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NEET CHEMISTRYEquilibriumMedium

Question

Consider the following reaction taking place in 1 L1\text{ L} capacity container at 300 K300\text{ K}: A+BC+D\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} If one mole each of A and B are present initially and at equilibrium 0.7 mol0.7\text{ mol} of C is formed, then the equilibrium constant (KcK_c) for the reaction is:

A

9.7

B

1.2

C

6.2

D

5.4

Step-by-Step Solution

For the given reversible reaction: A+BC+D\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}

Initial moles: nA=1n_A = 1 nB=1n_B = 1 nC=0n_C = 0 nD=0n_D = 0

Let xx be the number of moles of C formed at equilibrium. The moles of each species at equilibrium will be: nA=1xn_A = 1 - x nB=1xn_B = 1 - x nC=xn_C = x nD=xn_D = x

We are given that 0.7 mol0.7\text{ mol} of C is formed at equilibrium, so x=0.7x = 0.7. Since the volume of the container is 1 L1\text{ L}, the molar concentrations at equilibrium are equal to the number of moles : [A]=10.7=0.3 M[\text{A}] = 1 - 0.7 = 0.3\text{ M} [B]=10.7=0.3 M[\text{B}] = 1 - 0.7 = 0.3\text{ M} [C]=0.7 M[\text{C}] = 0.7\text{ M} [D]=0.7 M[\text{D}] = 0.7\text{ M}

The equilibrium constant expression (KcK_c) is: Kc=[C][D][A][B]K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}

Substituting the equilibrium concentrations into the expression: Kc=0.7×0.70.3×0.3=0.490.09=4995.44K_c = \frac{0.7 \times 0.7}{0.3 \times 0.3} = \frac{0.49}{0.09} = \frac{49}{9} \approx 5.44

Therefore, the equilibrium constant (KcK_c) is approximately 5.45.4.

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.

CHEMISTRYEquilibriumconsiderfollowingreactiontakingcapacity

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