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Chemical Kinetics

NEET Chemistry · 91 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
In a zero-order reaction, for every 10 o C rise of temperature, the rate is doubled. If the temperature is increased from 10 o C to 100 o C, the rate of the reaction will become
A 256 times
B 512 times
C 64 times
D 128 times
Correct Answer
Option B
Solution

For energy 10° rise in temperature the rate of reaction doubles.

So, rate = 2 n when, n = 1 rate = 2 1 = 2 when, temperature is increased from 10°C to 100°C change in temperature = 100 – 10 = 90°C i.e., n = 9 So, rate = 2 9 = 512 times

Q32
The half-life of a substance in a certain enzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L −-1 to 0.04 mg L −-1 is
A 414 s
B 552 s
C 690 s
D 276 s
Correct Answer
Option C
Solution

We know,

N=N0(12)nN = {N_0}{\left( {{1 \over 2}} \right)^n}

Give N 0 (original amount) = 1.28 mg L –1 N (amount of substance left after time T) = 0.04 mg L –1 ∴\therefore

0.041.28=(12)n{{0.04} \over {1.28}} = {\left( {{1 \over 2}} \right)^n}

⇒\Rightarrow

132=(12)n{1 \over {32}} = {\left( {{1 \over 2}} \right)^n}

⇒\Rightarrow n = 5 ∴\therefore Time required = 5 × t 1/2 = 5 × 138 = 690 s

Q33
The unit of rate constant for a zero order reaction is
A mol L −-1 s −-1
B L mol −-1 s −-1
C L 2 mol −-2 s −-1
D s −-1
Correct Answer
Option A
Solution

Rate = K[A] 0 Unit of k = mol L –1 sec –1

Q34
The rate of the reaction : 2N 2 O 5 → \to 4NO 2 + O 2 can be written in three ways. −d[N2O5]dt=k[N2O5]{{ - d\left[ {{N_2}{O_5}} \right]} \over {dt}} = k\left[ {{N_2}{O_5}} \right] d[NO2]dt=k′[N2O5];  {{d\left[ {N{O_2}} \right]} \over {dt}} = k'\left[ {{N_2}{O_5}} \right];\,\, d[O2]dt=k′′[N2O5]{{d\left[ {{O_2}} \right]} \over {dt}} = k''\left[ {{N_2}{O_5}} \right] The relationship between k and k' and between k and k'' are
A k′=2k;k′′=kk' = 2k ; k'' = k
B k′=2k;k′′=k/2k' = 2k ; k'' = k/2
C k′=2k;k′′=2kk' = 2k ; k'' = 2k
D k′=k;k′′=kk' = k ; k'' = k
Correct Answer
Option B
Solution

Rate of disappearance of reactants = Rate of appearance of products

−12- {1 \over 2}
−d[N2O5]dt{{ - d\left[ {{N_2}{O_5}} \right]} \over {dt}}

=

14{1 \over 4}
d[NO2]dt{{d\left[ {N{O_2}} \right]} \over {dt}}

=

d[O2]dt{{d\left[ {{O_2}} \right]} \over {dt}}

⇒\Rightarrow

12{1 \over 2}
k[N2O5]k\left[ {{N_2}{O_5}} \right]

=

14{1 \over 4}
k′[N2O5]k'\left[ {{N_2}{O_5}} \right]

=

k′′[N2O5]k''\left[ {{N_2}{O_5}} \right]

⇒\Rightarrow

k2=k′4=k′′{k \over 2} = {{k'} \over 4} = k''

⇒\Rightarrow

k′=2k;k′′=k/2k' = 2k ; k'' = k/2
Q35
Which one of the following statements for the order of a reaction is incorrect?
A Order can be determined only experimentally.
B Order is not influenced by stoichiometric coefficient of the reactants.
C Order of a reaction is sum of power to the concentration terms of reactants to express the rate of reaction.
D Order of reaction is always whole number.
Correct Answer
Option D
Solution

Order of a reaction is not always whole number. It can be zero, or fractional also.

Q36
The rate of the reaction, 2NO + Cl 2 → \to 2NOCl is given by the rate equation rate = k[NO] 2 [Cl 2 ]. The value of the rate constant can be increased by
A increasing the temperature
B increasing the concentration of NO
C increasing the concentration of the Cl 2
D doing all of these.
Correct Answer
Option A
Solution

The value of rate constant can be increased by increasing the temperature and is independent of the initial concerntration of the reactants.

Q37
For the reaction N 2 O 5(g) → \to 2NO 2(g) + 1/2O 2(g) the value of rate of disappearance of N 2 O 5 is given as 6.25 × \times 10 −-3 mol L −-1 s −-1 . The rate of formation of NO 2 and O 2 is given respectively as
A 6.25 × \times 10 −-3 mol L −-1 s −-1 and 6.25 × \times 10 −-3 mol L −-1 s −-1
B 1.25 × \times 10 −-2 mol L −-1 s −-1 and 3.125 × \times 10 −-3 mol L −-1 s −-1
C 6.25 × \times 10 −-3 mol L −-1 s −-1 and 3.125 × \times 10 −-3 mol L −-1 s −-1
D 1.25 × \times 10 −-2 mol L −-1 s −-1 and 6.25 × \times 10 −-3 mol L −-1 s −-1
Correct Answer
Option B
Solution

N 2 O 5(g) →\to 2NO 2(g) + 1/2O 2(g)

−d[N2O5]dt=12d[NO2]dt=2d[O2]dt- {{d\left[ {{N_2}{O_5}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{O_2}} \right]} \over {dt}} = 2{{d\left[ {{O_2}} \right]} \over {dt}}

⇒\Rightarrow

d[NO2]dt=−2d[N2O5]dt{{d\left[ {N{O_2}} \right]} \over {dt}} = - 2{{d\left[ {{N_2}{O_5}} \right]} \over {dt}}

= 2 ×\times 6.25 ×\times 10 mol l -1 sec -1 = 1.25 ×\times 10 −-2 mol L −-1 s −-1

d[O2]dt=−12d[N2O5]dt{{d\left[ {{O_2}} \right]} \over {dt}} = - {1 \over 2}{{d\left[ {{N_2}{O_5}} \right]} \over {dt}}

=

6.25×10−32{{6.25 \times {{10}^{ - 3}}} \over 2}

= 3.125 ×\times 10 −-3 mol L −-1 s −-1

Q38
During the kinetic study of the reaction, 2A + B → \to C + D, following results were obtained <br> <table class=tg> <tbody><tr> <th class=tg-de48>Run</th> <th class=tg-de48>[A]/mol L<sup>−-1</sup></th> <th class=tg-de48>[B]/mol L<sup>−-1</sup></th> <th class=tg-de48>Initial rate of formation <br>of D/mol L<sup>−-1</sup> min<sup>−-1</sup></th> </tr> <tr> <td class=tg-pjk9>I.</td> <td class=tg-wv9z>0.1</td> <td class=tg-wv9z>0.1</td> <td class=tg-wv9z>6.0× \times 10<sup>−-3</sup></td> </tr> <tr> <td class=tg-pjk9>II.</td> <td class=tg-wv9z>0.3</td> <td class=tg-wv9z>0.2</td> <td class=tg-wv9z>7.2× \times 10<sup>−-2</sup></td> </tr> <tr> <td class=tg-pjk9>III.</td> <td class=tg-wv9z>0.3</td> <td class=tg-wv9z>0.4</td> <td class=tg-wv9z>2.88× \times 10<sup>−-1</sup></td> </tr> <tr> <td class=tg-pjk9>IV.</td> <td class=tg-wv9z>0.4</td> <td class=tg-wv9z>0.1</td> <td class=tg-wv9z>2.40× \times 10<sup>−-2</sup></td> </tr> </tbody></table> <br>Based on the above data which one of the following is correct?
A Rate = k[A] 2 [B]
B Rate = k[A][B]
C Rate = k[A] 2 [B] 2
D Rate = k[A][B] 2
Correct Answer
Option D
Solution

Rate = k[A] x [B] y For the given situations (I) rate = k(0.1) x (0.1) y = 6.0×\times10 −-3 (II) rate = k(0.2) x (0.3) y = 7.2×\times10 −-2 (III) rate = k(0.3) x (0.4) y = 2.88×\times10 −-1 (IV) rate = k(0.4) x (0.1) y = 2.40×\times10 −-2 Dividing eq.

(I) by eq.

(IV) we get

(0.10.4)x(0.10.1)y=6.0×10−32.4×10−2{\left( {{{0.1} \over {0.4}}} \right)^x}{\left( {{{0.1} \over {0.1}}} \right)^y} = {{6.0 \times {{10}^{ - 3}}} \over {2.4 \times {{10}^{ - 2}}}}

⇒\Rightarrow

(14)x=(14)1{\left( {{1 \over 4}} \right)^x} = {\left( {{1 \over 4}} \right)^1}

⇒\Rightarrow x = 1 On dividing eq. (II) by eq. (III) we get

(0.30.3)x(0.20.4)y=7.2×10−22.88×10−1{\left( {{{0.3} \over {0.3}}} \right)^x}{\left( {{{0.2} \over {0.4}}} \right)^y} = {{7.2 \times {{10}^{ - 2}}} \over {2.88 \times {{10}^{ - 1}}}}

⇒\Rightarrow

(12)y=14{\left( {{1 \over 2}} \right)^y} = {1 \over 4}

⇒\Rightarrow y = 2 ∴\therefore Rate = k[A][B] 2

Q39
For the reaction, N 2 + 3H 2 → \to 2NH 3 , if d[NH3]dt{{d\left[ {N{H_3}} \right]} \over {dt}} = 2 × \times 10 −-4 mol L −-1 s −-1 , the value of −d[H2]dt{{ - d\left[ {{H_2}} \right]} \over {dt}} would be
A 4 × \times 10 −-4 mol L −-1 s −-1
B 6 × \times 10 −-4 mol L −-1 s −-1
C 1 × \times 10 −-4 mol L −-1 s −-1
D 3 × \times 10 −-4 mol L −-1 s −-1
Correct Answer
Option D
Solution

N 2 + 3H 2 →\to 2NH 3

−13d[H2]dt=12d[NH3]dt- {1 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}

⇒\Rightarrow

−d[H2]dt=32d[NH3]dt- {{d\left[ {{H_2}} \right]} \over {dt}} = {3 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}

=

32×2×10−4{3 \over 2} \times 2 \times {10^{ - 4}}

=

3×10−43 \times {10^{ - 4}}

mol L −-1 s −-1

Q40
In the reaction, BrO3(aq)−_{3(aq)}^ - + 5Br(aq)−_{(aq)}^ - + 6H + → \to 3Br 2(l) + 3H 2 O (l) . The rate of appearance of bromine (Br 2 ) is related to rate of disappearance of bromide ions as
A d[Br2]dt=−53d[Br−]dt{{d\left[ {B{r_2}} \right]} \over {dt}} = - {5 \over 3}{{d\left[ {B{r^ - }} \right]} \over {dt}}
B d[Br2]dt=53d[Br−]dt{{d\left[ {B{r_2}} \right]} \over {dt}} = {5 \over 3}{{d\left[ {B{r^ - }} \right]} \over {dt}}
C d[Br2]dt=35d[Br−]dt{{d\left[ {B{r_2}} \right]} \over {dt}} = {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}
D d[Br2]dt=−35d[Br−]dt{{d\left[ {B{r_2}} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}
Correct Answer
Option D
Solution

Rate =

13d[Br2]dt=−15d[Br−]dt{1 \over 3}{{d\left[ {B{r_2}} \right]} \over {dt}} = - {1 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}

∴\therefore

d[Br2]dt=−35d[Br−]dt{{d\left[ {B{r_2}} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}
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