Using Hess's Law To Determine K: Videos & Practice Problems

Understanding how alterations to a chemical reaction affect the equilibrium constant (K) is crucial. Multiplying the coefficients of a reaction will raise the K value exponentially; for instance, tripling the coefficients results in K being cubed. Conversely, reversing a reaction yields the inverse of K, changing the direction of the reaction and flipping the reactants and products. Dividing the reaction by a number raises K to the reciprocal of that number, such as halving the reaction coefficients resulting in the square root of K. These changes demonstrate that the relationship between K and the coefficients of a reaction is not linear but exponential, significantly impacting the equilibrium constant and the system's behavior.

Hess's Law and Equilibrium Constant

Hess's law illustrates that the enthalpy change of a reaction, denoted as ΔH, is directly proportional to the coefficients in a balanced chemical equation. This means that if the coefficients are multiplied by a factor, ΔH is also multiplied by that same factor. Conversely, if the reaction is reversed, the sign of ΔH changes accordingly.

In contrast, the relationship between the equilibrium constant (K) and the coefficients of a reaction is exponential. For example, consider the balanced chemical equation:

2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)

With an initial equilibrium constant value of K = 71.3, we can explore how changes to the reaction affect K.

1. **Multiplying the Reaction:** When the reaction is multiplied by a factor, K is raised to that same power. For instance, if we multiply the reaction by 3, the new equation becomes:

6 SO₂(g) + 3 O₂(g) ⇌ 6 SO₃(g)

In this case, the new equilibrium constant is K³ = 71.3³ ≈ 362467.097.

2. **Reversing the Reaction:** If the reaction is reversed, the new equilibrium constant becomes the inverse of the original K. For the reversed reaction:

2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g)

The new equilibrium constant is K^-1 = 71.3^-1 ≈ 0.0140.

3. **Dividing the Reaction:** When the reaction is divided by a factor, K is raised to the reciprocal of that factor. For example, if we divide the reaction by 2, the new equation becomes:

SO₂(g) + 0.5 O₂(g) ⇌ SO₃(g)

Here, the new equilibrium constant is K^1/2 = √71.3 ≈ 8.44.

In summary, the changes made to a chemical reaction—whether multiplying, reversing, or dividing—result in exponential changes to the equilibrium constant K. Understanding these relationships is crucial for predicting how alterations in a reaction will affect its equilibrium state.

Hess's Law and Kp Example

In chemical equilibrium, the equilibrium constant \( K_p \) is a crucial value that reflects the ratio of the concentrations of products to reactants at equilibrium. When manipulating a chemical reaction, such as through division, reversal, or multiplication of coefficients, the \( K_p \) value changes accordingly. Understanding how to calculate the new \( K_p \) is essential for predicting the behavior of the system.

For a given reaction with an original \( K_p \) value of \( 7.5 \times 10^{-2} \), we can derive new \( K_p \) values based on the changes made to the reaction:

1. **Dividing the Reaction by 2**: When the entire reaction is divided by a factor, the new \( K_p \) is calculated by taking the original \( K_p \) to the power of the reciprocal of that factor. In this case, dividing by 2 means:

\[ K_p' = (K_p)^{1/2} = (7.5 \times 10^{-2})^{1/2} \approx 0.27 \]

2. **Reversing the Reaction**: When the reaction is reversed, the new \( K_p \) is the inverse of the original \( K_p \). Thus:

\[ K_p' = (K_p)^{-1} = (7.5 \times 10^{-2})^{-1} \approx 13 \]

3. **Reversing and Multiplying by 4**: If the reaction is both reversed and the coefficients are multiplied (in this case, by 4), the new \( K_p \) is calculated by taking the inverse of the original \( K_p \) and raising it to the power of the new coefficient. Therefore:

\[ K_p' = (K_p)^{-1} \times 4 = (7.5 \times 10^{-2})^{-1} \] raised to the power of 4:

\[ K_p' = 13^4 \approx 3.2 \times 10^{4} \]

In summary, the manipulation of chemical reactions significantly impacts the equilibrium constant. Dividing, reversing, or multiplying coefficients leads to exponential changes in the \( K_p \) value, which can be calculated using the appropriate mathematical operations on the original \( K_p \).

K_c = 6.5 x 10² at a particular temperature for a reaction: 2 NO(g) + 2 H(g) ⇌ N₂(g) + 2 H₂O(g). Calculate K_c at same temperature for the following reaction: 1/3 N₂(g) + 2/3 H₂O(g) ⇌ 2/3 NO(g) + 2/3 H(g).

• A. 8.7
• B. 0.12
• C. 0.54
• D. 3.6×10^–9

Answer: (B)

Here’s what students ask on this topic:

How does multiplying the coefficients of a reaction affect the equilibrium constant (K)?

Multiplying the coefficients of a reaction affects the equilibrium constant (K) exponentially. If you multiply the coefficients by a factor, the new equilibrium constant (K') will be the original K raised to that factor. For example, if the original reaction has an equilibrium constant K and you multiply the coefficients by 3, the new equilibrium constant K' will be K³. This exponential relationship means that even small changes in the coefficients can significantly impact the value of K.

What happens to the equilibrium constant (K) when a reaction is reversed?

When a reaction is reversed, the equilibrium constant (K) becomes the inverse of the original K. This means that if the original reaction has an equilibrium constant K, the reversed reaction will have an equilibrium constant K' equal to 1/K. For example, if the original K is 71.3, the reversed reaction will have K' = 1/71.3, which is approximately 0.0140. Reversing the reaction flips the reactants and products, and this change is reflected in the new equilibrium constant.

How does dividing the coefficients of a reaction by a number affect the equilibrium constant (K)?

Dividing the coefficients of a reaction by a number affects the equilibrium constant (K) by raising it to the reciprocal of that number. For example, if you divide the coefficients by 2, the new equilibrium constant (K') will be the square root of the original K (K^1/2). If you divide by 3, K' will be K^1/3. This means that if the original K is 71.3 and you divide the coefficients by 2, the new K' will be √71.3, which is approximately 8.44.

Why is the relationship between the equilibrium constant (K) and the coefficients of a reaction considered exponential?

The relationship between the equilibrium constant (K) and the coefficients of a reaction is considered exponential because changes to the coefficients result in K being raised to a power. Multiplying the coefficients by a factor raises K to that factor, reversing the reaction takes the inverse of K, and dividing the coefficients raises K to the reciprocal of that number. These changes are not linear but involve exponents, which significantly impact the value of K and the behavior of the chemical system.

Can you provide an example of how altering a reaction affects its equilibrium constant (K)?

Sure! Consider the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) with an equilibrium constant K = 71.3. If we multiply the reaction by 3, the new reaction is 6SO₂(g) + 3O₂(g) ⇌ 6SO₃(g), and the new K' is K³ = 71.3³ ≈ 362467.097. If we reverse the original reaction, the new reaction is 2SO₃(g) ⇌ 2SO₂(g) + O₂(g), and the new K' is 1/K = 1/71.3 ≈ 0.0140. If we divide the original reaction by 2, the new reaction is SO₂(g) + 0.5O₂(g) ⇌ SO₃(g), and the new K' is √K = √71.3 ≈ 8.44.