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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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121 Medium
121 of 408
Application +4 | -1
pH of 0.2 M aqueous solution of sodium benzoate will be (PKaP{K_a} of benzoic acid = 4.2)
View answer and explanation
Correct option: B
8.758.75
Option B (8.75) is correct because it is calculated using the hydrolysis of salt of weak acid and strong base formula.
122 Hard
122 of 408
Analyzing +4 | -1
Assertion : In the titration of Na2CO3{Na}_2 {CO}_3 with HCl{HCl} using methyl orange indicator, the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator.

Reason : Two moles of HCl{HCl} are required for the complete neutralization of one mole of Na2CO3{Na}_2 {CO}_3
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) is correct. While both statements are true, the reason doesn't directly explain why methyl orange requires twice the volume of HCl compared to phenolphthalein.
123 Medium
123 of 408
Application +4 | -1
The concentration of CO2\mathrm{C}{\mathrm{O}_2} be in equilibrium with 2.5×10−2  mol  litre−1  of  CO  at  100∘C2.5 \times {10^{ - 2}}\;\mathrm{mol}\;\mathrm{ litr}{\mathrm{e}^{ - 1}}\;\mathrm{of }\;\mathrm{CO}\;\mathrm{at}\,\,100^\circ C for the reaction : FeO(s)  +  CO(g)   ⇋  Fe(s)+CO2(g);  Kc=5.0\mathrm{FeO}\left( s \right)\; + \,\,\mathrm{CO}\left( \mathrm{g} \right)\;\,\leftrightharpoons\;\mathrm{Fe}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\;{K_c} = 5.0
View answer and explanation
Correct option: D
0125  M0125\;M
Option D (0.125 M) is correct because it is calculated using the equilibrium constant expression and the given concentration of CO.
124 Medium
124 of 408
Analyzing +4 | -1
STATEMENT I : pH value of acidic buffer solution change, k if bufler solution is diluted upto very larger extent.

STATEMENT II : [H+]\left[H^{+}\right] decrease due to change in concentration as well as α\alpha and decrease in concentration is more as compare to increase in α\alpha.
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because upon significant dilution of a buffer, the pH can change because the concentrations of the acid and conjugate base become very low, affecting the buffer capacity and the degree of dissociation.
125 Hard
125 of 408
Application +4 | -1
The equilibrium concentrations of the species in the reaction A+B⇌C+DA+B \rightleftharpoons C+D are 2,3,102,3,10 and 6 mol L−16 \mathrm{~mol} \mathrm{~L}^{-1}, respectively at 300 K300 \mathrm{~K}. ΔG∘\Delta \mathrm{G}^{\circ} for the reaction is (R=2 cal/mol K)(R=2 \, \mathrm{cal} / \mathrm{mol} \, \mathrm{K})
View answer and explanation
Correct option: A
−1381.80 cal-1381.80 \, \mathrm{cal}
Option B (5.77%5.77 \%) is correct because it is calculated by setting up an ICE table and solving for the change in concentration using the equilibrium constant.
126 Easy
126 of 408
Remembering +4 | -1
Acidosis is diagnosed when blood pH\mathrm{pH}:
View answer and explanation
Correct option: A
Falls below 7.357.35
Option A (Falls below 7.357.35) is correct because acidosis is a condition where the blood pH falls below the normal range of 7.35-7.45.
127 Medium
127 of 408
Application +4 | -1
100  mL  of  0.01M100\;\mathrm{ }mL\;\mathrm{ }of\;\mathrm{ }0.01\mathrm{ }M solution of NaOH\mathrm{NaOH} is diluted to 1  dm31\;\mathrm{d}{\mathrm{m}^3}. What is the pH of the diluted solution?
View answer and explanation
Correct option: B
11
Option B (11) is correct because diluting the NaOH solution decreases the hydroxide concentration, leading to a pOH of 3 and therefore a pH of 11.
128 Medium
128 of 408
Analyzing +4 | -1
The decreasing order of strength of the bases,

OH−,NH2−,H−C≡C−  and  CH3−CH2−,  is\mathrm{O}{\mathrm{H}^ - },\mathrm{NH}_2^ - ,\mathrm{H}-\mathrm{C} \equiv {\mathrm{C}^ - }\;and\;\mathrm{C}{\mathrm{H}_3}-\mathrm{CH}_2^ - ,\;is
View answer and explanation
Correct option: A
CH3−CH2−>NH2−>H−C≡C−>OH−\mathrm{C}{\mathrm{H}_3}-\mathrm{CH}_2^ - > N\mathrm{H}_2^ - > H-C \equiv {\mathrm{C}^ - } > O{\mathrm{H}^ - }
Option A is correct. The order of base strength is determined by the stability of the conjugate acid. The weaker the conjugate acid, the stronger the base.
129 Hard
129 of 408
Understanding +4 | -1
(A): In polybasic acids Ka2K_{a_2} is always smaller than Ka1K_{a_1}

(R): Second dissociation takes place to a much smaller extent than the first dissociation.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A is correct because the second dissociation of a polybasic acid is always smaller than the first due to increased positive charge hindering proton removal.
130 Hard
130 of 408
Analyzing +4 | -1
H2S{\mathrm{H}_2}\mathrm{S} is passed into one dm3\mathrm{d}{\mathrm{m}^3} of a solution containing 0.1  mole  of  Zn3  and  0.01  mole  of  Cu20.1\;\mathrm{mole}\;\mathrm{ of }\;\mathrm{Z}{\mathrm{n}^3}\mathrm{ \;and}\;0.01\;\mathrm{mole }\;\mathrm{of}\;\mathrm{ C}{\mathrm{u}^2}  till the sulphide ion concentration reaches 8.1  ×  10108.1\; \times \;{10^{10}} moles. Which one of the following statements is true? [Ksp{K_{\mathrm{sp}}} of Zn\mathrm{Zn} and CuS\mathrm{CuS} are  3  ×1022  and  8  ×1036\mathrm{are}\;3\; \times {10^{22}}\;\mathrm{and}\;8\; \times {10^{36}} respectively]
View answer and explanation
Correct option: B
Both  CuS  and  ZnS  precipitates\mathrm{Both}\;\mathrm{CuS }\;\mathrm{and }\;\mathrm{ZnS }\;\mathrm{precipitates}
Option B (Both  CuS  and  ZnS  precipitates\mathrm{Both}\;\mathrm{CuS }\;\mathrm{and }\;\mathrm{ZnS }\;\mathrm{precipitates}) is correct because the ionic product for both CuSCuS and ZnSZnS exceeds their respective KspK_{sp} values, causing them to precipitate.
131 Medium
131 of 408
Remembering +4 | -1
The internal and external indicator among the following respectively are:
View answer and explanation
Correct option: A
KMnO4\mathrm{KMnO}_4, phenolphthalein
Option B (Phenolphthalein, KMnO4\mathrm{KMnO}_4) is correct because phenolphthalein is an internal indicator and KMnO4\mathrm{KMnO}_4 can act as a self-indicator.
132 Easy
132 of 408
Understanding +4 | -1
Which of the following will supress the ionisation of acetic acid is aqueous solution?
View answer and explanation
Correct option: B
HCl\mathrm{HCl}
Option B (HCl\mathrm{HCl}) is correct because it provides a common ion (H+) which suppresses the ionization of acetic acid due to the common ion effect.
133 Easy
133 of 408
Analyzing +4 | -1
Although CO\mathrm{CO} is neutral but it shows acidic nature on reaction with high PP and TT:
View answer and explanation
Correct option: B
NaOH\mathrm{NaOH}
The correct answer is B (NaOH) because CO reacts with strong bases like NaOH under high pressure and temperature to form sodium formate, exhibiting acidic behavior.
134 Easy
134 of 408
Remembering +4 | -1
The Henderson’s equation used to calculate the pOH\mathrm{pOH} of basic buffer is:
View answer and explanation
Correct option: A
pOH=pKb+log⁡[Conjugate  acid][  Base]\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}
Option A (pOH=pKb+log⁡[Conjugate  acid][  Base]\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}) is correct because it represents the Henderson-Hasselbalch equation for a basic buffer, relating pOH to pKb and the ratio of conjugate acid to base concentrations.
135 Easy
135 of 408
Understanding +4 | -1
Acidity of BF3B{F_3} can be explained on the basis of which of the following concept
View answer and explanation
Correct option: C
Lewis concept
Option C (Lewis concept) is correct because BF3BF_3 is an electron-deficient molecule that accepts a pair of electrons, fitting the definition of a Lewis acid.
136 Medium
136 of 408
Analyzing +4 | -1
Arrange NH4+,H2O,H3O+,HF  and  OH−\mathrm{NH}_4^ + ,{\mathrm{H}_2}\mathrm{O},{\mathrm{H}_3}{\mathrm{O}^ + },\mathrm{HF }\;\mathrm{and}\;\mathrm{ O}{\mathrm{H}^ - } in increasing order of acidic nature
View answer and explanation
Correct option: C
OH−<H2O<NH4+<HF<H3O+\mathrm{O}{\mathrm{H}^ - } < {\mathrm{H}_2}\mathrm{O} < N\mathrm{H}_4^ + < HF < {\mathrm{H}_3}{\mathrm{O}^ + }
Option C is correct. The acidity increases as follows: OH- < H2O < NH4+ < HF < H3O+. This is based on the stability of the conjugate base and electronegativity of the central atom.
137 Medium
137 of 408
Application +4 | -1
4 moles of A is mixed with 4 moles of B, at equilibrium for the reaction A+BC+DA + BC + D, 2 moles of C and D is formed. The equilibrium constant for the reaction will be
View answer and explanation
Correct option: C
1
Option C (1) is correct because the equilibrium constant is calculated as the ratio of products to reactants, and in this case, the change in moles is equal for all species.
138 Medium
138 of 408
Analyzing +4 | -1
In a reaction at equilibrium ′X′'X' mole of the reactant AA decompose to give 1  mole1\;mole each of C  and  DC\;and\;D . If the fraction of AA decomposed at equilibrium is independent of initial concentration of AA, then the value of ′X′'X' is :
View answer and explanation
Correct option: C
2
Option C (2) is correct because for the fraction of A decomposed to be independent of the initial concentration, the reaction must be first order, implying X=2.
139 Hard
139 of 408
Understanding +4 | -1
For exothermic reaction, if we change temperature from 70∘C70^{\circ} \mathrm{C} to 40∘C40^{\circ} \mathrm{C} then KC\mathrm{K}_{\mathrm{C}} will:
View answer and explanation
Correct option: A
Increases
Option A (Increases) is correct because for an exothermic reaction, decreasing the temperature shifts the equilibrium towards the product side, increasing the equilibrium constant Kc.
140 Medium
140 of 408
Application +4 | -1
Which of the following has PH{P^H} is equal to near about one
View answer and explanation
Correct option: D
75mℓ  M5  HCℓ+25ml  M5  NaOH75m\ell \,\,\frac{M}{5}\,\,HC\ell + 25ml\,\,\frac{M}{5}\,\,NaOH
Option D (75mℓM5\HCℓ+25mlM5\NaOH75m\ell \frac{M}{5}\HC\ell + 25ml\frac{M}{5}\NaOH) is correct because it results in an excess of HCl that gives a pH near 1 after calculation.
141 Medium
141 of 408
Remembering +4 | -1
What is equilibrium expression for the reaction, P4(s)+5O2(g)⇋P4O10(s)  ?{P_4}(s) + 5{O_2}(g)\leftrightharpoons{P_4}{O_{10}}(s)\,\,?
View answer and explanation
Correct option: B
Kc=1[O2]5{K_c} = \frac{1}{{{{\left[ {{O_2}} \right]}^5}}}
Option B (Kc=1[O2]5{K_c} = \frac{1}{{{{\left[ {{O_2}} \right]}^5}}}) is correct because solids are not included in the equilibrium expression.
142 Easy
142 of 408
Understanding +4 | -1
Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pHpH?
View answer and explanation
Correct option: A
BaCl2\mathrm{BaC}{\mathrm{l}_2}
Option A (BaCl2\mathrm{BaC}{\mathrm{l}_2}) is correct because barium hydroxide, formed by the hydrolysis of BaCl2, is the strongest base among the hydroxides formed by the other alkaline earth metals listed, leading to the highest pH.
143 Medium
143 of 408
Remembering +4 | -1
For the system;&nbsp;3A+2B⇋  C3A + 2B\leftrightharpoons\;C,the expression for equilibrium constant is
View answer and explanation
Correct option: B
[C][A]3[B]2\frac{{\left[ C \right]}}{{{{\left[ A \right]}^3}{{\left[ B \right]}^2}}}
Option B is correct because the equilibrium constant is the ratio of product concentration to reactant concentrations, each raised to the power of their stoichiometric coefficients.
144 Easy
144 of 408
Remembering +4 | -1
According to Arrhenius concept, base is a substance that:
View answer and explanation
Correct option: D
Gives OH−\mathrm{O}{\mathrm{H}^ - } ions in aqueous solution
Option D (Gives OH−\mathrm{O}{\mathrm{H}^ - } ions in aqueous solution) is correct because, according to the Arrhenius concept, a base is a substance that produces hydroxide ions in water.
145 Easy
145 of 408
Understanding +4 | -1
A higher value for equilibrium constant, KK shows that:
View answer and explanation
Correct option: A
The reaction has gone to near completion towards right
Option A (The reaction has gone to near completion towards right) is correct because a higher value of the equilibrium constant K indicates that the concentration of products is much greater than the concentration of reactants at equilibrium.
146 Medium
146 of 408
Understanding +4 | -1
A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour at 250C. Which operation result in a decrease in the equilibrium vapour pressure?
View answer and explanation
Correct option: D
Dissolving salt in the water
Option D is correct because dissolving salt in water lowers the vapor pressure according to Raoult's Law.
147 Easy
147 of 408
Analyzing +4 | -1
Assertion : SO2(g)+12O2(g)⇌SO3(g)+S O_2(g)+\frac{1}{2} O_2(g) \rightleftharpoons S O_3(g)+ heat.
Forward reaction is favoured at high temperature and low pressure.

Reason : Reaction is exothermic.
View answer and explanation
Correct option: D
If assertion is false but reason is true
Option D (If assertion is false but reason is true) is correct. The forward reaction is favored at low temperature and high pressure because it is exothermic and has fewer moles of gas on the product side.
148 Medium
148 of 408
Application +4 | -1
The equilibrium constant for the reaction between CH4(g)C{H_4}\left( g \right) and H2S(g){H_2}S\left( g \right) to form CS2(g)C{S_2}\left( g \right) and H2(g){H_2}\left( g \right) , at 1173 K is 3.6. For the following composition of the reaction mixture, [CH4]=1.0 M, [H2S]=1.20M   [CS2]=2.0M,  [H2]=1.8M\left[ {C{H_4}} \right] = 1.0\,M,\,\left[ {{H_2}S} \right] = 1.20M\,\,\,\left[ {C{S_2}} \right] = 2.0M,\,\,\left[ {{H_2}} \right] = 1.8M decide which of the following option is correct? [CH4(g)+2H2S(g)CS2(g)+4H2(g)]\left[ {C{H_4}\left( g \right) + 2{H_2}S\left( g \right)C{S_2}\left( g \right) + 4{H_2}\left( g \right)} \right]
View answer and explanation
Correct option: C
Reaction will shift to form more H2S{H_2}S
Option C is correct because the reaction quotient Q is greater than Kc, indicating that the reaction will shift to the left, forming more H2S.
149 Easy
149 of 408
Application +4 | -1
The pH\mathrm{pH} of a 0.02  M\mathrm{0}\mathrm{.02}\;\mathrm{M} solution of hydrochloric acid is:
View answer and explanation
Correct option: B
1.7\mathrm{1}\mathrm{.7}
Option B (1.7\mathrm{1}\mathrm{.7}) is correct because the pHpH of a 0.02 M solution of hydrochloric acid, a strong acid, is calculated using the formula pH=−log[H+]pH = -log[H^+].
150 Hard
150 of 408
Application +4 | -1
4.5 moles each of hydrogen and iodine heated in a sealed 10 L vessel. At equilibrium 3 moles of HI were found. The equilibrium constant for H2(g)+I2(g)⇋2HI(g){\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{HI}\left( \mathrm{g} \right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the equilibrium constant is calculated using the equilibrium concentrations of the reactants and products.
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