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If the reaction quotient is larger than the equilibrium constant, in what direction does the reaction proceed as it approaches equilibrium? What will happen if \(Q \leq K\) ?

Short Answer

Expert verified
If \(Q > K\), the reaction proceeds in the reverse direction. If \(Q \leq K\), it proceeds in the forward direction or stays at equilibrium.

Step by step solution

01

Understanding Reaction Quotient and Equilibrium Constant

The reaction quotient, denoted by \(Q\), and the equilibrium constant, \(K\), are measures of the relative concentrations of reactants and products at any given time. While \(K\) is specific to equilibrium conditions, \(Q\) can be calculated at any time to evaluate the progress of the reaction.
02

Analyzing when Q > K

When \(Q > K\), it means there are more products relative to reactants than at equilibrium. Consequently, the reaction will shift to convert some of the products back to reactants to achieve equilibrium. This implies the reaction needs to proceed in the reverse direction.
03

Analyzing when Q < K

If \(Q < K\), the situation is opposite: there are more reactants than products at that moment in comparison to the equilibrium state. Thus, the reaction will proceed in the forward direction, converting reactants into products to reach equilibrium.
04

Analyzing when Q = K

When \(Q = K\), it indicates that the reaction is at equilibrium, with no net change in the concentrations of reactants and products. Hence, the reaction will not proceed in either direction.

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Key Concepts

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

Reaction Quotient
The Reaction Quotient, symbolized as \( Q \), is an essential concept in understanding chemical reactions. It represents the ratio of the concentrations of products to reactants at any given point in a reaction. Unlike the equilibrium constant \( K \), which is fixed for a particular reaction at a given temperature, \( Q \) can vary as the reaction progresses. This makes \( Q \) a powerful tool to predict how a reaction will behave at a certain moment. To calculate \( Q \), you use the same formula as you would for \( K \), based on the balanced chemical equation:- Continue to divide the concentration of products by the concentration of reactants, each raised to the power of their coefficients in the equation.By comparing \( Q \) with \( K \), you can determine the direction in which the reaction will proceed to reach equilibrium:- If \( Q > K \), there are too many products, and the reaction will shift towards the reactants.- If \( Q < K \), there are too many reactants, and the reaction will shift towards the products.- If \( Q = K \), the reaction is already at equilibrium, so the concentrations remain constant.Understanding how to use \( Q \) can provide insights into reaction progress and what changes are necessary to reach equilibrium.
Equilibrium Constant
The Equilibrium Constant, denoted as \( K \), is a fundamental aspect of chemical equilibrium. This constant is unique for each reaction and applies only when the reaction is at equilibrium—meaning the rates of the forward and reverse reactions are equal, and no net change in concentrations occurs. The value of \( K \) provides critical insight into the nature of the reaction:- If \( K \) is large, the reaction favors the formation of products at equilibrium.- If \( K \) is small, the reaction favors the reactants, indicating that only a small portion of reactants are converted into products at equilibrium.Determining \( K \) involves the concentrations of products and reactants at equilibrium, each raised to the power of their stoichiometric coefficients in the balanced equation. This creates a numerical value allowing chemists to understand the balance between products and reactants at equilibrium.This constant does not change unless the temperature changes, which makes \( K \) a reliable indicator of a reaction's state under constant conditions. By observing \( K \) in relation to \( Q \), chemists can predict how the reaction proceeds as it approaches equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a key principle in understanding how equilibrium shifts in response to changes in concentration, temperature, or pressure. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will adjust to counteract the change, restoring a new equilibrium. Here's how Le Chatelier's Principle can be applied: - **Change in Concentration**: If the concentration of either reactants or products is altered, the equilibrium shifts in a way to oppose the change and restore equilibrium. For example: - Adding more reactants causes the equilibrium to shift towards the products, using up the excess reactants. - Similarly, removing products will also shift the equilibrium towards the products to replace the lost products. - **Change in Temperature**: Temperature changes can also affect equilibrium. The direction of the shift depends on whether the reaction is exothermic or endothermic. - Increasing temperature shifts equilibrium in the direction that absorbs heat. - Decreasing temperature favors the exothermic direction to release heat. - **Change in Pressure**: Changes in pressure, particularly in gaseous systems, can shift equilibrium as well. - Increasing pressure shifts equilibrium to the side with fewer moles of gas. - Decreasing pressure favors the side with more moles of gas. Le Chatelier’s Principle provides a predictive framework for understanding and controlling reaction conditions, which is especially beneficial in industrial chemical processes.

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Most popular questions from this chapter

For each of these reactions at \(25^{\circ} \mathrm{C}\), indicate whether the entropy effect, the energy effect, both, or neither favors the reaction. (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NF}_{3}(\mathrm{~g}) \quad \Delta_{1} H^{\circ}=-249 \mathrm{~kJ} / \mathrm{mol}\) (b) \(\mathrm{N}_{2} \mathrm{~F}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NF}_{2}(\mathrm{~g})\) \(\Delta_{t} H^{\circ}=93.3 \mathrm{~kJ} / \mathrm{mol}\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NCl}_{3}(\mathrm{~g}) \quad \Delta_{1} H^{\circ}=460 \mathrm{~kJ} / \mathrm{mol}\)

The vapor pressure of water at \(80 .{ }^{\circ} \mathrm{C}\) is \(0.467 \mathrm{~atm} .\) Determine the value of \(K_{\mathrm{c}}\) for the process $$ \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ at this temperature.

The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) is endothermic. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ (a) Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. (b) If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a sealed flask in which this equilibrium exists, how is the equilibrium affected? (c) What if some additional \(\mathrm{NH}_{3}\) is placed in a sealed flask containing an equilibrium mixture? (d) What will happen to the partial pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{~S}\) is removed from the flask?

Predict whether the equilibrium for the photosynthesis reaction described by the equation $$ \begin{array}{r} 6 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{~s})+6 \mathrm{O}_{2}(\mathrm{~g}) \\ \Delta_{1} H^{\circ}=2801.69 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ would (i) shift to the right, (ii) shift to the left, or (iii) remain unchanged for each of these changes: (a) decrease the concentration of \(\mathrm{CO}_{2}\) at constant volume. (b) increase the partial pressure of \(\mathrm{O}_{2}\) at constant volume. (c) remove one half of the \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (d) decrease the total pressure by increasing the volume. (e) increase the temperature. (f) introduce a catalyst into a constant-volume system.

Write equilibrium constant expressions, in terms of reactant and product concentrations, for each of these reactions. $$ \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \quad K_{\mathrm{c}}=1.0 \times 10^{-14} $$ \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) $$ \begin{array}{c} K_{\mathrm{c}}=1.8 \times 10^{-5} \\ \mathrm{~N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \end{array} $$ Assume that all gases and solutes have initial concentrations of \(1.0 \mathrm{~mol} / \mathrm{L}\). Then let the first reactant in each reaction change its concentration by \(-x\). (a) Using the reaction table (ICE table) approach, write equilibrium constant expressions in terms of the unknown variable \(x\) for each reaction. (b) Which of these expressions yield quadratic equations? (c) How would you go about solving the others for \(x ?\)

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