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NEET CHEMISTRYEquilibriumEasy

Question

The equilibrium reaction that doesn't have equal values for KcK_c and KpK_p is:

A

2NO(g)⇌N2(g)+O2(g)2NO(g) \rightleftharpoons N_2(g) + O_2(g)

B

SO2(g)+NO2(g)⇌SO3(g)+NO(g)SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)

C

H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

D

2C(s)+O2(g)⇌2CO2(g)2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)

Step-by-Step Solution

The relationship between KpK_p and KcK_c is given by Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, where Δn\Delta n is the difference between the number of moles of gaseous products and gaseous reactants (Δn=np−nr\Delta n = n_p - n_r). For KpK_p to be equal to KcK_c, Δn\Delta n must be zero. Let's calculate Δn\Delta n for the given reactions:

  1. 2NO(g)⇌N2(g)+O2(g)2NO(g) \rightleftharpoons N_2(g) + O_2(g); Δn=(1+1)−2=0\Delta n = (1 + 1) - 2 = 0
  2. SO2(g)+NO2(g)⇌SO3(g)+NO(g)SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g); Δn=(1+1)−(1+1)=0\Delta n = (1 + 1) - (1 + 1) = 0
  3. H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g); Δn=2−(1+1)=0\Delta n = 2 - (1 + 1) = 0
  4. 2C(s)+O2(g)⇌2CO2(g)2C(s) + O_2(g) \rightleftharpoons 2CO_2(g); Δn=2−1=1\Delta n = 2 - 1 = 1 (Carbon is in the solid state, so it is not considered in calculating Δn\Delta n). Since Δn≠0\Delta n \neq 0 for reaction 4, its KpK_p and KcK_c values are not equal.

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.

CHEMISTRYEquilibriumequilibriumreactiondoesntvalues

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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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