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Given the following descriptions of reversible reactions, write a balanced equation (smallest whole-number coefficients) and the equilibrium constant expression for each. (a) Nickel metal reacts with carbon monoxide to form nickel tetracarbonyl \(\left(\mathrm{Ni}(\mathrm{CO})_{4}\right)\) gas. (b) Aqueous nitrous acid in equilibrium with hydrogen and nitrite ions. (c) Chlorine gas and bromide ions in equilibrium with liquid bromine and chloride ions.

Short Answer

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Now, let's summarize the step by step solution. #Summary# The balanced chemical equations and the respective equilibrium constant expressions for the given descriptions are as follows: (a) Nickel metal with carbon monoxide: $$\mathrm{Ni(s)} + 4\mathrm{CO(g)} \rightleftharpoons \mathrm{Ni(CO)_4(g)}$$ Equilibrium constant expression: $$K = \frac{[\mathrm{Ni(CO)_4}]}{[\mathrm{Ni}][\mathrm{CO}]^4}$$ (b) Aqueous nitrous acid in equilibrium with hydrogen and nitrite ions: $$\mathrm{HNO_2(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{NO_2^-(aq)}$$ Equilibrium constant expression: $$K = \frac{[\mathrm{H^+}][\mathrm{NO_2^-}]}{[\mathrm{HNO_2}]}$$ (c) Chlorine gas and bromide ions in equilibrium with liquid bromine and chloride ions: $$\mathrm{Cl_2(g)} + 2\mathrm{Br^-(aq)} \rightleftharpoons \mathrm{2Cl^-(aq)} + \mathrm{Br_2(l)}$$ Equilibrium constant expression: $$K = \frac{[\mathrm{Cl^-}]^2[\mathrm{Br_2}]}{[\mathrm{Cl_2}][\mathrm{Br^-}]^2}$$.

Step by step solution

01

(a) Balanced equation for nickel metal with carbon monoxide

The reaction can be represented as follows: $$\mathrm{Ni(s)} + 4\mathrm{CO(g)} \rightleftharpoons \mathrm{Ni(CO)_4(g)}$$ Now, let's write the equilibrium constant expression.
02

(a) Equilibrium constant expression for nickel metal with carbon monoxide

The equilibrium constant expression can be written as: $$K = \frac{[\mathrm{Ni(CO)_4}]}{[\mathrm{Ni}][\mathrm{CO}]^4}$$
03

(b) Balanced equation for aqueous nitrous acid in equilibrium with hydrogen and nitrite ions

The reaction for aqueous nitrous acid can be represented as: $$\mathrm{HNO_2(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{NO_2^-(aq)}$$ Now, let's write the equilibrium constant expression.
04

(b) Equilibrium constant expression for aqueous nitrous acid

The equilibrium constant expression can be written as: $$K = \frac{[\mathrm{H^+}][\mathrm{NO_2^-}]}{[\mathrm{HNO_2}]}$$
05

(c) Balanced equation for chlorine gas and bromide ions in equilibrium with liquid bromine and chloride ions

The reaction can be represented as: $$\mathrm{Cl_2(g)} + 2\mathrm{Br^-(aq)} \rightleftharpoons \mathrm{2Cl^-(aq)} + \mathrm{Br_2(l)}$$ Now, let's write the equilibrium constant expression.
06

(c) Equilibrium constant expression for chlorine gas and bromide ions

The equilibrium constant expression can be written as: $$K = \frac{[\mathrm{Cl^-}]^2[\mathrm{Br_2}]}{[\mathrm{Cl_2}][\mathrm{Br^-}]^2}$$

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial for solving problems related to reversible reactions. It refers to a state in a reversible chemical process where the rate of the forward reaction equals the rate of the backward reaction. As a result, the concentrations of reactants and products remain constant over time, though not necessarily equal.

When writing equations for reactions at equilibrium, we use the double arrow symbol \rightleftharpoons to indicate that the reaction can proceed in both directions. The equilibrium constant expression, represented by K, quantifies the ratio of concentrations of products to reactants, each raised to the power of their stoichiometric coefficients from the balanced equation.

In our example of nickel metal reacting with carbon monoxide, the equilibrium constant expression is: $$K = \frac{[\mathrm{Ni(CO)_4}]}{[\mathrm{Ni}][\mathrm{CO}]^4}$$This expression helps predict the position of equilibrium and the extent to which reactants are converted to products.
Reversible Reactions
The concept of reversible reactions is an extension of the equilibrium principle and illustrates that many chemical reactions can go both ways, toward products or reactants. These reactions reach a point where their forward and reverse reactions occur at equal rates. This dynamic state is different from completion reactions that go to completion, where reactants are fully converted to products, and no backward reaction occurs.

An example we see here is the equilibrium involving aqueous nitrous acid: $$\mathrm{HNO_2(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{NO_2^-(aq)}$$In a reversible reaction, altering conditions such as concentration, temperature, or pressure can 'push' the equilibrium position favouring either the production of reactants or products, as predicted by Le Chatelier's principle.
Balanced Chemical Equations
Balanced chemical equations are vital for accurately representing chemical reactions. For equations to be balanced, they must show the same number of atoms for each element involved in the reaction on both sides of the equation. This is because the Law of Conservation of Mass dictates that matter cannot be created or destroyed in a chemical reaction.

For instance, the balanced equation for chlorine gas reacting with bromide ions is:$$\mathrm{Cl_2(g)} + 2\mathrm{Br^-(aq)} \rightleftharpoons 2\mathrm{Cl^-(aq)} + \mathrm{Br_2(l)}$$Balancing equations is also essential for determining the stoichiometry of the reaction which is necessary for the equilibrium constant expression.
Reaction Stoichiometry
Reaction stoichiometry delves into the quantitative relationship between reactants and products in a chemical reaction. This relationship is reflected in the balanced chemical equation, where stoichiometric coefficients indicate the proportion of moles of each substance involved. For instance, the coefficient '2' before \(\mathrm{Br^-}\) in our chlorine gas example indicates that two moles of \(\mathrm{Br^-}\) are required for every mole of \(\mathrm{Cl_2}\) reacting.

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Most popular questions from this chapter

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$ \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is 0.45. If \(20.0 \mathrm{~mL}(d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00-\mathrm{L}\) vessel and heated to \(180^{\circ} \mathrm{C},\) what percentage remains undissociated at equilibrium?

The following data are for the system $$ \mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ $$ \begin{array}{lcccccc} \hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}}(\text { atm }) & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{\text {B }} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\ \hline \end{array} $$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after \(30 \mathrm{~s}\) ? After \(90 \mathrm{~s}\) ?

The following data are for the system $$ \mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ $$ \begin{array}{lcccccc} \hline \text { Time (s) } & 0 & 30 & 45 & 60 & 75 & 90 \\ P_{\mathrm{A}} \text { (atm) } & 0.500 & 0.390 & 0.360 & 0.340 & 0.325 & 0.325 \\\ P_{\text {B }} \text { (atm) } & 0.000 & 0.220 & 0.280 & 0.320 & 0.350 & 0.350 \\\ \hline \end{array} $$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after \(45 \mathrm{~s}\) ? After \(90 \mathrm{~s}\) ?

A sealed flask has \(0.541 \mathrm{~atm}\) of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\). The following equilibrium is established. $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ At equilibrium, the partial pressure of oxygen is measured to be 0.216 atm. Calculate \(K\) for the decomposition of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\)

Consider the following reaction at \(75^{\circ} \mathrm{C}\) : $$ 3 \mathrm{R}(s)+2 \mathrm{Q}(g) \rightleftharpoons \mathrm{A}(g)+5 \mathrm{~B}(l) \quad K=9.4 $$ A \(10.0-\mathrm{L}\) sample contains \(0.30 \mathrm{~mol}\) of \(\mathrm{R}\) and \(\mathrm{Q}\) and \(0.50 \mathrm{~mol}\) of \(\mathrm{A}\) and \(\mathrm{B}\). In which direction will the reaction proceed?

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