back to directory
NEET CHEMISTRYEquilibriumMedium

Question

For the equilibrium 2NOCl(g)2NO(g)+Cl2(g)2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g) the value of the equilibrium constant is 3.0×1063.0 \times 10^{-6} at 1000 K1000\text{ K}. Find KpK_p for the reaction at this temperature (Given R=8.314 J K1mol1R = 8.314\text{ J K}^{-1}\text{mol}^{-1}):

A

1.493

B

2.494×1022.494 \times 10^{-2}

C

3.0×1063.0 \times 10^{-6}

D

2.494×1042.494 \times 10^{-4}

Step-by-Step Solution

For the given reaction: 2NOCl(g)2NO(g)+Cl2(g)2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g)

The relationship between the equilibrium constants KpK_p and KcK_c is given by the equation: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}

where Δn\Delta n is the difference between the sum of moles of gaseous products and the sum of moles of gaseous reactants. Δn=moles of gaseous productsmoles of gaseous reactants\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} Δn=(2+1)2=1\Delta n = (2 + 1) - 2 = 1

Given data: Kc=3.0×106K_c = 3.0 \times 10^{-6} R=8.314 J K1mol1R = 8.314\text{ J K}^{-1}\text{mol}^{-1} T=1000 KT = 1000\text{ K}

Substitute these values into the relationship (as directed by the explicitly given value of RR): Kp=(3.0×106)×(8.314×1000)1K_p = (3.0 \times 10^{-6}) \times (8.314 \times 1000)^1 Kp=3.0×106×8314K_p = 3.0 \times 10^{-6} \times 8314 Kp=24942×106K_p = 24942 \times 10^{-6} Kp=2.4942×102K_p = 2.4942 \times 10^{-2}

Thus, the value of KpK_p is 2.494×1022.494 \times 10^{-2}.

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.

CHEMISTRYEquilibriumequilibriumtextnoclgrightleftharpoonstextnogtextclg

More Equilibrium Questions

View all

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
MediumSolve

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.$\text{BF}_3$
EasySolve

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
MediumSolve

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.$100 \text{ mL } \frac{M}{10} \text{ HCl } + 100 \text{ mL } \frac{M}{10} \text{ NaOH}$
MediumSolve

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.$BF_3 > BCl_3 > BBr_3$
MediumSolve

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.$10^{-6} \text{ mol/L}$
EasySolve

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.$4 \times 10^{-4} \text{ mol/L}$
EasySolve

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.$[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \text{ M}$
MediumSolve

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →