ initial concentration
This corresponds to zero order reaction where $$t_{1/2}=\frac{a}{2k}$$
Question 2
1 / -0

A substance having rate constant $$k$$ and initial concentration $$a$$ reacts according to zero order kinetics. What will be the time for the reaction to go to completion?

A
$$\dfrac { a }{ k } $$

B
$$\dfrac { k }{ a } $$

C
$$\dfrac { a }{ 2k } $$

D
$$\dfrac { 2k }{ a } $$

Solution

For zero order reaction, $$t=\frac{a-at}{k}$$
where at=concentration of reactant at time t
For completion of reaction, at=0
$$t=\frac{a}{k}$$
Question 3
1 / -0

$$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+22$$ kcal. The activation energy for the forward reaction $$50$$ kcal. What is the activation energy for the backward reaction?

A
$$72$$ kcal

B
$$28$$ kcal

C
$$-72$$ kcal

D
$$-28$$ kcal

Solution

$$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+22$$ kcal
$$\Rightarrow \Delta H = -22\ kcal$$
$$\because$$ Activation energy for the forward reaction $$E_a=50$$ kcal
$$\therefore$$ Activation energy for the backward reaction $$E_b=E_a-\Delta H=50+22=72$$ kcal.
Question 4
1 / -0

Raw milk sours in about 4 h at $$27^ o C$$, but in about 48 h in a refrigerator at $$17^ o C$$. What is the activation energy for souring of milk ?

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

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

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

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

Solution

$$K\propto \cfrac {1}{t}$$

$$log \cfrac{K_2}{K_1}=log \cfrac {t_1}{t_2}=\cfrac {E_a}{R}\left ( \cfrac {T_2-T_1}{T_1 \times T_2} \right )$$

$$log \cfrac {48}{4}=\cfrac {E_a}{8.314}\left ( \cfrac {10}{300\times 290} \right )$$

$$E_a=78.3 \, kJ\, mol^{-1}$$
Question 5
1 / -0

$$N_2+3H_2\rightleftharpoons 2NH_3 +22$$kcal
The activation energy for the forward reaction is $$50$$kcal. What is the activation energy for the backward reaction?

A
$$-72$$ kcal

B
$$-28$$ kcal

C
$$+28$$ kcal

D
$$+72$$ kcal

Solution

$$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g) +22$$kcal

As there is 22Kcal heat is released so its exothermic.

$$\because$$ The activation energy for the forward reaction $$=50$$kcal

$$\therefore$$ The activation energy for the backward reaction $$=50+22=+72$$kcal.
Question 6
1 / -0

For the first order reaction, $$t_{99}$$% $$= x \times t_{90}$$%, the value of $$x$$ will be:

A
10

B
6

C
3

D
2

Solution

For $$t_{90\% }$$ ,$$a_t=\frac{10a_0}{100} = \frac{a_0}{10}$$
$$t_{90\% }=\frac{1}{k}In\frac{a-0}{a_0/10}=\frac{In10}{k}$$....(1)
For $$t_{99\% }$$, $$a_t=\frac{1}{100}a-0$$
$$t_{99\% }=\frac{1}{K} In\frac{a_0}{a_0/100}=\frac{2In10}{k}$$....(2)
From (1) and (2),
$$t_{99\% }=2\times t_{90\% }=> x=2$$
Question 7
1 / -0

$$A$$ and $$B$$ are two different chemical species undergoing first order decompositions with half-life periods as $$3$$ and $$4.5$$ minutes respectively. If the initial concentrations of $$A$$ and $$B$$ are in the ratio $$1 : 2$$, The ratio $$C_{t_{(A)}}/ C_{t_{(B)}}$$ after three half-lives of $$A$$ would be:

A
$$3 : 4$$

B
$$1 : 1$$

C
$$1 : 4$$

D
$$4 : 3$$

Solution

For the first order reaction, $$C_t=C_o\left (\cfrac {1}{2}\right)^n$$

where, $$n=\cfrac {t}{t_{1/2}}$$, $$\quad C_o=$$ Initial concentration

After $$3$$ half- lives of $$A$$, $${C_t}_{(A)}={C_o}_{(A)}\left(\cfrac {1}{2}\right)^\cfrac {3t_{1/2}}{t_{1/2}}=\cfrac {{C_o}_{(A)}}{8}$$

After $$3$$ half -lives of $$A$$, for $$B$$, $$\quad n=\cfrac {3\times 3}{4.5}=2$$

$$\therefore {C_t}_{(B)}={C_o}_{(B)}\left(\cfrac {1}{2}\right)^2=\cfrac {{C_o}_{(B)}}{4}$$

$$\therefore \cfrac {{C_t}_{(A)}}{{C_t}_{(B)}}=\left(\cfrac {{C_o}_{(A)}}{{C_o}_{(B)}}\right)\times \cfrac {4}{8}=\cfrac {1}{2}\times \cfrac {4}{8}=1:4$$
Question 8
1 / -0

A reaction takes place in three steps. The rate constants are $$k_{1}, k_{2}$$ and $$k_{3}$$. The over all rate constant $$k = \dfrac {k_{1}k_{3}}{k_{2}}$$. If (energy of activation) $$E_{1}, E_{2}$$ and $$E_{3}$$ are $$60, 30$$ and $$10\ kJ$$, the overall energy of activation is:

A
$$40$$

B
$$30$$

C
$$400$$

D
$$60$$

Solution

$$k=Ae^{-\dfrac{E_{a}}{RT}}$$
$$\therefore k=\dfrac{k_{1}k_{2}}{k_{2}}=\dfrac{A_{1}A_{3}}{A_{2}} \times e^{-\dfrac{E_a^{(1)}+E_a^{(3)}-E_a^{(2)}}{RT}}$$
$$E_{a}=E_a^{(1)}+E_a^{(3)}-E_a^{(2)}$$
$$=40KJ$$
Question 9
1 / -0

Graph between $$\log { k } $$ and $$\cfrac{1}{T}$$ is a straight line with $$OX=5,\tan{\theta}=\left( \cfrac { 1 }{ 2.303 } \right) $$. Hence $${E}_{a}$$ will be:

A
$$2.303\times 2$$ cal

B
$$\cfrac { 5 }{ 2.303 } cal$$

C
$$-2$$

D
None of these

Solution

$$\quad \log { k } =\log { A } -\cfrac { { E }_{ a } }{ 2.303RT } $$

Slope $$=\cfrac { -{ E }_{ a } }{ 2.303R } =\cfrac { 1 }{ 2.303 } (given)\quad $$

$$\quad { E }_{ a }=-2.303\times R\times slope=R=-2\ cal$$

Hence, the correct option is $$\text{C}$$.
Question 10
1 / -0

The activation energy for the forward reaction $$X\rightarrow Y$$ is $$60KJ$$ $${mol}^{-1}$$ and $$\Delta H$$ is $$-20KJ$$ $${mol}^{-1}$$. The activation energy for the backward reaction $$Y\rightarrow X$$ is:

A
$$80KJ$$ $${mol}^{-1}$$

B
$$40KJ$$ $${mol}^{-1}$$

C
$$60KJ$$ $${mol}^{-1}$$

D
$$20KJ$$ $${mol}^{-1}$$

Solution

$$ \implies\Delta H = E_a(f) - E_a(b) $$
$$\implies -20 = 60 - E_a(b) $$
$$ \implies E_a(b) = 60 + 20 $$
$$\implies 80 \space KJ mol^{-1}$$

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Re-Attempt Weekly Quiz Competition

Chemical Kinetics Test - 37

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Chemical Kinetics Test - 37

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 2

$$\dfrac { a }{ k } $$

$$\dfrac { k }{ a } $$

$$\dfrac { a }{ 2k } $$

$$\dfrac { 2k }{ a } $$

Solution

Question 3

$$72$$ kcal

$$28$$ kcal

$$-72$$ kcal

$$-28$$ kcal

Solution

Question 4

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

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

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

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

Solution

Question 5

$$-72$$ kcal

$$-28$$ kcal

$$+28$$ kcal

$$+72$$ kcal

Solution

Question 6

10

6

3

2

Solution

Question 7

$$3 : 4$$

$$1 : 1$$

$$1 : 4$$

$$4 : 3$$

Solution

Question 8

$$40$$

$$30$$

$$400$$

$$60$$

Solution

Question 9

$$2.303\times 2$$ cal

$$\cfrac { 5 }{ 2.303 } cal$$

$$-2$$

None of these

Solution

Question 10

$$80KJ$$ $${mol}^{-1}$$

$$40KJ$$ $${mol}^{-1}$$

$$60KJ$$ $${mol}^{-1}$$

$$20KJ$$ $${mol}^{-1}$$

Solution

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