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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Equilibria > Chemical Equilibria > The Reaction Quotient

The Reaction Quotient

The reaction quotient (Q) measures the relative amount of products and reactants present during a reaction at a particular point in time. The reaction quotient aids in figuring out which direction a reaction is likely to proceed, given either the pressures or the concentrations of the reactants and the products. The Q value can be compared to the Equilibrium Constant, K, to determine the direction of the reaction that is taking place.

K vs. Q

The main difference between K and \(Q\) is that K describes a reaction that is presently in equilibrium, while Q describes a reaction that is not at equilibrium. To determine \(Q\), you must know the concentrations of the reactants and products and this general equation:

\(Q_c = \dfrac{concentration \; of \; Products}{concentration \; of \; Reactants}\)

The first step is to write the \(Q\) equation based off of a general template. We are given this equation:

\(aA + bB \Longleftrightarrow cC + dD\)

We write the \(Q\) equation by multiplying the concentrations of the products and divide it by the concentrations of the reactants. If any component in the reaction has a coefficient, shown here with lower case letters, we raise the concentration to the power of the coefficient. The completed \(Q\) equation is 

\(Q_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\)

*Note: This equation only uses components in the gaseous or aqueous states. All pure liquids and solids have an activity of one and can be omitted. Equilibrium constants really contain a ratio of concentrations (actual concentration divided by the reference concentration that defines the standard state). Since the standard state for concentrations is usually chosen to be 1 mol/L, it is not written out in practical applications. Hence, the ratio of them does not contain units.

Based on how \(Q\) compares to K, we can tell which way the reaction will shift and which side of the reaction is favored. 

  • If Q>K, the reaction favors the reactants. This means that in the \(Q\) equation, the numerator, or the concentration or pressure of the products, is greater than the denominator, which is the concentration or pressure of the reactants. Since reactions always try to reach equilibrium (Le Châtelier's Principle), the reaction will produce more reactants from the excess products, therefore causing the system to shift to the LEFT. This allows the system to reach equilibrium.  
  • If Q<K, the reaction favors the products. This would mean that the denominator is greater that the numerator, giving us a small \(Q\) value. This translates to the concentration or the pressure of the reactants is greater than the products. Since the reaction will want to reach equilibrium, the system will shift to the RIGHT to make more products.
  • If Q=K, then the reaction is already at equilibrium. This is because the products are equal to the reactants and when placed in the \(Q\) equation, the answer is one. No side is favored and no shift occurs.

Summary

  • Q<K: Out of equilibrium: Reaction favors reactants and shifts Right.
  • Q>K: Out of equilibrium: Reaction favors Products and shifts Left.
  • Q=K: In equilibrium: No side favored and no shift. 

Activity

Another important concept that is used in the calculation of the reaction quotient is called an activity. For example, write the \(Q\) equation for this acid/base reaction: 

\(CH_3CH_2CO_2H{(aq)} + H_2O{(l)} \leftrightharpoons H_3O^+{(aq)} + CH_3CH_2CO_2^-{(aq)}\)

The Q equation is written as the concentrations of the products divided by the concentrations of the products, but only including components in the gaseous or aqueous states. We leave out pure liquid or solid states. The Q equation for this example would be 

\(Q_c = \dfrac{[H_3O^+{(aq)}][CH_3CH_2CO_2^-{(aq)}]}{[CH_3CH_2CO_2H{(aq)}]}\) 

Example 1

What is the Q value for this equation? Which direction will the reaction shift? 

Given: \(CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)\)

\(K_c\) = 1.0

[CO2(g)]= 2.0 M

[H2(g)]= 2.0 M

[CO(g)]= 1.0 M

[H2O(g)]= 1.0 M

Solution:

Step 1: Write the Q formula

\(Q_c = \dfrac{[CO_2][H_2]}{[CO][H_2O]}\)

Step 2: Plug in given concentration values.

\(Q_c = \dfrac{(2.0)(2.0)}{(1.0)(1.0)}\)

Q= 4.0

Step 3: Compare Q to K.

Since 4.0 is greater than 1.0, Q > K.  This means that the reaction shifts left.

Answer: Q= 4.0 and shifts left.

Example 2

Find the value of \(Q\) and determine which side of the reaction is favored.

Given K=0.5

\(HCl(g) + NaOH(aq) \rightleftharpoons  NaCl(aq) + H_2O(l)\)

[HCl]= 3.2

[NaOH]= 4.3

[NaCl]=6

Solutions

Step 1: Write \(Q\) formula. Since we know that the activity of a liquid is 1, we can omit the water component in the equation. 

\(Q_c = \dfrac{[NaCl{(aq)}]}{[HCl{(g)}][NaOH{(aq)}]}\)

Step 2: Plug in given concentrations into the \(Q\) formula.

\(Q_c = \dfrac{[6]}{[3.2][4.3]}\)

Step 3: Calculate using the given concentrations

Q = 0.436

Step 4: Compare \(Q\) to K. Now that \(Q\) is found to be 0.436, we can compare it to the given K value of 0.5, so \(Q\) is less than K. 

Since Q < K, the reaction is not at equilibrium and will proceed to the products side to reach dynamic equilibrium once again.

Answer: Q= 0.436 and the reaction favors the products.

Example 3

Given the equation, \(N_2(g) + 3H_2(aq) \rightleftharpoons  2NH_3(g)\) find Q and determine which direction the reaction will shift in order to reach equilibrium.

Given: \(N_2(g) + 3H_2(aq) \rightleftharpoons 2NH_3(g)\)

[N2]= 0.04M

[H2]= 0.09M

K= 0.040

Solution:

Step 1: Write the Q formula

\(Q_c = \dfrac{[NH_3{(g)}]^2}{[N_2{(g)}][H_2{(g)}]^3}\)

 
Step 2: Plug in Values.  Since the concentrations for N2 and H2 were given, we can plug those right in to the equation.  However, no concentration value was given for NH3 so we assume that there is no concentration, which is denoted with a zero. 
 
\(Q_c = \dfrac{(0)^2}{(.04)(.09)^3}\)
 
Step 3: Solve for Q.
 
Q=0
 
Step 4: Compare Q to K.  Since K=0.04 and Q is zero, K is greater than Q, meaning that the reaction will shift right in order to recreate equilibrium.
 
Answer: Q=0, shifts right.

References

  1. Petrucci, et al. General Chemistry Principles & Modern Applications. 9th ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2007

Contributors

  • Kellie Berman (UCD), Rebecca Backer (UCD), Deepak Nallur (UCD)

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Last Modified
14:41, 18 Nov 2013

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