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Chapter 12: Problem 16

At \(627^{\circ} \mathrm{C}, K=0.76\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ Calculate \(K\) at \(627^{\circ} \mathrm{C}\) for (a) the synthesis of one mole of sulfur trioxide gas. (b) the decomposition of two moles of \(\mathrm{SO}_{3}\).

Short Answer

Expert verified
Question: Calculate the equilibrium constants for the following derivative reactions, given the original reaction and K value: Original reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K = 0.76 Derivative reactions: (a) SO₂(g) + 1/2O₂(g) ⇌ SO₃(g) (b) 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) Answer: (a) For the synthesis of one mole of sulfur trioxide, K = 0.87. (b) For the decomposition of two moles of sulfur trioxide, K = 1.32.

Step by step solution

01

(Step 1: Identify the derivative reactions)

(The derivative reactions are: (a) Synthesis of one mole of sulfur trioxide: $$\mathrm{SO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)$$ (b) Decomposition of two moles of sulfur trioxide: $$2\mathrm{SO}_{3}(g) \rightleftharpoons 2\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)$$)
02

(Step 2: Manipulate the original reaction to find the derivative reaction)

(Compare the derivative reactions with the original reaction to determine the relationship between them: (a) The synthesis of one mole of sulfur trioxide is half of the original reaction. (b) The decomposition of two moles of sulfur trioxide is the reverse of the original reaction with no change in stoichiometry.)
03

(Step 3: Apply the rules for manipulating equilibrium constants)

(Using the relationships identified in Step 2, apply the rules for manipulating equilibrium constants: (a) If a reaction is halved, the equilibrium constant should be squared: $$K_{1} =\sqrt{K}$$ (b) If a reaction is reversed, the equilibrium constant should be inverted: $$K_{2} = \frac{1}{K}$$)
04

(Step 4: Calculate the new equilibrium constants)

(Using the original K value of 0.76, calculate the new equilibrium constants: (a) Synthesis of one mole of sulfur trioxide: $$K_{1} =\sqrt{0.76} = 0.87$$ (b) Decomposition of two moles of sulfur trioxide: $$K_{2} = \frac{1}{0.76} = 1.32$$) The equilibrium constants for the two derivative reactions are: (a) For the synthesis of one mole of sulfur trioxide, K = 0.87. (b) For the decomposition of two moles of sulfur trioxide, K = 1.32.

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Most popular questions from this chapter

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{1}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{P_{\mathrm{IF}}}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{C}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\) (d) \(K=\frac{\left(P_{\mathrm{NO}}\right)^{2}\left(P_{\mathrm{H}_{3} \mathrm{O}}\right)^{4}\left[\mathrm{Cu}^{2+}\right]^{3}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}\)

Predict the direction in which each of the following equilibria will shift if the pressure on the system is increased by compression. (a) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{SbCl}_{3}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) (c) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\)

Solid ammonium iodide decomposes to ammonia and hydrogen iodide gases at sufficiently high temperatures. $$\mathrm{NH}_{4} \mathrm{I}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HI}(g)$$ The equilibrium constant for the decomposition at \(673 \mathrm{~K}\) is \(0.215\). Fifteen grams of ammonium iodide are sealed in a \(5.0\) -L flask and heated to \(673 \mathrm{~K}\). (a) What is the total pressure in the flask at equilibrium? (b) How much ammonium iodide decomposes?

Given the following data at \(25^{\circ} \mathrm{C}\), $$\begin{array}{cl}2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K=1 \times 10^{-30} \\ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K=8 \times 10^{1} \end{array}$$ calculate \(K\) for the formation of one mole of NOBr from its elements in the gaseous state at \(25^{\circ} \mathrm{C}\).

A 1.0-L reaction vessel at \(90^{\circ} \mathrm{C}\) contains \(8.00 \mathrm{~g}\) of sulfur, hydrogen, and hydrogen sulfide gases with partial pressures of \(6.0 \mathrm{~atm}\) and \(0.40 \mathrm{~atm}\), respectively, at equilibrium: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ (a) Calculate \(K\) for the reaction at equilibrium. (b) The mass of sulfur is increased to \(10.0\) grams. What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished? (c) The pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is increased to \(1.0 \mathrm{~atm}\). What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished?
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