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Chapter 14: Problem 91

Hydrogen bromide decomposes when heated according to the equation $$ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) $$ The equilibrium constant \(K_{c}\) equals \(1.6 \times 10^{-2}\) at \(200^{\circ} \mathrm{C}\). What are the moles of substances in the equilibrium mixture at \(200^{\circ} \mathrm{C}\) if we start with \(0.010 \mathrm{~mol} \mathrm{HBr}\) in a 1.0 - \(\mathrm{L}\) vessel?

Short Answer

Expert verified
0.00126 mol H2, 0.00126 mol Br2, 0.00748 mol HBr.

Step by step solution

01

Write the Equilibrium Expression

For the reaction \(2 \text{HBr}(g) \rightleftharpoons \text{H}_2(g) + \text{Br}_2(g)\), the equilibrium constant expression \(K_c\) is given by:\[K_c = \frac{{[\text{H}_2][\text{Br}_2]}}{{[\text{HBr}]^2}}\]Here, \([\text{H}_2]\), \([\text{Br}_2]\), and \([\text{HBr}]\) are the molar concentrations of \(\text{H}_2\), \(\text{Br}_2\), and \(\text{HBr}\) at equilibrium.
02

Initial Moles and Changes at Equilibrium

Since we start with \(0.010 \text{ mol}\) of \(\text{HBr}\) in a 1.0 L vessel, the initial concentration of \(\text{HBr}\) is \(0.010 \text{ M}\). Assuming \(x\) moles of \(\text{H}_2\) and \(\text{Br}_2\) form at equilibrium, then the moles of \(\text{HBr}\) that decompose are \(2x\), leading to the following concentrations at equilibrium:- \([\text{H}_2] = x\)- \([\text{Br}_2] = x\)- \([\text{HBr}] = 0.010 - 2x\).
03

Set Up the Equation Involving Kc

Substitute the equilibrium concentrations into the expression for \(K_c\):\[K_c = \frac{{x \cdot x}}{{(0.010 - 2x)^2}} = \frac{{x^2}}{{(0.010 - 2x)^2}}\]Given that \(K_c = 1.6 \times 10^{-2}\).
04

Solve for x

Solving \(\frac{{x^2}}{{(0.010 - 2x)^2}} = 1.6 \times 10^{-2}\) involves assuming small x (small change approximation), meaning \(2x << 0.010\). Then,\[x^2 = 1.6 \times 10^{-2} \times (0.010)^2\]\[x^2 = 1.6 \times 10^{-6}\]\[x = \sqrt{1.6 \times 10^{-6}}\]\[x \approx 0.00126\].
05

Calculate Equilibrium Moles

Finally, calculate the equilibrium moles for each substance:- Moles of \(\text{H}_2\) at equilibrium: \(0.00126 \text{ mol}\).- Moles of \(\text{Br}_2\) at equilibrium: \(0.00126 \text{ mol}\).- Moles of \(\text{HBr}\) = \(0.010 - 2\times 0.00126 = 0.00748 \text{ mol}\).

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Key Concepts

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

Equilibrium Constant
The concept of the equilibrium constant, denoted as \(K_c\), is central to understanding how reactions behave at equilibrium. It provides a numeric measure of the ratio of the concentration of products to reactants for a chemical reaction at a given temperature. For the decomposition of hydrogen bromide (\(2\,\mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g)\)), the equilibrium constant \(K_c\) is given as \(1.6 \times 10^{-2}\) at \(200^{\circ} \mathrm{C}\). The expression for \(K_c\) is established based on the balanced chemical equation:
  • The concentration of \(\text{HBr}\) appears squared in the denominator because it has a stoichiometric coefficient of 2 in the reaction.
  • The concentrations of \(\text{H}_2\) and \(\text{Br}_2\) in the numerator are each raised to the power of 1, as they each have a stoichiometric coefficient of 1.
This standardized expression helps predict how a change in conditions might shift the concentrations of reactants and products at equilibrium.
Reaction Stoichiometry
In chemical reactions, stoichiometry is the calculation of reactants and products in chemical reactions. It is based on the conservation of mass where the total mass of reactants equals the total mass of products. In this particular decomposition reaction, it is vital to understand the stoichiometric relationships.For the reaction \(2\,\mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g)\), the stoichiometry is:
  • 2 moles of \(\text{HBr}\) decompose to form 1 mole of \(\text{H}_2\) and 1 mole of \(\text{Br}_2\).
  • Every time the reaction moves forward, the amount of \(\text{HBr}\) decreases by 2 moles for every mole each of \(\text{H}_2\) and \(\text{Br}_2\) formed.
To predict equilibrium concentrations, we use these stoichiometric relations to establish expressions for the change in moles, which helps in setting up the equilibrium expressions.
Decomposition Reaction
A decomposition reaction is a type of chemical reaction where one reactant breaks down into two or more products. In this context, hydrogen bromide (\(\text{HBr}\)) decomposes into hydrogen gas (\(\text{H}_2\)) and bromine gas (\(\text{Br}_2\)).The equation for the decomposition of \(\text{HBr}\) is dynamic and reversible, noted by the double arrow (\(\rightleftharpoons\)). This means that while \(\text{HBr}\) breaks down to form \(\text{H}_2\) and \(\text{Br}_2\), these products can also recombine to form \(\text{HBr}\), depending on the conditions. This reversibility is crucial as it allows the system to reach equilibrium, where the rates of the forward and reverse reactions are equal.
Equilibrium Concentrations
Equilibrium concentrations refer to the concentrations of the reactants and products once a chemical reaction has reached equilibrium. At this point, the concentration of each species remains constant over time because the forward and reverse reactions occur at the same rate.In the decomposition of \(\text{HBr}\), we started with 0.010 moles of \(\text{HBr}\) in a 1-liter vessel. Due to the stoichiometry and equilibrium considerations, we assigned an "x" to represent the change in moles of \(\text{H}_2\) and \(\text{Br}_2\) at equilibrium:
  • At equilibrium, the moles of \(\text{H}_2\) and \(\text{Br}_2\) are each \(x\).
  • The moles of \(\text{HBr}\) are \(0.010 - 2x\).
To find \(x\), we solve the equation derived from \(K_c\), giving us a deeper understanding of how the system balances itself under given conditions.

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Most popular questions from this chapter

A container with a volume of \(1.500 \mathrm{~L}\) was evacuated and then filled at low temperature with \(0.0500 \mathrm{~mol}\) liquid dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\). As the temperature was raised, this substance formed nitrogen dioxide, \(\mathrm{NO}_{2}\) : $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ If the pressure of this reaction mixture at equilibrium at \(25.0^{\circ} \mathrm{C}\) was \(0.971 \mathrm{~atm},\) what was the total moles of gas in this mixture? What are the moles of \(\mathrm{NO}_{2}\) and moles of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) What are the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) What is \(K_{p}\) at \(25.0^{\circ} \mathrm{C} ?\)

The equilibrium constant \(K_{c}\) for the equation $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ is 1.84 at \(425^{\circ} \mathrm{C}\). The equilibrium constant \(K_{c}\) for the equation $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ is 49.7 at \(458^{\circ} \mathrm{C}\). Is this second equation endothermic or exothermic? How do you know? Describe the effect on this second equation of an increase in temperature. What would be the effect of an increase in pressure?

What would you expect to be the effect of an increase of pressure on each of the following reactions? Would the pressure change cause the reaction to go to the right or left? $$ \begin{array}{l} \text { a } \mathrm{CH}_{4}(g)+2 \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \\ \text { b } \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) \\ \text { c } \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) \end{array} $$

The equilibrium-constant expression for a reaction is $$ K_{c}=\frac{\left[\mathrm{NO}_{2}\right]^{4}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]^{2}} $$ What is the equilibrium-constant expression when the equation for this reaction is halved and then reversed?

Nitrogen monoxide, \(\mathrm{NO},\) is formed in automobile exhaust by the reaction of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) (from air). $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ The equilibrium constant \(K_{c}\) is 0.0025 at \(2127^{\circ} \mathrm{C}\). If an equilibrium mixture at this temperature contains \(0.023 \mathrm{~mol} \mathrm{~N}_{2}\) and \(0.031 \mathrm{~mol} \mathrm{O}_{2}\) per liter, what is the concentration of \(\mathrm{NO}\) ?
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