Question

At a particular temperature, \(K_{c}=6.5 \times 10^{2}\) for $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ Calculate \(K_{c}\) for each of the following reactions: (a) \(\mathrm{NO}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{NO}(g)+4 \mathrm{H}_{2}(g)\)

Short answer

For reaction (a), \( K_c \) is approximately 25.5. For reaction (b), \( K_c \) is approximately \( 2.37 \times 10^{-6} \).

Step-by-step solution

- Understanding the given equilibrium

Given: \[2 \mathrm{NO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\] with \( K_c = 6.5 \times 10^2 \).This means the equilibrium constant for the initial reaction is \( K_c = 6.5 \times 10^2 \).

- Find the relationship for reaction (a)

Reaction (a): \[\mathrm{NO}(g) + \mathrm{H}_{2}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g)\].This is half of the original reaction. To find the equilibrium constant for this reaction, we take the square root of the provided \( K_c \).

- Calculate the equilibrium constant for reaction (a)

To find \( K_c \) for reaction (a), calculate the square root of the given \( K_c \):\[ K_c' = \sqrt{6.5 \times 10^2} \].

- Compute the square root

Calculate the square root of 650:\[ K_c' = \sqrt{650} \approx 25.5 \].

- Find the relationship for reaction (b)

Reaction (b): \[2 \mathrm{N}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 4 \mathrm{H}_{2}(g)\].This is the reverse of twice the original reaction. To find the equilibrium constant for this reaction, we take the reciprocal of the square of the provided \( K_c \).

- Calculate the equilibrium constant for reaction (b)

To find \( K_c \) for reaction (b), calculate the inverse of the square of the given \( K_c \):\[ K_c'' = \frac{1}{(6.5 \times 10^2)^2} \].

- Compute the value

Calculate \[ K_c'' = \frac{1}{(650)^2} = \frac{1}{422500} \approx 2.37 \times 10^{-6} \].

Key concepts

chemical equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant over time. It's important to remember that equilibrium does not mean the amounts of reactants and products are equal, but that their rates of formation are equal. Some key features include:

• Dynamic nature: Reaction continues but with no net change in concentrations.
• Constant concentrations: Reactants and products maintain their levels.
• Dependence on conditions: Equilibrium can shift if conditions like temperature or pressure change.

This forms the basis of calculating equilibrium constants, which quantify the ratio of product and reactant concentrations at equilibrium.

equilibrium constant

The equilibrium constant, denoted as K_c, is a crucial concept in chemical equilibrium. It is a numerical value that describes the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients from the balanced chemical equation. For a general reaction:
$$aA + bB \rightleftharpoons cC + dD$$
The equilibrium constant expression is:
$$ K_{c} = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
Here, squared brackets denote concentrations.
High K_c values (>1) suggest products are favored, while low K_c values (<1) indicate reactants are favored. The equilibrium constant depends only on temperature and is unaffected by the initial concentrations of reactants or products.

reaction stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It helps us understand how the moles of reactants convert to moles of products, guided by the balanced equation. For example, in the given problem:
$$ 2 \text{NO}(g) + 2 \text{H}_{2}(g) \rightleftharpoons \text{N}_{2}(g) + 2 \text{H}_{2} \text{O}(g)$$The coefficients (2 for NO and H₂, and so on) indicate the molar ratios of substances.
When making transformations (like halving the reaction for part (a) or reversing and doubling for part (b)), these stoichiometric coefficients guide how the equilibrium constant changes. Specifically:

• For halved reactions, take the square root of K_c.
• For reversed reactions, take the reciprocal of K_c.
• For doubled reactions, square the original K_c.

This ensures our equilibrium calculations stay accurate with the reaction's stoichiometry.

inverse and square root relations

Inverse and square root relations are critical when manipulating equilibrium constants for modified reactions. These mathematical operations adjust K_c to reflect changes in stoichiometry or reaction direction.
Here’s a detailed application:

• Square Root: When we halve the reaction, every coefficient in the balanced equation is halved. This adjustment means the new K_c is the square root of the original.
  Example: From the problem, halving the reaction:$$ \text{NO}(g) + \text{H}_2(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \text{H}_2 \text{O}(g)$$ New K_c: $$\text{K'} = \root{2}\frac{[N_2]^{0.5}[H_2O]}{[NO][H_2]} = \root{2} 6.5 \times 10^2 \text{ or approximately } 25.5$$
• Inverse: If a reaction is reversed, the products become reactants and vice versa, making the new K_c the reciprocal of the original.
  Example: Reverse of the initial reaction's doubled form leads to $$ 2 \text{N}_2(g) + 4 \text{H}_2 \text{O}(g) \rightleftharpoons 4 \text{NO}(g) + 4 \text{H}_2(g).$$New K_c = $$ \frac{1}{(6.5 \times 10^2)^2} \text{ or approximately } 2.37 \times 10^{-6}.$$
  This understanding of relationships ensures precise equilibrium constant manipulations for any given stoichiometric adjustments.