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NEET CHEMISTRYEquilibriumMedium

Question

Mark the conditions that favour the maximum product formation in the given reaction: A2(g)+B2(g)X2(g),ΔrH=X kJA_2(g) + B_2(g) \rightleftharpoons X_2(g), \Delta_r H = -X \text{ kJ}

A

Low temperature and high pressure.

B

Low temperature and low pressure.

C

High temperature and high pressure.

D

High temperature and low pressure.

Step-by-Step Solution

According to Le Chatelier's principle, a change in temperature, pressure, or concentration of a system at equilibrium will shift the equilibrium to counteract the change.

  1. Effect of Temperature: The given reaction is exothermic (ΔrH=X kJ\Delta_r H = -X \text{ kJ} is negative). For exothermic reactions, decreasing the temperature favours the forward reaction as the system shifts to produce more heat.
  2. Effect of Pressure: The forward reaction proceeds with a decrease in the number of gaseous moles (2 moles of reactants form 1 mole of product). An increase in pressure favours the direction that produces fewer moles of gas to reduce the pressure. Therefore, high pressure favours the forward reaction. Thus, low temperature and high pressure will favour maximum product formation.

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.

CHEMISTRYEquilibriumconditionsfavourmaximumproductformation

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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.$10^{-3} \text{ mol/L}$
B.$10^{-4} \text{ mol/L}$
C.$10^{-5} \text{ mol/L}$
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.$2 \times 10^{-4} \text{ mol/L}$
B.$16 \times 10^{-16} \text{ mol/L}$
C.$2 \times 10^{-16} \text{ mol/L}$
D.$4 \times 10^{-4} \text{ mol/L}$
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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 $\text{PbCl}_2 = 1.7 \times 10^{-5}$)

A.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-4} \text{ M}$
B.$[\text{Ag}^+] = 1.8 \times 10^{-7} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-6} \text{ M}$
C.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 8.5 \times 10^{-5} \text{ M}$
D.$[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \text{ M}$
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