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If \(K_{c}=0.042\) for \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g)\) at 500 \(\mathrm{K}\) , what is the value of \(K_{p}\) for this reaction at this temperature?

Short Answer

Expert verified
The value of Kp for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) at 500 K can be calculated using the relationship Kp = Kc(RT)^(Δn), where Kc = 0.042, R = 0.0821 L atm/mol K, T = 500 K, and Δn = -1 (calculated as Δn = moles of products - moles of reactants). Plugging in these values, we obtain Kp ≈ 0.001023.

Step by step solution

01

Calculate Δn

The balanced chemical equation is given as: PCl3(g) + Cl2(g) ⇌ PCl5(g) The change in moles of gas can be calculated as follows: Δn = moles of products - moles of reactants Δn = 1 (from PCl5) - (1 + 1) (from PCl3 and Cl2) Δn = 1 - 2 = -1
02

Use the Kc to Kp relationship

Now that we have Δn, we can use the relationship between Kc and Kp: Kp = Kc(RT)^(Δn) We were given the value Kc = 0.042, the temperature T = 500 K, and we calculated Δn = -1. The gas constant, R, is 0.0821 L atm/mol K. Plugging in these values, we get: Kp = (0.042)(0.0821 * 500)^(-1) Kp = 0.042 / (41.05) Kp = 0.001023
03

Write the final answer

The value of Kp for this reaction at 500 K is approximately 0.001023.

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Key Concepts

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

Kp and Kc Relationship
When studying chemical reactions that involve gases, understanding the relationship between the equilibrium constants Kp and Kc is crucial. Kp represents the equilibrium constant in terms of partial pressures of the gases involved, while Kc is expressed in terms of molar concentrations. These constants are related by the equation:

\[ K_{p} = K_{c}(RT)^{\Delta n} \]
Here, \(R\) is the ideal gas constant (0.0821 L atm/mol K), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas going from reactants to products. It's important to note that \(\Delta n\) is calculated by subtracting the sum of the moles of gaseous reactants from the sum of the moles of gaseous products. A negative \(\Delta n\) implies that there are fewer moles of gas in the products than reactants, leading to a decrease in pressure at equilibrium.
This equation applies to ideal gases and allows us to convert between Kp and Kc when the reaction involves changes in the number of moles of gases. Shorter, simpler sentences, such as 'Kp is used for pressure; Kc is used for concentration,' or 'Kp and Kc are connected by a simple formula involving the gas constant R and the temperature T,' can help solidify the concept for students.
Chemical Equilibrium
Whether you're looking at a color change in a solution or the production of a gas in a lab, chemical equilibrium plays a key role in these reactions. At chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and concentrations of reactants and products remain constant over time – they do not necessarily have to be equal, but they stop changing. This state can be quantified using the equilibrium constant, either Kc or Kp depending on the state of the reactants and products.

For the reaction \( \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g) \), Kc is given as 0.042 at 500 K, meaning that at this temperature, the ratio of the concentrations of the products to reactants is low, favoring the reactants slightly at equilibrium. Understanding this helps students predict reaction behavior such as which side is favored under certain conditions and how the reaction might respond to changes in conditions.
Le Chatelier's Principle
Le Chatelier's principle is like a guide to predicting how a system at equilibrium will respond to changes in concentration, temperature, or pressure. Simply put, if you change the conditions of a reaction at equilibrium, the system will adjust to counteract that change.

For instance, if you increase the concentration of reactants for the reaction \( \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g) \), the system will try to consume those extra reactants by making more products. This is the system's way of restoring equilibrium. Similarly, changes in temperature can shift the position of equilibrium. If the reaction is exothermic (releases heat), increasing the temperature will cause the system to favor the reverse reaction to absorb the excess heat.
A practical tip for students: Le Chatelier’s principle comes in handy for predicting shifts in reaction equilibria and can guide you when adjusting reaction conditions to obtain more products or reactants as needed.

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Most popular questions from this chapter

When 2.00 \(\mathrm{mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00 -L flask at 303 \(\mathrm{K}, 56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2} :\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at 303 \(\mathrm{K}\) . (c) According to Le Chatelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture were transferred to a \(15.00-\mathrm{L}\) . vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00-\mathrm{L}\) vessel at 303 \(\mathrm{K}\) .

For the equilibrium $$\mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g)$$ \(K_{p}=0.052\) at \(60^{\circ} \mathrm{C}\) (a) Calculate \(K_{C}\) (b) After 3.00 \(\mathrm{g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed 1.500 -L. vessel at \(60^{\circ} \mathrm{C}\) , the vessel is charged with 0.0500 \(\mathrm{g}\) of \(\mathrm{BCl}_{3}(g) .\) What is the equilibrium concentration of \(\mathrm{PH}_{3} ?\)

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

(a) If \(Q_{c} < K_{O}\) in which direction will a reaction proceed in order to reach equilibrium? (b) What condition must be satisfied so that \(Q_{c}=K_{c} ?\)

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2} :\) $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use thermochemical data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

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