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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. \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)\).

Short Answer

Expert verified
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 by step solution

01

Find the reaction relation with the given equation

Observe that this reaction is the reverse of the given reaction and divided by 2.
02

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).
03

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)
04

Find the reaction relation with the given equation

This reaction corresponds to the reverse of the given reaction.
05

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)
06

Find the reaction relation with the given equation

Observe that this reaction is the given reaction divided by 2.
07

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$.equilibrium constants \(\mathbf{r}_{\mathbf{p}}\).\(300^{\circ} \mathrm{C}, \quad 4.34 \times 10^{-3}\) \(500^{\circ} \mathrm{C}, \quad 1.45 \times 10^{-5}\) \(600^{\circ} \mathrm{C}, \quad 2.25 \times 10^{-6}\) Is the reaction exothermic or endothermic?

Suppose the reaction system $$\mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional UO \(_{2}(s)\) is added to the system. b. The reaction is performed in a glass reaction vessel; \(\mathrm{HF}(g)\) attacks and reacts with glass. c. Water vapor is removed.

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a 0.256 -mole sample of naphthalene has a mass of 32.8 g. What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene( \(s) \rightleftharpoons\) naphthalene \((g)$$$K=4.29 \times 10^{-6}(\text {at } 298 \mathrm{K})$$.If \)3.00 \mathrm{g}\( solid naphthalene is placed into an enclosed space with a volume of \)5.00 \mathrm{L}\( at \)25^{\circ} \mathrm{C},$ what percentage of the naphthalene will have sublimed once equilibrium has been established?

Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\text { Ni catalyst }}{=\frac{750^{\circ} \mathrm{C}}{7}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

For a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

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