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Chapter 15: Problem 26

A 2.50 -mol quantity of \(\mathrm{NOCl}\) was initially placed in a 1.50-L reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the NOCl had dissociated: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium constant \(K_{c}\) for the reaction.

Short Answer

Expert verified
The equilibrium constant (\(K_c\)) for the reaction is approximately 0.141

Step by step solution

01

Calculate initial concentration of NOCl

We start by calculating the initial concentration of NOCl in the reaction chamber. Concentration is defined as the amount of solute (in mol) divided by the volume of the solution (in L). Thus, the initial concentration of NOCl is given by \(\frac{2.50 \text{ mol}}{1.50 \text{ L}} \approx 1.67 \text{ M}\).
02

Calculate the change in concentration at equilibrium

Next, we calculate the change in concentration at equilibrium. We know that 28% of the initial NOCl dissociates. Therefore, at equilibrium, the change in concentration of NOCl is \(1.67 \text{ M} \times 0.28 = 0.467 \text{ M}\). According to the stoichiometry of the reaction, for every 2 mol of NOCl that dissociate, 2 mol of NO and 1 mol of Cl2 are formed. Therefore, at equilibrium, the concentration of NO increases by \(2 \times 0.467 \text{ M} = 0.934 \text{ M}\) and the concentration of Cl2 increases by \(\frac{1}{2} \times 0.467 \text{ M} = 0.2335 \text{ M}\).
03

Write the expression for \(K_c\) and substitute the equilibrium concentrations

The expression for the equilibrium constant \(K_c\) for the reaction is given by \[K_c = \frac{[NO]^2 \times [Cl_2]}{[NOCl]^2}\], where [NO], [Cl2], and [NOCl] are the concentrations of these substances at equilibrium. From step 2, we know these concentrations so we can substitute into the expression: \[K_c = \frac{(0.934)^2 \times 0.2335}{(1.67 - 0.467)^2}\].
04

Solve for \(K_c\)

Finally, we calculate the value of \(K_c\). Solving the equation we get \(K_c \approx 0.141\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium represents 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 a dynamic balance, not a static one, meaning that the reactants and products continuously convert into each other at the same rate.

For the equilibrium involving nitrogen monoxide chloride (NOCl), this implies that as NOCl decomposes into nitrogen monoxide (NO) and chlorine gas (Cl鈧�), some of the NO and Cl鈧� also recombine to form NOCl. Achieving this balance does not necessarily mean the reactants and products are present in equal concentrations, but that their concentrations have stabilized in a fixed ratio, characteristic of the particular reaction at a given temperature. This ratio is expressed by the equilibrium constant, symbolized as Kc.
Reaction Stoichiometry
Reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It is crucial for predicting how much product can be formed from a given amount of reactants or how much of one reactant is needed to react completely with another.

For the reaction involving NOCl dissociation, stoichiometry outlines that 2 moles of NOCl produce 2 moles of NO and 1 mole of Cl鈧�. This 2:2:1 ratio is imperative when calculating changes in concentration after the reaction reaches equilibrium. When it's given that 28% of NOCl dissociates, the stoichiometry helps us understand how this affects the amounts of NO and Cl鈧� formed. A clear grasp of stoichiometry ensures accurate predictions about concentration changes during reactions and is essential for the equilibrium constant calculation.
Equilibrium Concentration
Equilibrium concentration is the concentration of each reactant and product present when a chemical system is at equilibrium. Determining these concentrations allows chemists to calculate the equilibrium constant (Kc) for the reaction, which is a measure of the extent to which a chemical reaction occurs.

When the exercise states that 28% of the initial NOCl has dissociated, the equilibrium concentrations of NOCl, NO, and Cl鈧� are calculated based on this percentage. These concentrations are plugged into the Kc expression to solve for the equilibrium constant. Understanding the changes in equilibrium concentrations in response to the reaction progress and their substitution into the Kc expression is a key element in the equilibrium constant calculation. The proper calculation of equilibrium concentrations from initial conditions and the usage of stoichiometric ratios directly influences the accuracy of the determined Kc value.

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Most popular questions from this chapter

Consider the dissociation of iodine: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ A \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is heated at \(1200^{\circ} \mathrm{C}\) in a 500 -mL flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in problem \(15.65(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.

Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.

Write the expressions for the equilibrium constants \(K_{P}\) of these thermal decompositions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftharpoons 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\)

Consider the following reaction at \(1600^{\circ} \mathrm{C}\) : $$ \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) $$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 - \(\mathrm{L}\) flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.
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