Are these Chemistry questions free to attempt?

Yes. You can attempt every Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

Do I need an account to practise Chemistry MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Chemistry NEET Questions — Free MCQ Practice

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

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

● ● ● ● ●

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.

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

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

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

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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= ${(1 / 2)}^{4} \times 160 = \frac{160}{2 \times 2 \times 2 \times 2} = 10 g m$

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

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

● ● ● ● ●

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.

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

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

● ● ● ● ●

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.

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

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

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

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

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

● ● ● ● ●

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.

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

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

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

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

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

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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.

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

● ● ● ● ●

The half life period of gaseous substance undergoing thermal decomposition was measured for various initial pressure ‘P’ with the following result.

P(mm) 250 300 400 450

$t_{1 / 2}$ (min) 136 112.5 85 75.5

Calculate the order of reaction.

A.

2

B.

4

C.

6

D.

10

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(A) $t_{1 / 2} \propto \frac{1}{{(a)}^{n - 1}}$ where a is initial conc. or presuure or volume

$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{a_{2}}{a_{1}})}^{n - 1}$

from first & second data

$(\frac{136}{112.5}) = {(\frac{300}{250})}^{n - 1}; (1.2) = {(1.2)}^{n - 1} \Rightarrow n - 1 = 1$

n=2, Hence reaction is of second order.

Chemistry MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions