
$A+B \rightleftharpoons C+D$. If initially the concentration of $A$ and $B$ are both equal but at equilibrium, the concentration of $D$ will be twice that of $A$, then what will be the equilibrium constant of the reaction?
A. $\dfrac{4}{9}$
B. $\dfrac{9}{4}$
C. $\dfrac{1}{9}$
D. 4
Answer
233.1k+ views
Hint: If any reaction reaches the state of equilibrium, then with the help of the equilibrium constant you can find the amount of reactant with respect to the product or vice versa at the eqbm. position. The equilibrium constant (K) is defined as the ratio of the concentration of product raised to its stoichiometric coefficient to the product of the concentration of reactants each raise to power their stoichiometric coefficient.
For general reaction,
aA + bB ⇌ cC + dD, equilibrium constant is defined as
Kc= [C]c[D]d/[A]a[B]b, Where [C]c and [D]d are the molar concentration of products, C and D both raised to the power of their stoichiometric coefficients, c and d (number of moles) and similarly for reactant
Complete Step by Step Answer:
In the given question, the nitrogen molecule $\left(\mathrm{N}_{2}\right)$ reacts with three moles of hydrogen molecule $\left(\mathrm{H}_{2}\right)$. to give two moles of ammonia $\left(\mathrm{NH}_{3}\right)$ in a gaseous state, which is a reversible reaction such that at any instant time both reactant and product will attain eqbm. State such as
N2(g) + H2(g) ⇌ 2NH3(g)
Now, to find the value of the equilibrium constant (tell about the relative amount of product with respect to reactant), calculate the ratio of concentration of ammonia with raise to the power of its stoichiometric coefficient (number of moles) to the product’s concentration of $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$ with raise to the power of its stoichiometric coefficient such as
$K=\left[\mathrm{NH}_{3}\right]^{2} /\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{2}$; where the concentration of ammonia, a product, is raised to the power of 2 (number of moles) and the product of the concentration of nitrogen and hydrogen molecule, reactants and concentration of hydrogen molecule is raised to the power of 3 (number of moles).
Thus, the correct option is A.
Note: Stoichiometric coefficients are not actual coefficients. The stoichiometric coefficients are used before reactant or product species to balance the chemical equation for any chemical reaction whereas the actual coefficients are the coefficients that are present before balancing a chemical equation. Stoichiometric coefficients can be whole numbers as well as fractional numbers.
For general reaction,
aA + bB ⇌ cC + dD, equilibrium constant is defined as
Kc= [C]c[D]d/[A]a[B]b, Where [C]c and [D]d are the molar concentration of products, C and D both raised to the power of their stoichiometric coefficients, c and d (number of moles) and similarly for reactant
Complete Step by Step Answer:
In the given question, the nitrogen molecule $\left(\mathrm{N}_{2}\right)$ reacts with three moles of hydrogen molecule $\left(\mathrm{H}_{2}\right)$. to give two moles of ammonia $\left(\mathrm{NH}_{3}\right)$ in a gaseous state, which is a reversible reaction such that at any instant time both reactant and product will attain eqbm. State such as
N2(g) + H2(g) ⇌ 2NH3(g)
Now, to find the value of the equilibrium constant (tell about the relative amount of product with respect to reactant), calculate the ratio of concentration of ammonia with raise to the power of its stoichiometric coefficient (number of moles) to the product’s concentration of $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$ with raise to the power of its stoichiometric coefficient such as
$K=\left[\mathrm{NH}_{3}\right]^{2} /\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{2}$; where the concentration of ammonia, a product, is raised to the power of 2 (number of moles) and the product of the concentration of nitrogen and hydrogen molecule, reactants and concentration of hydrogen molecule is raised to the power of 3 (number of moles).
Thus, the correct option is A.
Note: Stoichiometric coefficients are not actual coefficients. The stoichiometric coefficients are used before reactant or product species to balance the chemical equation for any chemical reaction whereas the actual coefficients are the coefficients that are present before balancing a chemical equation. Stoichiometric coefficients can be whole numbers as well as fractional numbers.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions (2025-26)

Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 4 The d and f Block Elements (2025-26)

Biomolecules Class 12 Chemistry Chapter 10 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 10 Biomolecules (2025-26)

