Q.Equilibrium constant, Kc for the reaction N2

Concept: Reaction Quotient vs. Equilibrium Constant

To decide whether the system is at equilibrium, calculate the reaction quotient $Q_c$ using the given concentrations and compare it with $K_c$.

Step 1. Write the expression for $Q_c$:

Step 2. Substitute the given concentrations:

Step 3. Compare $Q_c$ with $K_c$: …

Why we compare $Q_c$ and $K_c$

The equilibrium constant $K_c$ tells us the ratio of product to reactant concentrations when a system is at equilibrium—it's a fixed number at a given temperature. But what if we're given a snapshot of concentrations at some arbitrary moment? We need a way to ask: "Is this mixture already at equilibrium, or does it need to shift?"

That's where the reaction quotient $Q_c$ comes in. It has the exact same mathematical form as $K_c$, but we plug in the current concentrations, not equilibrium ones. By comparing $Q_c$ to $K_c$, we learn which way the reaction must shift:

• If $Q_c = K_c$: the system is at equilibrium.
• If $Q_c < K_c$: too few products (or too many reactants) relative to equilibrium → reaction shifts forward (right).
• If $Q_c > K_c$: too many products relative to equilibrium → reaction shifts backward (left).

Step-by-step solution

1. Write the expression for $Q_c$

For the reaction

the reaction quotient is

Notice the exponents match the stoichiometric coefficients, and products go in the numerator.

2. Substitute the given concentrations

At the particular time in question:

• $[\ce{N2}] = 3.0 \, \text{mol L}^{-1}$
• $[\ce{H2}] = 2.0 \, \text{mol L}^{-1}$
• $[\ce{NH3}] = 0.5 \, \text{mol L}^{-1}$

Plugging these into $Q_c$:

3. Compare $Q_c$ with $K_c$

We're told $K_c = 0.061$ at 500 K. Now:

Since $Q_c < K_c$, the ratio of products to reactants is currently smaller than it should be at equilibrium. …