Understanding the Problem

We are given the equilibrium constant K_C = 4 × 10^–4 for the reaction:

$N2\left(g\right)+O2\left(g\right)\rightleftharpoons 2\mathrm{NO}\left(g\right)$

We need to find the value of K_C' for the reaction:

$\mathrm{NO}\left(g\right)\rightleftharpoons \frac{1}{2}N2\left(g\right)+\frac{1}{2}O2\left(g\right)$

at the same temperature.

Step-by-Step Solution

Step 1: Analyze the Relationship Between the Reactions

The second reaction is the exact reverse of the first reaction, but it is also divided by 2 (i.e., all stoichiometric coefficients are halved).

Let's denote the original reaction and its constant:

Reaction 1: A + B ⇌ 2C;  K_C1 = 4 × 10^–4

The reaction we want is the reverse of Reaction 1, divided by 2.

Reaction 2: C ⇌ (1/2)A + (1/2)B

Step 2: Find the Constant for the Reverse Reaction

If you reverse a reaction, the new equilibrium constant (K_reverse) is the reciprocal of the original constant.

$K_{reverse}=\frac{1}{K_{original}}=\frac{1}{4\times {10}^{–4}}=2500$

So, for the reaction 2C ⇌ A + B, K_C = 2500.

Step 3: Find the Constant for the Reaction Divided by 2

If you multiply a reaction by a factor n, the new equilibrium constant (K_new) is the original constant raised to the power n (K_originalⁿ).

Our target reaction is exactly half of the reversed reaction (n = 1/2).

$K_{final}={\left(K_{reverse}\right)}^{1/2}=\sqrt{K_{reverse}}=\sqrt{2500}$

$K_{final}=50$

Step 4: Combine the Steps into One Formula

We can combine both operations into a single step. The new reaction is the original reaction multiplied by -1/2 (reverse and halve).

$K_{final}={\left(K_{original}\right)}^{-1/2}=\frac{1}{\sqrt{K_{original}}}=\frac{1}{\sqrt{4\times {10}^{–4}}}=\frac{1}{0.02}=50$

Final Answer

The value of K_C for the reaction NO(g) ⇌ ½N₂(g) + ½O₂(g) is 50.0.

Related Concepts & Formulae

1. The Equilibrium Constant (K_C)

For a general reaction: $aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is defined as: $K_C=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

where [X] denotes the molar concentration of X at equilibrium.

2. Manipulating Equilibrium Constants

• Reversing a Reaction: The new constant is the reciprocal of the original. $K_{new}=\frac{1}{K_{old}}$
• Multiplying a Reaction by a factor n: The new constant is the old constant raised to the power n. $K_{new}={\left(K_{old}\right)}^n$
• Adding Reactions: If a reaction is the sum of two or more reactions, its equilibrium constant is the product of the constants of the individual reactions. $K_{total}=K_1\times K_2$

3. Key Theory

The equilibrium constant is a measure of the extent to which a reaction proceeds before reaching equilibrium. A large K (>1) favors products, while a small K (<1) favors reactants. It is crucial to remember that the equilibrium constant is only dependent on temperature. It does not change with initial concentrations, pressure, volume, or the presence of a catalyst.