Chemical Kinetics Test

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Chemical Kinetics Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The half year period for a zero order reaction is equal to:

A
$$2K/[A]o$$

B
$$\cfrac{[A]_o}{2k}$$

C
$$\cfrac{0.693}{k}$$

D
$$\cfrac{0.693}{k[A]_o}$$

Solution

Half Year Period for a Zero order reaction is $$\dfrac{[{A}_{0}]}{2k}$$
Where $$[A_0]=$$ initial concentration
$$k=$$ rate constant
Question 2
1 / -0

$$ 2A \rightarrow B + C $$ It would be a zero order reaction when:

A
the rate of reaction is proportional to square of conc. of A

B
the rate of reaction remains same at any conc of A

C
the rate remains unchanged at any conc. of B and C

D
the rate of reaction doubles if conc. of B is increased to double

Solution

It Would be Zero Order Reaction, when the Rate is independent of the Reactant Concentration.

So, the rate of reaction remains same at any conc of A.
Question 3
1 / -0

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $$20^0C$$ to $$35^0C$$? (R = 8.314 J $$mol^{-1}K^{-1}$$)

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

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

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

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

Solution

$$log \dfrac{{k}_{2}}{{k}_{1}}$$ = $$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$$

Substituting the values

$$log\space {2}$$ = $$\dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{293} - \dfrac{1}{308})$$

$${{E}_{a}} = 34.7\space kJ\space {mol}^{-1}$$
Question 4
1 / -0

99% of a first order reaction, was completed in 32 minute. When will 99.9% of the reaction complete?

A
50 minute

B
46 minute

C
49 minute

D
48 minute

Solution

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

Substituting the values,

$$k = \dfrac{2.303}{32} log \dfrac{100}{1}$$

$$k = 0.144\space {min}^{-1}$$

$$t = \dfrac{2.303}{0.144} log \dfrac{100}{0.1}$$

$$t = 48\space min$$
Question 5
1 / -0

For a first order reaction involving decomposition of $$N_2O_5$$ the following information is available
$$2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)$$ rate $$=k[N_2O_5]$$
$$N_2O_5(g)\rightarrow 2NO_2(g)+1/2 O_2(g)$$ rate $$=k^\prime[N_2O_5]$$

A
$$k=k^\prime$$

B
$$k^\prime=2k$$

C
$$k^\prime=1/2k$$

D
$$k>k^\prime$$

Solution

$${2N_2O_5}_{(g)}\longrightarrow {4NO_2}_{(g)}+{O_2}_{(g)}$$
$$\Rightarrow k=\cfrac {[NO_2]^4[O_2]}{[N_2O_5]^2}\longrightarrow (1)$$
$${N_2O_5}_{(g)} \longrightarrow {2NO_2}_{(g)}+{1/2O_2}_{(g)}$$
$$\Rightarrow k'= \cfrac {[NO_2]^2[O_2]^{1/2}}{[N_2O_5]}$$
$$\Rightarrow [N_2O_5]=\cfrac {[NO_2]^2[O_2]^{1/2}}{k'}\longrightarrow (2)$$
$$(2)$$ in $$(1) \Rightarrow k=\cfrac {[NO_2]^4[O_2]}{\cfrac {[NO_2]^4[O_2]}{k'^2}}=k'^2$$
$$\Rightarrow k >k'$$
Question 6
1 / -0

The rate of first order reaction, $$A \rightarrow$$ Products, is $$7.5$$ M/s. If the concentration of $$A$$ is 0.5 M. The rate constant is:

A
3.75

B
2.51

C
15.0

D
8.01

Solution

As we know that,
$$r\ (rate) = K \left[ A \right]$$ ------ 1

where,
$$r =$$ Rate of reaction $$= 7.5 \; {mol}/{(L.s)}$$
$$K =$$ Rate constant $$ = \ ?$$
$$\left[ A \right] =$$ Initial concentration of reactant $$= 0.5 \; {mol}/{L}$$

Therefore,
$$K = \cfrac{7.5}{0.5} = 15.0 \; {s}^{-1}$$

Hence the rate constant for the given reaction is $$15.0\; {s}^{-1}$$.
Question 7
1 / -0

Milk turns sour at $$40^0C$$ three times as faster as at $$0^0C$$. The energy of activation for souring of milk is:

A
4.693 kcal

B
2.6 kcal

C
6.6 kcal

D
none of these

Solution

$$log\dfrac{{k}_{2}}{{k}_{1}} = \dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$$

Milk turns sour at 313 K 3 times as faster as at 273 K.Substituting the values

$$log{1}{3} = \dfrac{{E}_{a}}{2.303 \times 2}(\dfrac{1}{313} - \dfrac{1}{273})$$

$${E}_{a} = 4.693\space kcal$$
Question 8
1 / -0

If a first order reaction is completed to the extent of 75% and 50% in time intervals, $$t_1$$ and $$t_2$$, what is the ratio, $$t_1 : t_2$$?

A
In 2

B
$$\frac{In(3/4)}{In 2}$$

C
2

D
1/2

Solution

Its a 1st Order Reaction

$$\dfrac{{t}_{1}}{{t}_{2}}$$ = $$\dfrac{log \dfrac{[{100}]}{[100 - 75]}}{log \dfrac{[{100}]}{[100 - 50]}}$$

$$\dfrac{{t}_{1}}{{t}_{2}}$$ = $$2$$
Question 9
1 / -0

A $$1^{st}$$ order reaction is $$50\%$$ complete in $$30$$ minutes at $$27^oC$$ and in $$10$$ minutes at $$47^oC$$. Calculate energy of activation for the reaction.
(Assume log $$3$$=$$0.48$$)

A
$$46.87 \, KJ \, mol^{-1}$$

B
$$42.21 \, KJ \, mol^{-1}$$

C
$$44.11 \, KJ \, mol^{-1}$$

D
$$49.20 \, KJ \, mol^{-1}$$

Solution

$${k}_{1} = \dfrac{0.693}{10}$$

$${k}_{2} = \dfrac{0.693}{30}$$

$$log \dfrac{{k}_{2}}{{k}_{1}}$$ = $$\dfrac{{E}_{a}}{2.303 \times R} (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$$

$$log3 = \dfrac{{E}_{a}}{2.303 \times 8.314} (\dfrac{1}{300} - \dfrac{1}{320})$$

$${E}_{a} = 44.11\space KJ\space {mol}^{-1}$$
Question 10
1 / -0

The decomposition $$NH_3$$ gas on a heated tungsten surface gave the following results:
Initial pressure (mm) 65 105 y 185
Half-life (sec) 290 x 670 820
Calculate approximately the values of x and y.

A
x = 410 sec, y = 115 mm

B
x = 467 sec, y = 150 mm

C
x = 490 sec, y = 120 mm

D
x = 430 sec, y = 105 mm

Solution

$${t}_{1/2}\space \alpha\space {P}^{1-n}$$

$$\dfrac{290}{820} = \dfrac{65}{185}$$, Which Implies

$$1 -n = 1$$

$$n = 0$$

$$k =\dfrac{[{P}_{0}]}{2{t}_{1/2}}$$

$$k = 0.112\space mol{L}^{-1}\space {t}^{-1}$$

$$x = \dfrac{105}{2 \times {0.112}}$$ = $$467\space {sec}$$

$$y = 0.112 \times 2\times {670}$$ = $$150\space mm$$

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Initial pressure (mm)	65	105	y	185
Half-life (sec)	290	x	670	820

Chemical Kinetics Test - 49

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Chemical Kinetics Test - 49

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Question 1

$$2K/[A]o$$

$$\cfrac{[A]_o}{2k}$$

$$\cfrac{0.693}{k}$$

$$\cfrac{0.693}{k[A]_o}$$

Solution

Question 2

the rate of reaction is proportional to square of conc. of A

the rate of reaction remains same at any conc of A

the rate remains unchanged at any conc. of B and C

the rate of reaction doubles if conc. of B is increased to double

Solution

Question 3

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

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

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

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

Solution

Question 4

50 minute

46 minute

49 minute

48 minute

Solution

Question 5

$$k=k^\prime$$

$$k^\prime=2k$$

$$k^\prime=1/2k$$

$$k>k^\prime$$

Solution

Question 6

3.75

2.51

15.0

8.01

Solution

Question 7

4.693 kcal

2.6 kcal

6.6 kcal

none of these

Solution

Question 8

In 2

$$\frac{In(3/4)}{In 2}$$

2

1/2

Solution

Question 9

$$46.87 \, KJ \, mol^{-1}$$

$$42.21 \, KJ \, mol^{-1}$$

$$44.11 \, KJ \, mol^{-1}$$

$$49.20 \, KJ \, mol^{-1}$$

Solution

Question 10

x = 410 sec, y = 115 mm

x = 467 sec, y = 150 mm

x = 490 sec, y = 120 mm

x = 430 sec, y = 105 mm

Solution

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