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For the reaction \(\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}\) the value of \(K_{c}\) at \(800^{\circ} \mathrm{C}\) is \(0.1 .\) What is the value of \(K_{p}\) at this temperature? (a) \(0.5\) (b) \(0.01\) (c) \(0.05\) (d) \(0.1\)

Short Answer

Expert verified
The value of \(K_p\) at \(800^\circ C\) is \(0.1\).

Step by step solution

01

- Understand the Relationship Between Kc and Kp

The relationship between the equilibrium constants Kc and Kp is given by the equation: \( K_p = K_c(RT)^{\Delta n} \), where \( R \) is the gas constant (in units that are consistent with the units of pressure), \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas between the products and reactants.
02

- Calculate the Change in Moles of Gas, Delta n

For the reaction \(\mathrm{N}_{2(g)} + \mathrm{O}_{2(g)} \rightleftharpoons 2\mathrm{NO}_{(g)}\), \( \Delta n \) is the difference in moles of gaseous products and reactants. This is calculated as \(\Delta n = 2 - (1 + 1) = 0\).
03

- Apply the Relationship between Kc and Kp

Since \( \Delta n = 0 \), the relationship simplifies to \( K_p = K_c(RT)^{0} = K_c \) because \((RT)^{0} = 1\). Therefore, \( K_p = K_c = 0.1 \).

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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a crucial concept in chemistry, which signifies a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of the reactants and products over time. It's vital to recognize that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their ratios remain constant.

At equilibrium, the system is said to be dynamically balanced since the conversion of reactants to products and vice versa continues, albeit at an identical rate. One way to express the equilibrium position quantitatively is through the equilibrium constant, known as the Kc value for reactions in solution. It's calculated by taking the concentration of the products raised to the power of their coefficients, divided by the concentration of the reactants raised to the power of their coefficients.
The Kc and Kp Relationship
Diving into the relationship between Kc and Kp, or the equilibrium constants for concentration and pressure, we come across an essential formula:
\[ K_p = K_c(RT)^{\Delta n} \]
where R is the ideal gas constant, T is the temperature in Kelvin, and \(\Delta n\) represents the change in moles of gaseous products minus gaseous reactants. This formula allows us to convert the equilibrium constant in terms of concentration (Kc) to the one in terms of pressure (Kp) when dealing with gaseous reactions.

In practical applications, you need to ensure that R is expressed in the correct units, often liters·atm/mol·K for this context, and T is the absolute temperature. The exponents in this relationship, given by \(\Delta n\), reflect the net change in the number of moles of gas and can drastically influence the relationship between Kc and Kp.
Applying Le Chatelier's Principle
Le Chatelier's principle is a straightforward yet powerful tool in predicting the effect of changes in conditions on a system at equilibrium. It states that if an external change is imposed on a system at equilibrium, the system will adjust itself to counteract that change and establish a new equilibrium.

There are several ways that a system can be stressed, including changes in concentration, temperature, and pressure. For instance, if the concentration of a reactant is increased, the system will shift towards producing more products to re-establish equilibrium. Similarly, changing the temperature will benefit either the endothermic or exothermic reaction direction, subsequently shifting the equilibrium position. For pressure changes, particularly for gases, increasing the pressure will generally favor the reaction direction that results in fewer moles of gas.

Understanding how these shifts occur helps chemists control the outcomes of reactions, making Le Chatelier's principle a fundamental concept for both theoretical and practical chemistry.

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Most popular questions from this chapter

If the value of equllibrium constant \(K_{r}\) for the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(7 .\) The equilibrium constant for the reaction \(2 \mathrm{~N}_{2}+6 \mathrm{H}_{2} \rightleftharpoons 4 \mathrm{NH}_{3}\) will be (a) 49 (b) 7 (c) 14 (d) 28

1 mole of \(\mathrm{NO}\) and 1 mole of \(\mathrm{O}_{3}\) are taken in a \(10 \mathrm{~L}\) vessel and heated. At equilibrium, \(50 \%\) of \(\mathrm{NO}\) (by mass) reacts with \(\mathrm{O}_{3}\) according to the equation: $$ \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} $$ What will be the equilibrium constant for this reaction? (a) 1 (b) 2 (c) 3 (d) 4

What is the minimum concentration of \(\mathrm{SO}_{4}^{2-}\) required to precipitate \(\mathrm{BaSO}_{4}\) in a solution containing \(1 \times 10^{-4}\) mole of \(\mathrm{Ba}^{2+} ?\left(K_{i p}\right.\) for \(\left.\mathrm{BaSO}_{4}=4 \times 10^{-10}\right)\) (a) \(4 \times 10^{-10} \mathrm{M}\) (b) \(2 \times 10^{-10} \mathrm{M}\) (c) \(4 \times 10^{-6} \mathrm{M}\) (d) \(2 \times 10^{-3} \mathrm{M}\)

Which of the following species can act both as an acid as well as a base? (a) \(\mathrm{SO}_{4}^{2-}\) (b) \(\mathrm{HSO}_{4}^{-}\) (c) \(\mathrm{PO}_{4}^{3-}\) (d) \(\mathrm{OH}\)

What will be the solubility of \(\mathrm{AgCl}\) in \(0.05 \mathrm{M} \mathrm{NaCl}\) aqueous solution if solubility product of \(\mathrm{AgCl}\) is \(1.5 \times 10^{-10} ?\) (a) \(3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(0.05 \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(3 \times 10^{9} \mathrm{~mol} \mathrm{~L}^{-1}\)

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