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NEET CHEMISTRYEquilibriumMedium

Question

For the reaction 2AB+C2\text{A} \rightleftharpoons \text{B} + \text{C}; Kc=4×103K_c = 4 \times 10^{-3}. At a given time, the composition of the reaction mixture is: [A]=[B]=[C]=2×103 M[\text{A}] = [\text{B}] = [\text{C}] = 2 \times 10^{-3}\text{ M}. In light of the above facts, which of the following is correct?

A

Reaction has a tendency to go in the forward direction.

B

Reaction has a tendency to go in the backward direction.

C

Reaction has gone to completion in the forward direction.

D

Reaction is at an equilibrium.

Step-by-Step Solution

For the reaction 2AB+C2\text{A} \rightleftharpoons \text{B} + \text{C}, the reaction quotient (QcQ_c) at the given time is calculated using the expression: Qc=[B][C][A]2Q_c = \frac{[\text{B}][\text{C}]}{[\text{A}]^2}

Substituting the given concentrations ([A]=[B]=[C]=2×103 M[\text{A}] = [\text{B}] = [\text{C}] = 2 \times 10^{-3}\text{ M}): Qc=(2×103)(2×103)(2×103)2=1Q_c = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = 1

We are given the equilibrium constant Kc=4×103K_c = 4 \times 10^{-3} (or 0.0040.004). Comparing QcQ_c and KcK_c, we find that Qc>KcQ_c > K_c (1>0.0041 > 0.004). When the reaction quotient is greater than the equilibrium constant (Qc>KcQ_c > K_c), the reaction will proceed in the reverse (backward) direction to reach equilibrium .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumreactionrightleftharpoonscompositionreactionmixture

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Consider the following reaction: $\text{A}_2(g) + \text{B}_2(g) \rightleftharpoons 2\text{AB}(g)$. At equilibrium, the concentrations of $[\text{A}_2] = 3.0 \times 10^{–3} \text{ M}$; $[\text{B}_2] = 4.2 \times 10^{–3} \text{ M}$ and $[\text{AB}] = 2.8 \times 10^{–3} \text{ M}$. The value of $K_c$ for the above-given reaction in a sealed container at $527^\circ\text{C}$ is:

A.3.9
B.0.6
C.4.5
D.2
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Amongst the given options, which of the following molecules/ions acts as a Lewis acid?

A.$\text{OH}^-$
B.$\text{NH}_3$
C.$\text{H}_2\text{O}$
D.$\text{BF}_3$
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Boric acid is an acid because its molecule

A.contains replaceable H⁺ ion
B.gives up a proton
C.accepts OH⁻ from water releasing proton
D.combines with proton from water molecule
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The following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. pH of which one of them will be equal to 1?

A.$60 \text{ mL } \frac{M}{10} \text{ HCl } + 40 \text{ mL } \frac{M}{10} \text{ NaOH}$
B.$55 \text{ mL } \frac{M}{10} \text{ HCl } + 45 \text{ mL } \frac{M}{10} \text{ NaOH}$
C.$75 \text{ mL } \frac{M}{5} \text{ HCl } + 25 \text{ mL } \frac{M}{5} \text{ NaOH}$
D.$100 \text{ mL } \frac{M}{10} \text{ HCl } + 100 \text{ mL } \frac{M}{10} \text{ NaOH}$
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The tendency of $BF_3$, $BCl_3$ and $BBr_3$ to behave as Lewis acid decreases in the sequence:

A.$BCl_3 > BF_3 > BBr_3$
B.$BBr_3 > BCl_3 > BF_3$
C.$BBr_3 > BF_3 > BCl_3$
D.$BF_3 > BCl_3 > BBr_3$
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A compound $\text{BA}_2$ has $K_{sp} = 4 \times 10^{-12}$. Solubility of this compound will be:

A.$10^{-3} \text{ mol/L}$
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D.$10^{-6} \text{ mol/L}$
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What is the molarity of the saturated solution if the solubility product for a salt of type AB is $4 \times 10^{-8}$?

A.$2 \times 10^{-4} \text{ mol/L}$
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C.$2 \times 10^{-16} \text{ mol/L}$
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In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\text{Ag}^+$ and $\text{Pb}^{2+}$ at a concentration of $0.10 \text{ M}$. Aqueous $\text{HCl}$ is added to this solution until the $\text{Cl}^-$ concentration is $0.10 \text{ M}$. What will the concentration of $\text{Ag}^+$ and $\text{Pb}^{2+}$ at equilibrium? ($K_{sp}$ for $\text{AgCl} = 1.8 \times 10^{-10}$, $K_{sp}$ for $\text{PbCl}_2 = 1.7 \times 10^{-5}$)

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