Suppose that the gas-phase reactions A → B and B → A are both elementary reactions with rate constants of 4.7×10 −3 s −1 and 5.8×10 −1 s −1 , respectively. (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

Hello everyone today. We are being given the following problem and asked us to solve for it. So this is a hypothetical gas phase reaction. C to D has a forward rate constant of 4.22 times to the negative second in inverse seconds. The reverse reaction D two C has a rate constant of 1.74 times 10 to the negative fourth. The inverse of seconds. Which component C. R. D has a smaller partial pressure on equilibrium. So the reaction of C. T. D. Or the forward reaction can be denoted as K. C is equal to the constant of the forward reaction divided by the constant of the reverse reaction. And so using this equation we can plug in some numbers here. Our constant for the forward reaction is 4.2, 2 times 10 to the -7 inverse seconds Over the constant for the reverse reaction which is 1. times 10 to the negative 4th in verse seconds. And we plug that in, we get a value of 242.5. Now note that this is a gas phase reaction With the number of moles in equaling zero. Total. We have our K. P. Or our Constant for the pressure of 2 4, 2.5. So there are some rules in relation to the value of K. For example, when K is less than one, there are more reactant than products at equilibrium. When K equals one, the number of reactant and products are equal at equilibrium and when the value of K is greater than one, the there are going to be more products than reactive at equilibrium. And so we see here that our value of KP is 242, which is way greater than one. Therefore there must be more product then reacting at equilibrium. So I'm gonna see more product than the reactant at equilibrium. This would denote that c our reactant has a smaller partial pressure at equilibrium, since there will be more product form or deformed. So with that we have solved our problem. I hope this helped, and until next time.

Ch.15 - Chemical Equilibrium

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Brown 14th Edition

Ch.15 - Chemical Equilibrium

Problem 13b

Chapter 15, Problem 13b

Suppose that the gas-phase reactions A → B and B → A are both elementary reactions with rate constants of 4.7×10⁻³ s⁻¹ and 5.8×10⁻¹ s⁻¹, respectively. (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

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Identify the rate constants for the forward reaction (A → B) and the reverse reaction (B → A). The rate constant for A → B is 4.7×10−3 s−1 and for B → A is 5.8×10−1 s−1.

Understand that at equilibrium, the rate of the forward reaction (A → B) equals the rate of the reverse reaction (B → A). This can be expressed as kf[A] = kr[B], where kf and kr are the rate constants for the forward and reverse reactions, respectively, and [A] and [B] are the concentrations of A and B.

Rearrange the equilibrium expression to find the ratio of the concentrations of B to A at equilibrium: [B]/[A] = kf/kr.

Substitute the given rate constants into the expression to find the ratio: [B]/[A] = (4.7×10−3 s−1) / (5.8×10−1 s−1).

Compare the ratio [B]/[A] to determine which is greater at equilibrium. If [B]/[A] > 1, then [B] is greater; if [B]/[A] < 1, then [A] is greater.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant

The equilibrium constant (K) for a reaction at equilibrium is defined as the ratio of the concentrations (or partial pressures) of the products to the reactants, each raised to the power of their stoichiometric coefficients. For the reactions A ⇌ B, K can be calculated using the rate constants of the forward (k₁) and reverse (k₂) reactions: K = k₁/k₂. This constant helps determine the relative amounts of A and B at equilibrium.

Rate Constants

Rate constants (k) are specific to each reaction and indicate the speed at which a reaction proceeds. In this case, the rate constant for the forward reaction (A → B) is 4.7×10⁻³ s⁻¹, while for the reverse reaction (B → A) it is 5.8×10⁻¹ s⁻¹. The magnitude of these constants influences the equilibrium position, with larger constants favoring the formation of products or reactants.

Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is disturbed, the system will adjust to counteract the disturbance and restore a new equilibrium. In the context of the given reactions, knowing the rate constants allows us to predict which species will be favored at equilibrium, as the system will shift towards the side with the lower energy or higher stability based on the rate constants.

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The diagram shown here represents the equilibrium state for the reaction A2(𝑔) + 2B(𝑔) ⇌ 2AB(𝑔). (a) Assuming the volume is 2 L, calculate the equilibrium constant 𝐾𝑐 for the reaction.

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Write the expressions for 𝐾_𝑐 for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous.

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Suppose that the gas-phase reactions A → B and B → A are both elementary reactions with rate constants of 4.7×10−3 s−1 and 5.8×10−1 s−1, respectively. (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

Verified Solution

Key Concepts

Equilibrium Constant

Rate Constants

Le Chatelier's Principle

Suppose that the gas-phase reactions A → B and B → A are both elementary reactions with rate constants of 4.7×10⁻³ s⁻¹ and 5.8×10⁻¹ s⁻¹, respectively. (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?