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Chemistry NEET Questions — Free MCQ Practice

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Three gaseous equilibria have values of equilibrium constants as $k_1 , k_2, k_3 $ . resp. (i) $ A +B \rightleftharpoons C $ (ii) $ B+C \rightleftharpoons P+Q $ (iii) $ A+2B \rightleftharpoons P+Q $. What is the relation between $k_1 ,k_2, k_3$ . (?)

A. K_3 = K_1 / K_2 B. $ K_3 = { K_1 \times K_2 } $ C. $ K_1 = K_2 \times K_3 $ D. $ K_1 \times K_2 \times K_3 = 1 $

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Explanation

Addition of (1) and (2) gives (3) then $ K_1 \times K_2 = { [C] \over [A] [ B] ] } \times { [P] [Q] \over [ B] [C] } = { [P] [Q] \over [ A] [ B ] ^2 } = K_3 $ $ \therefore K_3 = K_1 \times K_2 $ When the addition of equilibria leads to another equilibria then the product of their equilibria constants gives the equilibria constant of the resultant equilibrium.

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Three gaseous equilibria have value of equilibrium constants as $K_1, K_2 , K_3$ respectively. (i)$ N_2 + O_2 \rightleftharpoons 2 NO (K_1) $ (ii) $ N_2 + 2O_2 \rightleftharpoons 2 NO_2 (K_2 ) $ (iii) $ 2NO +2 O_2 \rightleftharpoons 2 NO_2 (K_3 ) $ What is the relation between $ K_1, K_2 and K_3 (?) $

A. $ K_3 = K_1 \times K_2 $ B. $ K_1 = K_2 \times K_3 $ C. $ K_2 = K_1 \times K_3 $ D. $ K = K_1 \times K_2 \times K_3 $

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Explanation

The given equilibria can be analyzed to find the relationship between the equilibrium constants. Rearranging and combining the given reactions, we get the relation $ K_2 = K_1 imes K_3 $. This comes from the fact that if you add the first equilibrium (i) and the third equilibrium (iii), you get the second equilibrium (ii), and the equilibrium constant for the added reactions is the product of their individual constants.

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The equilibrium constant for the reaction, $ N_{2(g)} + O_{2(g) } \rightleftharpoons 2 NO_{(g)} is 4.4 \times 10 ^ {-4 } $ at 2000k temp In presence ofa catalyst, equilibrium is attained ten times faster. Therefore the equilibrium constant, in presence of catalyst at 2000K is........

A. $ 4.4 \times 10 ^ {-4} $ B. $ 4.4 \times 10 ^ {-5} $ C. $ 4.4 \times 10 ^ {-3} $ D. difficult to compute

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Explanation

The value of equilibrium constant does not change in presence of catalyst.

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For the following reaction in gaseous phase $ CO + 1/2 O_2 \rightleftharpoons CO_2 $ $ KC/ KP $ is

A. $ ( RT) ^ {1/2} $ B. $ ( RT) ^ { -1/2} $ C. RT D. $ ( RT)^{-1} $

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Explanation

Formula: $ Kp = Kc(RT) ^ {\triangle n_{ (g)} } $ $ \triangle_{(g)} = n_p - n_r $

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For the reaction equilibrium, $ N_2O_{4(g )} \rightleftharpoons 2NO_{( g)} the concentration of N_2O_4 and NO_2 at equilibrium are 4.8 \times 10 ^ {-2} and 1.2 \times 10 ^ {-2} molL^{-1} respectively. The value of K_C $ for the reaction is :

A. $ 3.3\times 10^2 mol L ^{-1} $ B. $ 3\times 10^1 mol L ^{-1} $ C. $ 3 \times 10^{-3 }mol L ^{-1}$ D. $ 3\times 10^3 mol L ^{-1} $

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Explanation

For the given reaction equilibrium $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$, the equilibrium constant $K_C$ is calculated using the formula:

$$ K_C = rac{[ NO_2]^2}{[ N_2O_4]} $$

Given the concentrations:

$$ [NO_2] = 1.2 imes 10^{-2} ext{ mol L}^{-1} $$

$$ [N_2O_4] = 4.8 imes 10^{-2} ext{ mol L}^{-1} $$

Substitute these values into the formula:

$$ K_C = rac{(1.2 imes 10^{-2})^2}{4.8 imes 10^{-2}} = rac{1.44 imes 10^{-4}}{4.8 imes 10^{-2}} = 3 imes 10^{-3} ext{ mol L}^{-1} $$

Therefore, the correct answer is $3 imes 10^{-3} ext{ mol L}^{-1}$.

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Which one of the following statements is not true ?

A.

The conjugate base of $ H_2 PO_4 ^ - is HPO_4 ^ {2-} $

B.

$ P ^ H + P ^ {OH} = 14 for all aqueous solutions at 298 K $

C.

$ P ^ H of 1 \times 10 ^ {-8} M HCl is 8 $

D.

$ NH_3 $ is a lewis base

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Explanation

$ H_2O is also present in 1 \times 10^{-8} M HCl solution So due to self ionisation of H_2O $ , $ H_2 O \rightleftharpoons H^+ + OH^- $ $ 1 \times 10 ^ {-7} $ M at 298 K

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What is the equilibrium expression for the reaction. $ P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)} $

A. $ Kc = {{ [ P_4 O_{10}] }\over [ P_4] [ O_2] ^ 5 }$ B. $ Kc = {1 \over [ O_2] ^ 5 }$ C. $ Kc = {{ [ P_4 O_{10}] }\over 5[ P_4] [ O_2] }$ D. $ Kc = { [ O_2] ^ 5 }$

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Explanation

For the reaction $P_{4(s)} + 5O_{2(g)} ightleftharpoons P_4O_{10(s)}$, the equilibrium constant expression $K_C$ involves only the concentrations of gases and aqueous species. Solids are not included in the equilibrium expression. Thus, the expression for $K_C$ is:

$$ K_C = rac{1}{[O_2]^5} $$

Therefore, the correct answer is $K_C = rac{1}{[O_2]^5}$.

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For the reaction CO(g) + Cl2(g) ----> COCl2(g) then Kp/ Kc is equal to

A.

$ { 1 \over RT} $

B.

RT

C.

$ \sqrt { RT} $

D.

1

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The equilibrium constant for the reaction $ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} at temperature. 300 k is 4 \times 10^{-4 } $.The value of Kc for the reaction $ NO_{(g)} \rightleftharpoons {1 /2 } N_{2(g) } + { 1/2} O_{2(g) } $ at the same temperature is

A. $ 2.5 \times 10 ^ 2 $ B. $ 4 \times 10 ^ {-4} $ C. 0.02 D. 50

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Explanation

$ N_{2 (g)} + O_{ 2 (g) } \rightleftharpoons 2 NO_{(g)} $ $ K_p = { P^2 NO \over P_ {N_2} \times P_{O_2} $ For reaction $ NO(g) \rightleftharpoons {1 \over 2} O_{2 (g) } + { 1 \over 2 } N_2 $ $ K_p = { P ^ {1/2} _{O_2} \times P^ {NO} = { 1 \over ( Kp ) ^ {1/2} }= { 1 \over ( 4 \times 10 ^ {-4} ) ^ {1/2} } = 50 $

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Hydrogen ion concentration in mol/ L in a solution of $ P^H = 5.4 $ will be,

A. $ 3.98 \times 10 ^ {-6} $ B. $ 3.68 \times 10 ^ {-6} $ C. $ 3.88 \times 10 ^ 6 $ D. $ 3.98 \times 10 ^ 8 $

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Explanation

$ P ^ H = 5.4 $ $ P ^ H = - log [ H^+] $ $ 5.4 = - - log [ H ^ + ] $ $ 6.6 = 3.981 \times 10 ^ {-6} = [ H^ + ] $

Chemistry MCQs for NEET — Practice Questions with Answers

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