For the reaction; \(\mathrm{N}_{2 \mathrm{k} \mathrm{g}}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) At \(400 \mathrm{~K}, K_{\mathrm{p}}=41 \mathrm{~atm}^{-2}\). Find the value of \(K_{\mathrm{p}}\) for each of the following reactions at the same temperature: (i) \(2 \mathrm{NH}_{3(\mathrm{~g})} \Longrightarrow \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})}\) (ii) \({ }_{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{3}{2} \mathrm{H}_{2(g)} \Longrightarrow \mathrm{NH}_{3(\mathrm{~g})}\); (iii) \(2 \mathrm{~N}_{2(\mathrm{~g})}+6 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{~g})}\)

Short Answer

Expert verified
\( K_{p\text{(i)}} = \frac{1}{41} \, atm^2, \, K_{p\text{(ii)}} = \sqrt{\frac{1}{41}} \, atm, \, K_{p\text{(iii)}} = 41^2 \, atm^{-4}. \)

Step by step solution

01

Understanding the Given Reaction and Equilibrium Constant

The given reaction is \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) and the equilibrium constant for this reaction at 400 K is \( K_{p} = 41 \, atm^{-2} \). This information will be used to determine the equilibrium constants for the other given reactions.
02

Finding Kp for Reaction (i)

The first reaction (i) is the reverse of the given reaction, so the equilibrium constant \( K_{p} \) for reaction (i) is the reciprocal of the equilibrium constant of the given reaction. Therefore, \( K_{p\text{(i)}} = 1/K_{p} = 1/41 \, atm^2 \).
03

Finding Kp for Reaction (ii)

The second reaction (ii) is obtained by reversing the given reaction and then dividing it by 2. The equilibrium constant for a reaction that is reversed is the reciprocal of the original, and for a reaction that is multiplied or divided by a number, the exponent in the equilibrium constant changes by that number. Thus, \( K_{p\text{(ii)}} = (1/K_{p})^{1/2} = (1/41)^{1/2} \, atm \).
04

Finding Kp for Reaction (iii)

The third reaction (iii) is obtained by doubling the given reaction. The equilibrium constant for a reaction that is multiplied by 2 is raised to the power of 2. Therefore, \( K_{p\text{(iii)}} = (K_{p})^2 = (41)^2 \, atm^{-4} \).

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Key Concepts

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

Equilibrium Constant Kp
The equilibrium constant, represented as Kp, is a crucial concept in understanding the balance that exists in chemical reactions that occur in the gaseous phase. It helps us quantify the concentration of reactants and products at equilibrium. Kp is specifically used when dealing with partial pressures of gases.

For a general reaction of the form \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\), Kp is defined by the equation \[ K_p = \frac{(P_C)^c \cdot (P_D)^d}{(P_A)^a \cdot (P_B)^b} \] where \(P_X\) represents the partial pressure of the substance X, and the lowercase letters (a, b, c, d) are the stoichiometric coefficients from the balanced chemical equation. If the reaction is reversed, Kp becomes its reciprocal, and if the stoichiometry is changed, Kp is raised to the power corresponding to the change.

In the given exercise, to find the Kp for different variations of the reaction, we manipulated the original Kp according to these rules. For instance, we reversed the reaction in (i) yielding a Kp of \(1/41 \, atm^2\), reflecting the dynamic nature of chemical equilibrium.
Le Chatelier's Principle
Le Chatelier's principle is a fantastic tool for predicting how a system at equilibrium reacts to external changes. It states that if an external condition (such as concentration, temperature, or pressure) is altered, the system will adjust to minimize that change.

For example, if the pressure is increased by decreasing the volume of the container for a gaseous reaction at equilibrium, the system will shift in the direction that produces fewer moles of gas, thus lowering the pressure. Similarly, if the temperature is changed, the system shifts in a direction to absorb or release heat, depending on whether the temperature is increased or decreased.

This principle is valuable for understanding how changes can affect the position of equilibrium and is indirectly linked to the Kp by dictating how equilibrium shifts in response to changes, although it does not change the value of Kp itself.
Chemical Reaction Reversibility
Chemical reaction reversibility is an inherent aspect of many chemical processes where the reactants form products, which in turn can recreate the original reactants. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, meaning the concentration of reactants and products remains constant over time, though not necessarily equal.

It's important to understand that not all reactions are reversible in practical terms. A reversible reaction is one that can reach equilibrium, and it is these reactions we describe using the equilibrium constant Kp. This concept was demonstrated in the exercise when the direction of the equation was reversed, significantly affecting the Kp value.
Stoichiometry in Chemical Reactions
Stoichiometry in chemical reactions deals with the quantitative relationships between the reactants and products. The coefficients in a balanced equation tell us in what ratios the substances react and are formed.

In terms of equilibrium, the stoichiometric coefficients directly influence the equilibrium constant's exponents as highlighted in our calculations. Doubling the coefficients in reaction (iii) required us to square the original Kp value. Understanding the stoichiometry is essential for accurate calculations and predicting how the system responds to changes, a concept closely tied to Le Chatelier's Principle.

By comprehending stoichiometry, students can improve their understanding of chemical concepts significantly, enabling them to solve equilibrium problems and predict the outcome of chemical reactions more effectively.

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Most popular questions from this chapter

\(100 \mathrm{~g}\) of \(\mathrm{NaCl}\) is stirred in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) till the equilibrium is attained : (a) How much \(\mathrm{NaCl}\) goes into the solution and how much of it is left undissolved at equilibrium? The solubility of \(\mathrm{NaCl}\) at \(20^{\circ} \mathrm{C}\) is \(6.15\) mol/litre. (b) What will be the amount of \(\mathrm{NaCl}\) left undissolved if the solution is diluted to \(200 \mathrm{~mL} ?\)

A compound formed by elements \(A\) and \(B\) crystallizes in cubic structure where \(A\) atoms are at the corners of a cube and \(B\) atoms are at the face centre. What is the formula of the compound?

Calculate the concentration of \(\mathrm{CO}_{2}\) in a soft drink that is bottled with a partial pressure of \(\mathrm{CO}_{2}\) of \(4.0 \mathrm{~atm}\) over the liquid at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{CO}_{2}\) in water is \(3.1 \times 10^{-2} \mathrm{~mol} /\) litre-atm at this temperature.

State which one is homogeneous or heterogeneous? (a) \(\quad \overline{\text { C }_{\text {Diamond }}} \rightleftharpoons \bar{C}_{\text {graphite }}\) (b) \(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}=\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\) (c) \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) (d) \(\quad \mathrm{MgCO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{MgO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}\) (e) \(\mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}=\mathrm{PCl}_{5(\mathrm{~g})}\)

\(\Delta G^{\circ}\) for \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) is \(-16.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). Find out \(K_{\mathrm{p}}\) for the reaction. Also report \(K_{\mathrm{p}}\) and \(\Delta G^{\circ}\) for: $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \quad \text { at } 25^{\circ} \mathrm{C} $$

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