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Chapter 15: Problem 80

A \(0.831-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 - \(\mathrm{L}\) container and heated to \(1100 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) 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 \(1.300 \mathrm{~atm}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Short Answer

Expert verified
The equilibrium constants for the given reaction at 1100 K are \(K_c = 17.64\) and \(K_p = 15.60 \: atm^{-1}\).

Step by step solution

01

Determine the initial moles of SO₃

First, convert the given mass (0.831 g) of SO₃ into moles using the formula: moles = (mass)/(molar mass) The molar mass of SO₃ = S (32.06 g/mol) + 3 × O (16.00 g/mol) = 80.06 g/mol. Thus, moles of SO₃ = (0.831 g) / (80.06 g/mol) = 0.0104 mol.
02

Find initial concentrations

We are given that the container has a volume of 1.00 L. We can find the initial concentrations of the reactants and products as follows: [SO₃]₀ = moles of SO₃ / volume = 0.0104 mol / 1.00 L = 0.0104 M Initially, there are no products formed, so [SO₂]₀ = 0 and [O₂]₀ = 0.
03

Set up ICE (Initial, Change, Equilibrium) table

Set up an ICE table to keep track of the changes in concentrations as the reaction proceeds towards equilibrium. ``` [SO₃] [SO₂] [O₂] Initial 0.0104 M 0 0 Change -2x +2x +x Equilibrium 0.0104-2x 2x x ```
04

Use total pressure to find equilibrium concentrations

We are given that the total pressure at equilibrium is 1.300 atm. We can use the relationship between partial pressures and concentrations (PV=nRT) to find the partial pressures of the gases. Let P_SO₃, P_SO₂, and P_O₂ be the partial pressures of SO₃, SO₂, and O₂ at equilibrium, respectively. Then: P_SO₃ = x_SO₃ * R * T / V = (0.0104 - 2x) * (0.0821 L atm / K mol) * 1100 K / 1.00 L P_SO₂ = x_SO₂ * R * T / V = (2x) * (0.0821 L atm / K mol) * 1100 K / 1.00 L P_O₂ = x_O₂ * R * T / V = (x) * (0.0821 L atm / K mol) * 1100 K / 1.00 L Total pressure at equilibrium = P_SO₃ + P_SO₂ + P_O₂ = 1.300 atm (0.0104 - 2x) * 90.71 + 2x * 90.71 + x * 90.71 = 1.300
05

Solve for x

Solve the above equation to find the value of x. After solving, x = 0.0042
06

Calculate equilibrium concentrations

Substitute the value of x into the equilibrium concentrations from the ICE table. [SO₃] = 0.0104 - 2x = 0.0104 - 2 * 0.0042 = 0.0020 M [SO₂] = 2x = 2 * 0.0042 = 0.0084 M [O₂] = x = 0.0042 M
07

Calculate Kc

Calculate Kc using the equilibrium concentrations. Kc = ([SO₂]^2 * [O₂]) / ([SO₃]^2) = (0.0084^2 * 0.0042) / (0.0020^2) = 17.64
08

Calculate Kₚ

Since Kc and Kₚ are related through the equation Kₚ = Kc(RT)^(Δn), calculate Kₚ using Kc and the given temperature. Δn = (moles of gaseous products - moles of gaseous reactants) = (2 + 1) - (2) = 1 Kₚ = Kc(RT)^Δn = 17.64 * (0.0821 L atm / K mol * 1100 K)^1 = 15.60 atm^{-1} The equilibrium constants for the given reaction at 1100 K are Kc = 17.64 and Kₚ = 15.60 atm^{-1}.

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Most popular questions from this chapter

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\) If you start with \(25.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(500.0 \mathrm{~mL}\), what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) at equilibrium?

Water molecules in the atmosphere can form hydrogenbonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSEPR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at \(300 \mathrm{~K}\) and 0.020 at \(350 \mathrm{~K}\). Is water dimer formation endothermic or exothermic?

Consider the following equilibrium, for which \(K_{p}=0.0752\) at $$\begin{array}{l} 480{ }^{\circ} \mathrm{C}: \\ \quad 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \end{array}$$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathrm{c})\) What is the value of \(K_{c}\) for the reac- tion in part (b)?

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of \(\mathrm{A}\) or the partial pressure of \(\mathrm{B}\) ? Explain.

The equilibrium constant \(K_{c}\) for \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is 1.9 at \(1000 \mathrm{~K}\) and 0.133 at \(298 \mathrm{~K}\). (a) If excess \(\mathrm{C}\) is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a \(3.00-\mathrm{L}\) vessel at \(1000 \mathrm{~K}\), how many grams of \(\mathrm{CO}\) are produced? (b) How many grams of \(\mathrm{C}\) are consumed? \((\mathrm{c})\) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?
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