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Chemistry MCQs for NEET — Practice Questions with Answers

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For a single step reaction $ A + 2B ---> Products $ , the molecularity is

A. zero B. 1 C. 2 D. 3

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Explanation

3

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Which of the following statement is false ?

A. For a zero order reaction, the rate changes with temperature. B. Both order and molecularity of a reaction are always the same. C. Active mass of 128 g of HI present in a two litre flask is 0.5. D. For the first order reaction, the rate of reaction halved as the concentration of a reactant halved.

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Explanation

Both order and molecularity of a reaction are always the same.

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If â€˜aâ€™ is the initial concentration of the reactant, the time taken for completion of the reaction, it if is of zero order, will be

A. K / 2a B. a / 2k C. a/ k D. k /a

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Explanation

a/k For Zero order reaction $ t = { 1 \over k} { [A]_o - [A] } But [A]_o = a $ And when reaction complete $ [A] = 0 \therefore t = { a \over k } $

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The reaction $ 2O_3 ---> 3O_2 $ proceeds in two steps as follows. (i) $ O3 --> O_2 + O (fast) $ (ii) $ O + O_3 --> 2O_2 (slow) $ The rate law expression should be...

A. $ r = K [O_3 ] ^ 2 $ B. $ r = K [ O_3 ] ^2 [ O_2 ]^ { -1 } $ C. $ r = K [ O_3 ] [O_2 ] $ D. $ r = K [O_3 ] ^ 2 [O_2 ] ^ 1 $

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Explanation

$ r = k [O_3]^2 [O_2] ^ {-1} From slowest step r = k [ O_3 ] [O] $ $ From eq (i) keq = [O_2] [O] / [O_3 ] $ $ \therefore [O] = Keq [O_3] / [O_2] _2 $ $ \therefore r = k [O_3] Keq [O_2] = K^1 [O_3]^2 [O_2]^{-1} $

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For the reaction Zero order

A. $ t _ { 1 /2} \alpha Co^2 $ B. $ t _ { 1 /2} \alpha \sqrt {Co} $ C. $ t _ { 1 /2} \alpha Co^{-1} $ D. $ t _ { 1 /2} \alpha Co $

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Explanation

$ t _ { 1/2} \alpha C_o $

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For reaction first order

A. $ t _ {1/2} = { 0.693 \over k } $ B. $ t _ {1/2} = { 0.693 \over Co } $ C. $ t _ {1/2} \alpha { 0.693 \over k } $ D. $ t _ {1/2} \alpha { 0.693 \over Co } $

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Explanation

$ t_ { 1/2} = { 0.693 \over k } $

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Which of the following represents the expression for Â¾th life of a first order reaction

A. $ { k \over 2.303} log { 4 \over 3} $ B. $ { 2.303 \over k } log { 3 \over 4} $ C. $ { 2.303 \over k } log 4 $ D. $ { 2.303 \over k } log 3 $

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Explanation

$ {2.303 \over k } log 4 $ $ t_{3/4} = { 2.303 \over k} log { ao \over ao -ao \times {3 \over 4} } = { 2.303 \over k } log { ao \over { ao /4 } } $

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If initial concentration is doubled, the time for half reaction is also doubled. The order of reaction is ...

A. First B. second C. Third D. zero

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Explanation

Zero for zero order reaction $ t_ { 1/2} \alpha C_o $

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If a is the initial concentration of the reactant, the half life period of the reaction of the $n^{th} $ order is proportional to ...

A. $ a ^ {n+1 } $ B. $ a ^ {1-n} $ C. $ a ^ n $ D. $ a ^ { n -1 } $

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Explanation

$ a^ {1-n} t_{1/2} \alpha a^ {1-n} $

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For the first order reaction, half life is 14 s. The time required for the initial concentration to reduce to 1/8th of its value is ...

A. 28 s B. 42 s C. $ ( 14 ) ^ 2 s $ D. $ ( 14) ^ 3 s $

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Explanation

42 S $ Ao \rightarrow {Ao \over 2} \rightarrow { Ao \over 4} \rightarrow { Ao \over 8} $ $ \therefore 3 \times t _ {1/2} $

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