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.

$$Q = \frac{1}{25} = 0.04$$