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Chapter 7: Problem 151

The equilibrium constants for the reactions \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) and \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) are \(K_{1}\) and \(K_{2}\) respectively. The correct relationship between \(\mathrm{K}_{1}\) and \(\mathrm{K}\), is (a) \(\mathrm{K}_{1}=\frac{\mathrm{K}_{2}}{2}\) (b) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}_{\mathrm{t}}}\) (c) \(\mathrm{K}_{2}=\mathrm{K}_{1}^{2}\) (d) \(\mathrm{K}_{1}=\mathrm{K}_{2}\)

Short Answer

Expert verified
The correct relationship is (b) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}_{1}}\).

Step by step solution

01

Understanding Equilibrium Constants

Equilibrium constants describe the ratio of the concentration of reactants and products at equilibrium for a given reaction at a specific temperature. Changing the coefficients in the balanced reaction affects the equilibrium constant for the reaction.
02

Reaction Comparison

The reaction \( \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \) has the equilibrium constant \( K_1 \). The second reaction \( \frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3} \) is simply the first reaction divided by 2. This means the second reaction has half the number of moles of reactants and products.
03

Relation Between Reactions

When a reaction is multiplied or divided by a coefficient in a balanced equation, the equilibrium constant changes. Specifically, when a reaction is divided by a coefficient, the new equilibrium constant is the original constant raised to the inverse of that coefficient. In this case, since the second reaction is the first reaction divided by 2, \( K_2 = (K_1)^{1/2} \).
04

Answer Interpretation

By comparing the relationship we derived (\(K_2 = \sqrt{K_1} \)) to the given options, it corresponds to option (b): \( \mathrm{K}_{2}=\sqrt{\mathrm{K}_{1}} \).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to the state in a chemical reaction where the concentrations of reactants and products become constant. This occurs when the forward and reverse reactions proceed at the same rate. It's important to note that at equilibrium, the reaction doesn't stop. Rather, it continues with no net change in the concentrations of substances.

  • At the start, reactants are converted into products until equilibrium is reached.
  • At equilibrium, reactions proceed in both directions, maintaining a constant ratio of reactants to products.
Reaching the equilibrium state varies for different reactions. It can depend on factors like temperature and pressure. Understanding equilibrium helps in predicting how changes in conditions affect the balance of chemical systems.
Reaction Coefficients
Reaction coefficients are the numbers placed in front of compounds in a balanced chemical equation. They indicate the relative amounts of reactants and products involved in the reaction. Changing these coefficients affects the equilibrium constant of a reaction.

  • If a reaction is multiplied by a factor, the equilibrium constant is raised to the power of that factor.
  • If a reaction is divided by a factor, the equilibrium constant is taken to the root equivalent to that factor.
Understanding reaction coefficients is crucial because they don't just affect the quantitative relationships in the reaction, but they also influence how the equilibrium constant is calculated.
Equilibrium Expressions
An equilibrium expression is a formula that relates the concentrations of reactants and products of a reaction at equilibrium. For a general reaction:\[aA + bB \rightleftharpoons cC + dD\]the equilibrium expression is:\[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]

  • Concentrations of products are in the numerator, and reactants in the denominator.
  • Each concentration is raised to the power of its coefficient in the balanced equation.
These expressions are pivotal for calculating equilibrium constants, which help predict the extent of a reaction and the position of equilibrium in a chemical process. They provide a mathematical way to understand and quantify the state of balance in chemical reactions.

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Most popular questions from this chapter

\(\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{c}}\) for the reaction \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) will be (a) \(\sqrt{\mathrm{RT}}\) (b) RT (c) \(\frac{1}{\sqrt{\mathrm{RT}}}\) (d) 1

Assume that the decomposition of \(\mathrm{HNO}_{3}\) can be represented by the following equation \(4 \mathrm{HNO}_{3}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) and the reaction approaches equilibrium at \(400 \mathrm{~K}\) temperature and 30 atm pressure. At equilibrium, partial pressure of \(\mathrm{HNO}_{3}\) is \(2 \mathrm{~atm} .\) Calculate \(\mathrm{K}_{\mathrm{c}}\) in \((\mathrm{mol} / \mathrm{L})^{3}\) at \(400 \mathrm{~K}\) : (Use : \(\mathrm{R}=0.08 \mathrm{~atm}-\mathrm{L} / \mathrm{mol}-\mathrm{K})\) (a) 4 (b) 8 (c) 16 (d) 32

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at \(300 \mathrm{~K}\) is kept in a closed container under one atmospheric pressure. It is heated to \(600 \mathrm{~K}\) when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}(\mathrm{~g})\). The resultant pressure is (a) \(1.0 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.2 \mathrm{~atm}\)

Consider an endothermic reaction \(\mathrm{X} \longrightarrow \mathrm{Y}\) with the activation energies \(\mathrm{E}_{\mathrm{b}}\) and \(\mathrm{E}_{\mathrm{f}}\) for the backward and forward reactions, respectively. In general \([\mathbf{2 0 0 5}]\) (a) \(\mathrm{E}_{b}<\mathrm{E}_{\mathrm{f}}\) (b) \(E_{b}>E_{f}\) (c) \(E_{b}=E_{f}\) (d) there is no definite relation between \(\mathrm{E}_{\mathrm{b}}\) and \(\mathrm{E}_{\mathrm{f}}\)

Match the following Column-I (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) Column-II (p) Unaffected by inert gas addition at constant volume (q) Forward shift by rise in pressure (r) Unaffected by increase in pressure (s) Backward shift by rise in pressure (t) reaction has \(\Delta n_{g}>0\)
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