KEY2CHEM

Reaction Quotient (\(Q\))

 

The reaction quotient (\(Q\)) is formulated as a ratio of product to reactant for a particular point in a reversible reaction. Specifically, the format of reaction quotient based on molar concentrations for a reaction

\(\require{mhchem}\ce{aA + bB <=> cC + dD}\)

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

When more products are present, the magnitude of \(Q\) is larger.  Modification of the chemical equation results in a change in the value of \(Q\), since its format is derived for a particular chemical equation.

 


Example 1.

What is the reaction quotient expression for the equation \(\require{mhchem}\ce{cC + dD <=> aA + bB}\)?

 

 

A. \(Q = \frac{[A]^a[B]^b}{[C]^c[D]^d}\)

B. \(Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

C. \(Q = \frac{[C][D]}{[A][B]}\)

 

 

 

 

 

Solution

 

B. \(Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

The general format of \(Q\) is products divided by reactants, with each component raised to its respective coefficient. Notice that reversing the reaction (from \(\require{mhchem}\ce{aA + bB <=> cC + dD}\) to \(\require{mhchem}\ce{cC + dD <=> aA + bB}\)) results in a reaction quotient which is the reciprocal of the original value.

 


Example 2.

 

What is the missing value of \(K\)?

 

\(\require{mhchem}\ce{N2 + O2 <=> 2 NO}\;\;\;\;\;\;\;\;\;\;K_1 = 3.2 \times 10^ {-24}\)

\(\require{mhchem}\ce{2NO + O2 <=> 2NO2}\;\;\;\;\;K_2 = 1.6 \times 10^ {9}\)

\(\require{mhchem}\ce{N2 + 2O2 <=> 2NO2}\;\;\;\;\;\;\;K_{missing} = ?\)

 

 

A. \(5.1 \times 10 ^{-15}\)

B. \(1.6 \times 10^{9}\)

C. \(1.6 \times 10^{-15}\)

 

 

 

 

Solution

 

A. \(5.1 \times 10 ^{-15}\)

Adding chemical equations together requires multiplication of their equilibrium constants. The format of the equilibrium constant is the same as the reaction quotient.

\(K_{missing} = K_1 \times K_2 = (3.2 \times 10^ {-24}) \times (1.6 \times 10^ {9}) = 5.1 \times 10 ^{-15}\)

 


Example 3.

 

A reaction has a reaction quotient of \(25\). What is the value of the reaction quotient of the reverse reaction under the same conditions?

A. \(25\)

B. \(2.5\)

C. \( 0.04\)

 

 

 

 

 

 

 

 

 

Solution

C. \(0.04\)

The reverse reaction has a reaction quotient that is the reciprocal of the original reaction quotient.

\(Q = \frac{1}{25} = 0.04\)