Question

For the reaction $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3}\) \(M,\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{~N}_{2}(g)\right]=5.3 \times 10^{-2} M\), and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Short answer

The value of the equilibrium constant (K) for the reaction at this temperature is approximately \(4.04 \times 10^{6}\).

Step-by-step solution

Identify the given concentrations

We are given the equilibrium concentrations for each substance as follows: $$[\text{NO}] = 8.1 \times 10^{-3}\,M$$ $$[\text{H}_2] = 4.1 \times 10^{-5}\,M$$ $$[\text{N}_2] = 5.3 \times 10^{-2}\,M$$ $$[\text{H}_2\text{O}] = 2.9 \times 10^{-3}\,M$$

Apply the equilibrium constant formula

Now we will use the formula for the equilibrium constant (K) and plug in the given concentrations: $$K = \frac{[\text{N}_2][\text{H}_2\text{O}]^2}{[\text{NO}]^2[\text{H}_2]^2}$$

Substitute the given concentrations

Replace the substance concentrations with their given values: $$K = \frac{(5.3 \times 10^{-2})(2.9 \times 10^{-3})^2}{(8.1 \times 10^{-3})^2(4.1 \times 10^{-5})^2}$$

Calculate the equilibrium constant (K)

Perform the calculations to find the value of K: $$K = \frac{(5.3 \times 10^{-2})(8.41 \times 10^{-6})}{(6.561 \times 10^{-5})(1.681 \times 10^{-9})}$$ $$K = \frac{4.4553 \times 10^{-7}}{1.10261 \times 10^{-13}}$$ $$K \approx 4.04 \times 10^{6}$$ The value of the equilibrium constant (K) for the reaction at this temperature is approximately \(4.04 \times 10^{6}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. At this point, the concentrations of each reactant and product remain constant over time. This doesn't mean that the reactions have stopped; rather, the rate at which the reactants are converted to products is equal to the rate at which products are converted back to reactants. It is a dynamic balance and can be disturbed by changing conditions like temperature, pressure, or concentration. However, if left undisturbed, a system at equilibrium will remain unchanged.

Reaction Concentrations

The concentration of a substance in a reaction gives an idea of how much of that substance is present in a given volume. In a chemical equation, these concentrations are often denoted in molarity (M), which is moles per liter.

For a reaction at equilibrium, the concentrations of the reactants and products are crucial for determining the position of equilibrium. Higher concentrations of reactants might suggest that the reaction can proceed further to produce more products, reaching equilibrium again. In our exercise, the equilibrium concentrations are:

• [NO] = 8.1 × 10^-3 M
• [H₂] = 4.1 × 10^-5 M
• [N₂] = 5.3 × 10^-2 M
• [H₂O] = 2.9 × 10^-3 M

Equilibrium Expression

The equilibrium expression is a mathematical representation of a chemical reaction at equilibrium. It uses the equilibrium concentrations of the reactants and products to express the equilibrium constant, denoted as K.

In our example, the chemical equilibrium expression is derived from:

• The chemical formula for the reaction: \[ \text{2NO(g) + 2H}_2\text{(g) } \rightleftharpoons \text{ N}_2\text{(g) + 2H}_2\text{O(g)} \]
• The equilibrium expression: \[ K = \frac{[\text{N}_2][\text{H}_2\text{O}]^2}{[\text{NO}]^2[\text{H}_2]^2} \]This expression shows us how the equilibrium constant (K) depends on the concentrations of products raised to the power of their coefficients, divided by the concentrations of reactants raised to their coefficients. Each concentration is placed inside brackets, signaling they are equilibrium concentrations.

Reaction Quotient

The reaction quotient, Q, is a ratio of the concentration of products to reactants at any point in time during a reaction that may or may not be at equilibrium.

The expression to calculate Q is similar to the equilibrium expression:

• Apply the form of the equilibrium expression using current concentrations rather than those at equilibrium.
• Compare Q with the equilibrium constant K to understand the status of the reaction.

If Q < K, the reaction will continue to produce more products to reach equilibrium. If Q > K, the reaction will proceed in reverse, producing more reactants to regain equilibrium. When Q = K, the system is at equilibrium, indicating that the forward and reverse reactions occur at the same rate.