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If the value of equllibrium constant \(K_{r}\) for the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(7 .\) The equilibrium constant for the reaction \(2 \mathrm{~N}_{2}+6 \mathrm{H}_{2} \rightleftharpoons 4 \mathrm{NH}_{3}\) will be (a) 49 (b) 7 (c) 14 (d) 28

Short Answer

Expert verified
The equilibrium constant for the reaction \(2 \text{N}_{2}+6 \text{H}_{2} \rightleftharpoons 4 \text{NH}_{3}\) will be 49.

Step by step solution

01

Understand the Relationship of Equilibrium Constants

The equilibrium constant expression for a chemical reaction is dependent on the stoichiometry of the reaction. If an equation is multiplied by a factor, the equilibrium constant for the new equation is the original equilibrium constant raised to the power of that factor.
02

Identify the Multiplication Factor

Comparing the original reaction with the new reaction, we can see that the coefficients in the second reaction are exactly twice those in the first. Therefore, the multiplication factor is 2.
03

Apply the Multiplication Factor to the Equilibrium Constant

Since the original equilibrium constant (\(K_{r}\)) for the first reaction is given as 7, the equilibrium constant for the new reaction will be \(K_{r}(new) = K_{r}^{factor} = 7^{2} = 49\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Understanding chemical equilibrium is essential for a thorough grasp of how reactions occur. It represents the state of a reaction when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products. This dynamic state might seem static, but it is actually a result of continuous and opposing processes.

To quantify this state, we use the equilibrium constant, denoted as \( K \). It is a mathematical ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective coefficients from the balanced chemical equation. For example, for the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), the equilibrium constant (\(K_{r}\)) can be written as:\[K_{r} = \frac{[\mathrm{NH}_{3}]^{2}}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^{3}}\]
where the square brackets signify concentration.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between the amounts of reactants and products in a chemical reaction. It's similar to a recipe, but instead of cups and teaspoons, chemists use moles to measure the reactants and products. When dealing with equilibrium, the stoichiometry of the reaction is particularly crucial because the power to which the concentration of each substance is raised in the equilibrium expression is based on the stoichiometric coefficients from the balanced equation.

For example, if we consider the reaction \( \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), and we double all the coefficients to get \( 2\mathrm{N}_{2}+6 \mathrm{H}_{2} \rightleftharpoons 4 \mathrm{NH}_{3}\), we are effectively altering the ratio of reactants to products. The new equilibrium constant will be the old constant raised to the power equal to the factor of change in the stoichiometric coefficients, which in our case, is 2: \(K_{r}(new) = K_{r}^{2} = 7^{2} = 49\).
Le Chatelier's Principle
Le Chatelier's principle provides insight into how a system at equilibrium responds to a change in conditions. If an external stress, such as a change in concentration, pressure, or temperature, is applied to a system in equilibrium, the system will adjust to counteract the change and restore a new equilibrium.

For instance, if we increase the concentration of \(\mathrm{N}_{2}\) in the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), according to Le Chatelier's principle, the system will shift to the right to reduce the concentration of \(\mathrm{N}_{2}\) by producing more \(\mathrm{NH}_{3}\). This behavior helps to understand and predict the effects of different changes on the system and is a fundamental concept in chemical reactions and processes.

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Most popular questions from this chapter

A reaction is said to be in equilibrium when (a) the rate of transformation of reactants to products is equal to the rate of transformation of products to the reactants (b) \(50 \%\) of the reactants are converted to products (c) the reaction is near completion and all the reactants are converted to products (d) the volume of reactants is just equal to the volume of the products.

What is the minimum concentration of \(\mathrm{SO}_{4}^{2-}\) required to precipitate \(\mathrm{BaSO}_{4}\) in a solution containing \(1 \times 10^{-4}\) mole of \(\mathrm{Ba}^{2+} ?\left(K_{i p}\right.\) for \(\left.\mathrm{BaSO}_{4}=4 \times 10^{-10}\right)\) (a) \(4 \times 10^{-10} \mathrm{M}\) (b) \(2 \times 10^{-10} \mathrm{M}\) (c) \(4 \times 10^{-6} \mathrm{M}\) (d) \(2 \times 10^{-3} \mathrm{M}\)

For a reaction, \(A_{x} B_{y} \rightleftharpoons x A^{y+}+y B^{x-}, K_{s p}\) can be represented as (a) \(\left[A^{y+}\right]^{x}\left[B^{x-}\right]^{y}\) (b) \([A]^{y}[B]^{x}\) (c) \([A]^{x}[B]^{y}\) (d) \([A]^{x+y}[B]^{x-y}\)

If the equilibrium constant for the given reaction is \(0.25\) \(\mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2}\), then the equilibriumconstant for the reaction \(\frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) will be (a) 1 (b) 2 (c) 3 (d) 4

What will be the solubility of \(\mathrm{AgCl}\) in \(0.05 \mathrm{M} \mathrm{NaCl}\) aqueous solution if solubility product of \(\mathrm{AgCl}\) is \(1.5 \times 10^{-10} ?\) (a) \(3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(0.05 \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(3 \times 10^{9} \mathrm{~mol} \mathrm{~L}^{-1}\)

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