Problem 24 - Chemistry An Atoms First Approach

Question

For the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$.\(K_{\mathrm{p}}=3.5 \times 10^{4}\) at \(1495 \mathrm{K} .\) What is the value of \(K_{\mathrm{p}}\) for the following reactions at \(1495 \mathrm{K} ?\) a. \(\operatorname{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\). b. \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) c. \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)\).

Step-by-Step Solution

Answer

For the three reactions at 1495 K, the values of Kp are: a. Kp = 0.00535 b. Kp = 2.857 × 10^(-5) c. Kp = 187.082

Step 1: Find the reaction relation with the given equation

Observe that this reaction is the reverse of the given reaction and divided by 2.

Step 2: Calculate the equilibrium constant for the reverse reaction

Since the reaction is reversed, we need to find the reciprocal of the equilibrium constant of the given reaction Kp_reverse = 1/Kp = 1/(3.5 × 10^4) = 2.857 × 10^(-5).

Step 3: Calculate the equilibrium constant for the new reaction

Next, raise the Kp_reverse to the power of the fraction by which the reaction has been divided: Kp_new = (Kp_reverse)^(1/2) = (2.857 × 10^(-5))^(1/2) = 0.00535. Answer: Kp = 0.00535 for reaction a at 1495 K. b. 2 HBr(g) <=> H2(g) + Br2(g)

Step 1: Find the reaction relation with the given equation

This reaction corresponds to the reverse of the given reaction.

Step 2: Calculate the equilibrium constant for the reversed reaction

Since the reaction is reversed, we need to find the reciprocal of the given equilibrium constant: Kp_reverse = 1/Kp = 1/(3.5 × 10^4) = 2.857 × 10^(-5). Answer: Kp = 2.857 × 10^(-5) for reaction b at 1495 K. c. (1/2) H2(g) + (1/2) Br2(g) <=> HBr(g)

Step 1: Find the reaction relation with the given equation

Observe that this reaction is the given reaction divided by 2.

Step 2: Calculate the equilibrium constant for the new reaction

Raise the given equilibrium constant Kp to the power of the fraction by which the reaction has been divided: Kp_new = (Kp)^(1/2) = (3.5 × 10^4)^(1/2) = 187.082. Answer: Kp = 187.082 for reaction c at 1495 K.

Key Concepts

Kp calculation

In the world of chemistry, the equilibrium constant, often denoted as \(K_p\), plays a crucial role in defining the equilibrium state of gaseous reactions. At a given temperature, \(K_p\) tells us about the extent to which a reaction moves towards products or remains more reactants. The calculation of \(K_p\) can vary depending on the nature of the chemical equation being considered. For example, consider a reaction that is reversed or one that is multiplied by a coefficient:

• When a chemical reaction is reversed, the \(K_p\) value of the original equation is inversed. For instance, if the \(K_p\) is \(3.5 \times 10^4\), its reverse will be \(\frac{1}{3.5 \times 10^4}\).
• If the balanced chemical equation is multiplied by a factor, the \(K_p\) is raised to the power of that factor. This means dividing a reaction by 2 results in taking the square root of the original \(K_p\).

Through understanding these adjustments, calculating \(K_p\) under varying conditions becomes a straightforward process which is fundamental in analyzing chemical equilibria.

Chemical reactions

Chemical reactions are processes where substances, also known as reactants, transform into different substances, called products. Understanding the nature of these reactions and how they achieve equilibrium is foundational to chemistry.Reactions proceed forward or backward based on several factors, such as concentration, temperature, and pressure changes. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, establishing a dynamic balance.In gaseous reactions, equilibrium can be described by the equilibrium constant \(K_p\), which is expressed in terms of the partial pressures of the gases involved:

• Gaseous reactants and products can adjust their states to reach equilibrium.
• \(K_p\) helps predict which direction a reaction will proceed to achieve equilibrium based on initial pressures.

This understanding helps chemists control and optimize reactions for desired outcomes.

Reaction quotient

The reaction quotient, represented as \(Q\), is a tool chemists use to determine the direction in which a reaction will proceed to reach equilibrium. It is calculated in a similar way to \(K_p\), but it applies to non-equilibrium conditions.By comparing \(Q\) to \(K_p\), you can predict changes in the reaction:

• If \(Q < K_p\), the forward reaction will proceed, increasing the concentrations of the products.
• If \(Q > K_p\), the reverse reaction will dominate, converting products back into reactants.
• When \(Q = K_p\), the system is at equilibrium, and no net change occurs.

This makes \(Q\) an essential concept for understanding how a system reaches its equilibrium state, guiding adjustments in reaction conditions to drive a process favorably.