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Chemical Kinetics NEET Questions — Free MCQ Practice

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The rate constant , the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are $3.0 \times 10^{- 4} s^{- 1}$ $, 104.4 k J m o l^{- 4}$ and $6.0 \times 10^{14}$ s–1 respectively the value of the rate constant as T $\to \infty$ is :

A.

$2.0 \times 10^{18} s^{- 1}$

B.

$6.0 \times 10^{- 14} s^{- 1}$

C.

$\infty$

D.

$3.6 \times 10^{30} s^{- 1}$

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Explanation

(B) $K = A e^{- E a / R T} W h e n T \to \infty K \to A A = 6 \times 10^{14} s^{- 1}$

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If in the fermentation of sugar in an enzymatic solution that is 0.12 M, the concentration of the sugar is reduced to 0.06 M in 10h and to 0.03 M in 20h, what is the order of the reaction-

A.

1

B.

2

C.

3

D.

0

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Explanation

(A) Half life is independent to the initial concentration of the reactant.

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In a Ist order reaction A $\to$ products, the concentration of the reactant decreases to 6.25% of its initial value in 80 minutes. What is (i) the rate constant and (ii) the rate of the reaction. 100 minutes after the starts, if the initial concentration is 0.2 mole/litre ?

A.

$2.17 \times 10^{- 2} m i n^{- 1}, 3.47 \times 10^{- 4} m o l . l i t r e^{- 1} m i n^{- 1}$

B.

$3.465 \times 10^{- 2} m i n^{- 1}, 2.166 \times 10^{- 4} m o l . l i t r e^{- 1} m i n^{- 1}$

C.

$3.465 \times 10^{- 3} m i n^{- 1}, 2.17 \times 10^{- 3} m o l . l i t r e^{- 1} m i n^{- 1}$

D.

$2.166 \times 10^{- 3} m i n^{- 1}, 2.667 \times 10^{- 4} m o l . l i t r e^{- 1} m i n^{- 1}$

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Explanation

(B). (i) Given $[A_{0}] = 0.2 M & [A] = 0.2 \times \frac{6.25}{100} = 0.0125$

$K = \frac{2.303}{t} \log \frac{[A_{0}]}{[A]} = \frac{2.303}{80} \log \frac{0.2}{0.0125} = 0.0346 m i n^{- 1}$

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The half life time for the decomposition of a substance dissolved in $C C l_{4}$ is 2.5 hour at $30 °$ C. How much of the substance will be left after 10 hours, if the initial weight of the substance is 160 gm ?

A.

20 gm

B.

30 gm

C.

40 gm

D.

10 gm

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Explanation

(D) No of half life periods

$= \frac{T o t a l t i m e}{T i m e o f o n e h a l f l i f e p e r i o d} = \frac{10}{2.5} = 4$

Now we know that the quantity= ${(1 / 2)}^{4} \times 160 = \frac{160}{2 \times 2 \times 2 \times 2} = 10 g m$

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Find out the percentage of the reactant molecules crossing over the energy bariier at 325 K, given that $∆ H_{325} = 0.12 kcal, E_{a (b)} = + 0.02 kcal$

A.

80.62 %

B.

85.23 %

C.

89.27 &

D.

None of the above

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Explanation

Given,

$∆ H_{325} = 0.12 \times 10^{3} kcal, E_{a (b)} = + 0.02 \times 10^{3} kcal (E_{a} can never be negative) ∆ H = E_{a (f)} - E_{a (b)} E_{a (f)} = 0.12 \times 10^{3} + 0.02 \times 10^{3} cal = 0.14 \times 10^{3} cal$

% of molecule crossing over the barrier

$= 100 \times e^{- E_{a (f)} / R T} = 100 \times e^{- 140 / (2 \times 325)} = 100 \times \times 0.8062 = 80.62 %$

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1 mole of gas changes linearly from initial state (2 atm, 10 lt) to final state (8 atm, 4 lt). Find the value of rate constant, at the maximum temperature, that the gas can attain. Maximum rate constant is equal to 20 $s e c^{- 1}$ and value of activation energy is 40 kJ $m o l e^{- 1}$ , assuming that activation energy does not change in this temperature range.

A.

$0.56 \times 10^{- 3} s e c^{- 1}$

B.

$3.16 \times 10^{- 3} s e c^{- 1}$

C.

$1.56 \times 10^{- 3} s e c^{- 1}$

D.

$5.12 \times 10^{- 3} s e c^{- 1}$

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Explanation

(C) Let the equation of straight line is P=mV+c

Now putting the values, we get P+V=12

From the ideal gas equation,

$T = \frac{P V}{n R} = \frac{P V}{R}; T = \frac{(12 - V) V}{R}$

For T to be maximum $\frac{d T}{d V}$ =0

V=6 lt, P=6 atm.

Value of $T_{m a x} = \frac{36}{0.082} = 439 K$

Now Putting the values of $A = 20 s e c^{- 1}$

$E_{a} = 40 \times 10^{3} J m o l e^{- 1}$

we get, $k = A e^{- E_{a} / R T}$

$k = 20 e^{- \frac{40 \times 10^{3}}{8.314 \times 439}}; k = 1.56 \times 10^{- 3} s e c^{- 1}$

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A first order reaction was started with a decimolar solution of the reactant. After 8 minutes and 20 seconds, its concentration was found to be M/100. Determine the rate constant of the reaction.

A.

$4.6 \times 10^{- 3} s e c^{- 1}$

B.

$16.6 \times 10^{- 3} s e c^{- 1}$

C.

$24.6 \times 10^{- 3} s e c^{- 1}$

D.

$40.6 \times 10^{- 3} s e c^{- 1}$

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Explanation

(A) Here a= 0.1 M

$a - x = \frac{M}{100} = 0.01 M$

t=8 minutes 20 seconds=500 seconds

Substituting the values in the first order reaction

$k = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{500} \log \frac{0.1}{0.01} = \frac{2.303}{500} \log 10 = \frac{4.606}{1000} \times 1 = 4.6 \times 10^{- 3} s e c^{- 1}$

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87.5% of a radioactive substance disintegrates in 40 minutes. What is the half life of the substance ?

A.

13.58 min

B.

135.8 min

C.

1358 min

D.

None of these

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Explanation

(A) Radioactive decay follow first order kinetics Determination K by substituting the respective values.

$K = \frac{2.303}{t} \log \frac{a}{a - x} = \frac{2.303}{40} \log \frac{a}{a - 0.875 a} = \frac{2.303}{40} \log \frac{a}{0.125 a} = \frac{2.303}{40} \log 8 = 0.051 m i n^{- 1} ∴ t_{1 / 2} = \frac{0.693}{K} = \frac{0.693}{0.051} = 13.58 m i n .$

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The inversion of cane sugar proceeds with half life of 500 minute at pH 5 for any concentration of sugar. However if pH = 6, the half life changes to 50 minute. The rate law expression for the sugar inversion can be written as

A.

$r = K [s u g a r]^{2} [H]^{6}$

B.

$r = K [s u g a r]^{1} [H]^{0}$

C.

$r = K [s u g a r]^{0} [H^{+}]^{6}$

D.

$r = K [s u g a r]^{0} [H^{+}]^{1}$

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Explanation

(B). Since $t_{1 / 2}$ does not depends upon the sugar concentration means it is first order respect to sugar concentration. $t_{1 / 2} \propto [s u g a r]^{1} .$

$t_{1 / 2} \times a^{n - 1} = k$

$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = \frac{{[H^{+}]}_{1}^{1 - n}}{{[H^{+}]}_{2}^{1 - n}}; \frac{500}{50} = {(\frac{10^{- 5}}{10^{- 6}})}^{1 - n}$

$10 = (10)^{1 - n} H e n c e n = 0$

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Decomposition on $N H_{3}$ on heated tungsten yields the following data :

Initial pressure (mm) 65 105 y 185

Half-life (s) 290 x 670 820

What are the values of x and y in that order ?

A.

420 s, 110 mm

B.

500 s, 160 mm

C.

520 s, 170 mm

D.

460 s, 150 mm

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Explanation

(D) $\frac{H a l f l i f e}{I n i t i a l p r e s s u r e} = \frac{290}{65} = 4.46$

$\frac{820}{185} = 4.43$ The values nearly agree.

$∴$ the reaction is of zero order.

$∴ \frac{x}{105} \approx 4.4 ∴ x \approx 460 s; \frac{670}{y} = 4.4 ∴ y \approx 150 m m .$

Chemical Kinetics MCQs for NEET — Chemistry Questions with Answers

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