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Chapter 15: Problem 36

A mixture of 1.374 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) and 70.31 \(\mathrm{g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00 -L vessel at 700 \(\mathrm{K}\) . These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain 0.566 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2},\) and \(\mathrm{HBr}\) . (b ) Calculate \(K_{c} .\)

Short Answer

Expert verified
The equilibrium concentrations of H₂, Br₂, and HBr are 0.140 mol/L, 0.020 mol/L, and 0.400 mol/L, respectively. The equilibrium constant Kc for the given reaction is approximately 57.14.

Step by step solution

01

Calculate the initial moles of H2 and Br2

To find the initial moles of H2 and Br2, we need to use their masses and molar masses. The molar mass of H2 is 2.02 g/mol, and for Br2, it's 159.8 g/mol. Moles of H2 = mass / molar mass = 1.374 g / 2.02 g/mol ≈ 0.680 mol Moles of Br2 = mass / molar mass = 70.31 g / 159.8 g/mol ≈ 0.440 mol
02

Calculate the moles of H2 and Br2 reacted, and the moles of HBr formed

To find the moles of H₂ and Br₂ reacted, we can subtract the moles of H₂ at equilibrium from the initial moles of H₂. Moles of H₂ reacted = Initial moles of H₂ - moles of H₂ at equilibrium Moles of H₂ reacted = 0.680 mol - (0.566 g / 2.02 g/mol) ≈ 0.680 mol - 0.280 mol = 0.400 mol Since the stoichiometry of the reaction shows that one mole of H₂ and one mole of Br₂ react to give two moles of HBr, the moles of Br₂ reacted and the moles of HBr formed can be determined. Moles of Br₂ reacted = 0.400 mol (same as moles of H₂ reacted) Moles of HBr formed = 2 * moles of H₂ reacted = 2 * 0.400 mol = 0.800 mol
03

Calculate the equilibrium moles of H2, Br2, and HBr

Now, we can find the equilibrium moles of all species. Moles of H₂ at equilibrium = 0.680 mol - 0.400 mol = 0.280 mol Moles of Br₂ at equilibrium = 0.440 mol - 0.400 mol = 0.040 mol Moles of HBr at equilibrium = 0 + 0.800 mol = 0.800 mol
04

Calculate the equilibrium concentrations of H2, Br2, and HBr

We can now find the equilibrium concentrations by dividing the moles by the volume of the vessel (2.00 L). [H₂] = 0.280 mol / 2.00 L = 0.140 mol/L [Br₂] = 0.040 mol / 2.00 L = 0.020 mol/L [HBr] = 0.800 mol / 2.00 L = 0.400 mol/L
05

Calculate the equilibrium constant Kc

Using the equilibrium concentrations, we can now calculate the equilibrium constant Kc using the formula: Kc = ([HBr]²) / ([H₂] * [Br₂]) Kc = (0.400 mol/L)² / (0.140 mol/L * 0.020 mol/L) ≈ 57.14 The equilibrium constant Kc for the given reaction is approximately 57.14.

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Most popular questions from this chapter

At \(800 \mathrm{K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{A}(g)\) is \(K_{c}=3.1 \times 10^{-4}\) . (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r}\) ? (b) If the value of \(k_{f}=0.27 \mathrm{s}^{-1},\) what is the value of \(k_{r}\) at 800 \(\mathrm{K} ?(\mathrm{c})\) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to \(1000 \mathrm{K},\) will the reverse rate constant \(k_{r}\) increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f} ?\)

A \(0.831-\) g sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 -L container and heated to 1100 \(\mathrm{K}\) . The SO \(_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the total pressure in the container is 1.300 atm. Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at 1100 \(\mathrm{K}\) .

Consider the following equilibrium for which \(\Delta H<0\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) O \(_{2}(g)\) is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) \(\mathrm{SO}_{3}(g)\) is removed from the system?

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? $$\begin{array}{ll}{\text { (a) } 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)} & {K_{p}=5.0 \times 10^{12}} \\\ {\text { (b) } 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)} & {K_{c}=5.8 \times 10^{-18}}\end{array}$$

Two different proteins \(X\) and \(Y\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\) . The proteins bind in a \(1 : 1\) ratio to form \(X Y . A\) solution that is initially 1.00 \(\mathrm{mM}\) in each protein is allowed to reach equilibrium. At equilibrium, 0.20 \(\mathrm{mM}\) of free \(\mathrm{X}\) and 0.20 \(\mathrm{mM}\) of free Y remain. What is \(K_{c}\) for the reaction?
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