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Equilibrium MCQs for NEET — Chemistry Questions with Answers

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Ionisation constant of $ CH_3COOH is 1.7 \times10 ^{-5} and concentration of H ^+ ions is 3.4 \times 10 ^ {-4} The initial concentration of CH_3COOH$ molecules is

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Explanation

$ CH_3 COOH \rightleftharpoons CH_3 COO ^ - + H^+ $ $ at eq (a - 3.4 \times 10 ^ {-4} ) 3.4 \times 10 ^ {-4} 3.4 \times 10 ^ {-4} $ $ { (3.4 \times 10 ^ {-4} ) (3.4 \times 10 ^ {-4} ) \over ( a - 3.4 \times 10 ^ {-4} ) } = 1.7 \times 10 ^ {-5} (Given) $ $ \therefore a = 6.8 \times 10 ^ {-3} $

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In the reversible reaction $ A + B \rightleftharpoons C + D $ , the concentration of each C and D at equilibrium was 0.8 mol litre, then the equilibrium constant $ K_C $ will be.

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Explanation

$ A+ B \rightleftarrow C + D $ initial 1 1 0 0 conc. Ateqm(1 - 0.8) 0.8 0.8 conc.(1- 0.8) = 0.2 = 0.2 $ Kc = { [C] [D] \over [A] [D] } = { 0.8 \times 0.8 \over 0.2 \times 0.2 } = 16.0 $

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A reversible chemical reaction having two reactants in equilibrium. If the concentration of the reactants are doubled, then equilibrium constant will

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Explanation

$ K_C remains same beacause K_C $ is a characteristic constant.

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Two moles of PCls are heated in a closed vessel of 2L capacity. At equilibrium, 40 % of $ PCl_5 is dissociated in to PCl_3 \& Cl_2 $ . The value of equilibrium constant is,

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Explanation

$ PCl_5 \rightleftharpoons PCl_3 + Cl_2 $ 2 0 0 $ {2 \times 60 \over 60 } { 2 \times 40 \over 100 } { 2 \times 40 \over 100} $ moles = 1.2 1.8 0.8 conc. = {mol \lit } = 1.2 / 2 0.8 / 2 0.8 /2 $ \therefore Kc = { [PCl_3 ] [Cl_2] \over [PCl_5] } = {0.4 \times 0.4 \over 0.6 } = 0.266 $

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The dissociation constant for acetic acid and HCN at $ 25 ^\circ C are 1.5 \times 10 ^{-5} and 4.5 \times 10 ^{-10} respectively. the equilibrium constant for reaction CN^- + CH_3 COOH \rightleftharpoons HCN + CH_3COO ^ - $

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Explanation

$ Dissociation of CH_3 COOH CH_3 COOH \rightleftharpoons H ^ + + CH_3 COO ^ - K_{a1} = 1.5 \times 10 ^ {-3} $ $ Dissociation of HCN : HCN \rightleftharpoons H ^ = +CN^- k_{a2} = 4.5 \times 10 ^ {-3} $ For a reaction $ CN ^ - + CH_3 COOH \rightleftharpoons CH_3 COO ^ - + HCN is Ka = { Ka_1 \over Ka_2 } = { 1.5 \times 10 ^ {-3} \over 4.5 \times 10 ^ {-10} }= 3.33 \times 10 ^ 4 $

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If in the reaction $ N_2O_4 \rightleftharpoons 2NO_2 , \alpha is that part of N_2O_4$ will dissociate, then the number of moles at equilibrium will be,

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Explanation

$ N_2 O_4 \rightleftarrow 2NO_2 $ 1 0 $ (1 - \alpha) $ $ 2 \alpha $ Total moles = $ 1- \alpha +2 \alpha =1+ \alpha from equation 2x=3 \therefore x = 3 /2 = 1.5 $

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4.5 moles eachof hydrogen and iodine heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for $ H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI(g) $ is

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Explanation
            $ H_2 + I_2 \rightlleftarrow  2 HI $

Initial 4.5 4.5 0 conc. at eqm. (4.5-x)(4.5-x) 2x $ \therefore [H_2] = 4.5 -1.5 = 3 $ $ Kc = { [HI] ^2 \over [H_2 ] [I_2] } = { 3 \times 3 \over 3 \times 3 } = 1 $

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The rate constant for forward and backward reaction of hydrolysis of ester are $ 1.1 \times 10 ^{-2 } \& 1.5 \times 10 ^ {-3 } $ per minute respectively. Equilibrium constant for reaction is, $ CH_3 COOC_2 H_5 + H_2 O \rightleftharpoons CH_3 COOH + C_2 H_5 OH $

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Explanation

$ K_f = 1.1 \times 10 ^ {-2} $ $ K_b = 1.5 \times 10 ^{-3} $ $ kc = { kf \over k_b } = { 1.1 \times 10 ^ {-2} \over 1.5 \times 10 ^ {-3} } = 7.33$

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What is equilibrium expression for the reaction $ P_4 + 5O_2 \rightleftharpoons P_4O_{10} $

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Explanation

$ P_{4(S)}+ 5O_{2(g)} \rightleftharpoons P_4O_{10(s)} $ $ K_C = { [P_4 O_{10} \over [P_{4(s) ] [ O_{2(g)] ^ 5 $ we know that concentration of a solid component is always taken as a unity $ \therefore Kc = { 1 \over [ O_2 ] ^ 5 $

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Partial pressure of $ O_2$ in $ 2Ag_2 O(g) \rightleftharpoons 4 Ag(s) + O_{2 (g) } $ is

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Explanation

$ 2Ag_2O_{(s)} \rightleftharpoons 4 Ag_{(s)} + O_{2_{(g)} } $ for this reaction $ Kp = Po_2 $

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