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

● ● ● ● ●

$ In a reaction, CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_{(g)} OH \triangle H^ \circ = -92 KJ mol ^ {-1 } $ concentration of hydrogen,carbon monoxide and methanol become constant at equilibrium. what willhappen if an inert gas is added to the system.?

A. reaction becomes fast B. reaction becomes slow C. equilibrium state disturbs D. equilibrium state remains undisturbed

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

Explanation

When an inert gas is added to a system at constant volume, the total pressure increases but the partial pressures of the reacting gases remain the same. Therefore, the equilibrium state remains undisturbed as the concentrations of the reactants and products do not change.

● ● ● ● ●

At 473 k, equilibrium constant $ K_C for the reaction PCl_{5(s)} \rightleftharpoons PCl_{3(g)}+ P Cl_{2(g)} is 8.3 \times 10 ^ {-3} what is the value of K_C $ for the reverse reaction at the same temperature ?

A. $ 8.3\times 10 ^ {-3} $ B. 120.48 C. $ 16.6 \times 10 ^ {-3} $ D. $ 4.15 \times 10 ^ {-3} $

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

Explanation

The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Hence, if $ K_C $ for the forward reaction is $ 8.3 imes 10^{-3} $, then $ K_C $ for the reverse reaction is $ \frac{1}{8.3 imes 10^{-3}} = 120.48 $.

● ● ● ● ●

At 473 k, equilibrium constant KC for a reaction, $ PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)} is 8.3 \times 10 ^ {-3 } ( \triangle H = 124 Kjmol ^{-1 }) $ What would be the effect on $ K_C $ if (i) more PCls is added (ii) the pressure is increased (iii) the temp is increased

A. $ K_c $ remains unchanged, unchanged, increase. B. $ K_c $ increases, increases, decreases. C. $ K_c $ remains unchanged, increase, unchanged. D. $ K_c $ decreases, increases, unchanged.

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

Explanation

For the reaction $ PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)} $, the equilibrium constant $ K_C $ is only affected by temperature. Therefore, if more $ PCl_5 $ is added or if the pressure is increased, $ K_C $ remains unchanged. However, since the reaction is endothermic (Î”H > 0), increasing the temperature will increase $ K_C $.

● ● ● ● ●

The equilibrium constant for the reaction $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} is 4 \times 10 ^ {-4} at 2000K temperature. what is the value of kc for the reaction 3/2 N_{2(g)} + 3/2 O_{2 (g)} \rightleftharpoons 3 NO_{(g)} $

A. $ 4 \times 10 ^ {-4} $ B. $ 8 \times 10 ^ {-6} $ C. $ 8 \times 10 ^ {-4} $ D. $ 16 \times 10 ^ {-4} $

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

Explanation

To find the equilibrium constant for the reaction $ \frac{3}{2} N_{2(g)} + \frac{3}{2} O_{2(g)} \rightleftharpoons 3 NO_{(g)} $, we need to relate this reaction to the given reaction $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} $. The given equilibrium constant for $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} $ is $ K_c = 4 \times 10^{-4} $. If we multiply the coefficients of the given reaction by $ \frac{3}{2} $, we get $ \frac{3}{2} N_{2(g)} + \frac{3}{2} O_{2(g)} \rightleftharpoons 3 NO_{(g)} $. The new equilibrium constant is related to the original one by raising it to the power of $ \frac{3}{2} $. Therefore, $ K_c' = (4 \times 10^{-4})^{\frac{3}{2}} = 8 \times 10^{-6} $.

● ● ● ● ●

For the reversible reaction $ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} at 500 \circ c the value of K_P is 1.44 \times 10 ^{-5} $ . what would be the value of KC for the same reaction

A. $ 1.44 \times 10 ^ 5 / (0.082 \times 500)^{-2} $ B. $ 1.44 \times 10 ^{- 5} / (8.314 \times 773)^{-2} $ C. $ 1.44 \times 10 ^{- 5} / (0.082 \times 773)^{2} $ D. $ 1.44 \times 10 ^ {-5 } / (0.082 \times 773)^{-2} $

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

● ● ● ● ●

The following equilibria are given $ N2 + 3H_2 \rightleftharpoons 2NH_3 ---K_1 N_2 + O_2 \rightleftharpoons ----K_2 H_2 + 1/2 O_2 \rightleftharpoons H_2 O ---K_3 2NH_3 + 5/2 O_2 \rightleftharpoons 2 NO + 3 H_2 O . The equilibrium constant of the reaction in terms of k_1 , k_2 and k_3 $ is

A. $ k_1 k_2 / k_3 $ B. $ k_1 k_3 ^ 2 / k_2 $ C. $ k_2 k_3 ^ 3 / k_1 $ D. $ k_1 k_2 k_3 $

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

● ● ● ● ●

The $ P^{Ka} of a weak acid HA is 4.80 The p^{Kb} of a weak base BOH is 4.78 The P^H $ of an aqueous solution ofthe corrosponding salt BAwillbe

A. 9.58 B. 4.79 C. 7.01 D. 9.22

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

Explanation

For salt of weak acid and weak base $ P^ H = - { 1 \over 2} [ log Ka + log Kw - log K_b ] $ $ = + { 1 \over 2} pKa + { 1 \over 2 } pKw - { 1 \over 2 } pK_b $ $ = { 1 \over 2 } \times 4.8 + { 1 \over 2 } \times 14 - { 1 \over 2 } \times 4.78 $ = 7.01

● ● ● ● ●

The equilibrium constant for the reaction $ SO_{3(g)} \rightleftharpoons SO_{2(g)} + 1/2 O_{2(g)} Kc = 4.9 \times 10 ^ {-2} $ The value for the $K_C$ of the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} $ will be

A. 416 B. $ 2.40 \times 10 ^ {-3} $ C. $ 9.8 \times 10 ^ {-2} $ D. $ 4.9 \times 10 ^ {-2} $

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

Explanation

Equilibrium constant for the reaction $ SO_{ 2 (g) } + { 1 \over 2 } O_{ 2 (g) } \rightleftharpoons SO_{3(g)} $ and for reaction $ 2 SO_{ 2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} $ $ K_C = ( { 1 \over 4.9 \times 10 ^ {-2} } ) ^ 2 = 416 .49 $

● ● ● ● ●

$ P^H$ of 0.1 M solution of weak acid is 3. The value of ionisation constant Ka of acid is

A. $ 3 \times 10 ^ {-1} $ B. $ 1 \times 10 ^ {-3} $ C. $ 1 \times 10 ^ {-5} $ D. $ 1 \times 10 ^ {-7} $

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

Explanation

$ P^H = 3 \therefore [ H^+_3 ] = 1 \times 10 ^ {-3} $ $ Ka = { [ H_3 ^ + O ] ^2 \over C } = { ( 1 \times 10 ^ {-3} ) ^2 \over 0.1 } = 1 \times 10 ^ {-5} $

● ● ● ● ●

Three reactions involving $ H_2 PO^- _ 4 $ are given below.

(i) $ H_3 PO_4 + H_2O \rightleftharpoons H_3 O^+ + H_2 PO^- _4 $ (ii) $ H_2 PO^- _4 + H_2 O \rightleftharpoons HPO^{-2} _4 + H^+ _3 O$
(iii) $ H_2 PO^- _4 + OH^- \rightleftharpoons H_3 PO_4 + O^{2-} $ In which of the above does $H_2PO^- _ 4 $ act as an acid.

A. (i) only B. (ii) only C. (i) and (ii) D. (iii) only

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

Explanation

According to lowry-bronsted acid base theory in (ii) reaction $ H_2 PO_4 ^- donates H^+ ion to H_2 O$ it acts as an acid.

Chemistry MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions