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The equilibrium constant for a reaction is \(K\), and the reaction quotient is \(Q .\) For a particular reaction mixture, the ratio \(\frac{K}{Q}\) is \(0.33 .\) This means that : (a) the reaction mixture will equilibrate to form more reactant species (b) the reaction mixture will equilibrate to form more product species (c) the equilibrium ratio of reactant to product concentrations will be 3 (d) the equilibrium ratio of reactant to product concentrations will be \(0.33\)

Short Answer

Expert verified
The reaction mixture will equilibrate to form more reactant species.

Step by step solution

01

- Understanding the Reaction Quotient, Q

The reaction quotient, denoted as Q, indicates the relative amounts of products and reactants present during the chemical reaction at a given moment. It is calculated using the same formula as the equilibrium constant (K) but with the current concentrations instead of those at equilibrium.
02

- Comparing Q to K

The value of Q compared to K determines the direction the reaction will shift to reach equilibrium. If Q < K, the reaction will shift to the right to form more products and if Q > K, the reaction will shift to the left to form more reactants.
03

- Analyzing the Given Ratio of K/Q

Since we are given that K/Q = 0.33, it means that K < Q (because the ratio K/Q is less than 1). Thus, the reaction is currently further towards the product side compared to the equilibrium position.
04

- Determining Reaction Shift

Given that K < Q, the reaction will shift to the left to achieve equilibrium, which corresponds to the formation of more reactant species.

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Key Concepts

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

Equilibrium Constant (K)
The equilibrium constant (K) is a fundamental concept in chemical reactions, encapsulating the relationship between the concentrations of reactants and products in a reaction at equilibrium. It is expressed as a ratio where the concentrations of the products are in the numerator, and those of the reactants are in the denominator, raised to the power of their respective stoichiometric coefficients.

For example, the equilibrium constant for a reaction of type \(aA + bB \leftrightarrow cC + dD\) is given by the formula \[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\], where square brackets denote concentrations. Importantly, K is temperature-dependent and remains constant at a given temperature, unless the reaction conditions change. Understanding K helps predict whether a reaction mixture is at equilibrium and if not, in which direction it will shift to reach equilibrium.
Reaction Quotient (Q)
While the equilibrium constant (K) is a measure of a system at equilibrium, the reaction quotient (Q) serves a similar purpose but for a system that is not at equilibrium. Q is calculated with the same formula as K, using the current concentrations of reactants and products during a reaction: \[Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]. By comparing the values of Q and K, you can predict the direction in which the reaction will shift to achieve equilibrium. This is vital for students to understand because it lets them predict the reaction's behavior at any point in time, not just when it's at equilibrium.
Shifts in Chemical Reactions
Chemical reactions are dynamic and can shift to balance the changes that occur in a system. The direction in which a reaction will shift to attain equilibrium is predicted by comparing the reaction quotient (Q) to the equilibrium constant (K).

  • If \(Q < K\), there are more reactants than required at equilibrium, so the reaction will shift right to produce more products.
  • If \(Q > K\), there are more products than at equilibrium, so the reaction will shift left to form more reactants.

Le Châtelier’s Principle

This principle further explains that if a stress is applied to a reaction mixture at equilibrium (such as changing concentration, pressure, or temperature), the system will adjust to counteract the stress and reestablish equilibrium. Understanding these shifts is crucial for students who are mastering the ability to predict changes in reaction conditions.
Reactant and Product Concentration
The concentrations of reactants and products play a pivotal role in the behavior of a chemical reaction. At equilibrium, these concentrations do not change because the rate of the forward reaction is equal to the rate of the reverse reaction. However, when a system is not at equilibrium, concentrations will change in a way to reach that balance.

Clarity about the stoichiometry of the reaction and the initial concentrations of reactants and products is essential to determine the direction in which these concentrations will shift in response to a disturbance. This is directly related to the concept of Q and K as these concentrations dictate their values and thus the reaction's behavior until equilibrium is achieved.

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Most popular questions from this chapter

At \(273 \mathrm{~K}\) and 1 atm, 10 litre of \(\mathrm{N}_{2} \mathrm{O}_{4}\) decomposes to \(\mathrm{NO}_{2}\) according to equation $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ What is degree of dissociation \((\alpha)\) when the original volume is \(25 \%\) less than that of existing volume? (a) \(0.25\) (b) \(0.33\) (c) \(0.66\) (d) \(0.5\)

\(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) For the reaction initially the mole ratio was \(1: 3\) of \(\mathrm{N}_{2}: \mathrm{H}_{2} .\) At equilibrium \(50 \%\) of each has reacted. If the equilibrium pressure is \(p\), the partial pressure of \(\mathrm{NH}_{3}\) at equilibrium is : (a) \(\frac{P}{3}\) (b) \(\frac{P}{4}\) (c) \(\frac{P}{6}\) (d) \(\frac{P}{8}\) \(\therefore\)

The value of \(\Delta G^{\circ}\) for a reaction \(^{\circ}\) in aqueous phase having \(K_{c}=1\), would be : (a) \(-R T\) (b) \(-1\) (c) 0 (d) \(+R T\)

Consider the reactions \(\alpha\) (i) \(2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2}(g)\); Eqm. Constant \(=K_{1}\) (ii) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) ;\) Eqm. Constant \(=K_{2}\) (iii) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) ;\) Eqm. Constant \(=K_{3}\) Which of the following relation is correct? (a) \(K_{3}=\frac{K_{1}}{K_{2}}\) (b) \(K_{3}=\frac{K_{1}^{2}}{K_{2}^{2}}\) (c) \(K_{3}=K_{1} K_{2}\) (d) \(K_{3}=\sqrt{K_{1}} \cdot K_{2}\)

The equilibrium constant \(\left(K_{c}\right)\) for the reaction $$ 2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is \(4 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). What is the equilibrium constant for the reaction? $$ \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{HCl}(g) $$ (a) \(2 \times 10^{-17}\) (b) \(2.5 \times 10^{33}\) (c) \(5 \times 10^{6}\) (d) None of these

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