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Chapter 15: Problem 68

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00 -L flask. At \(648 \mathrm{~K}\), there is 0.0345 mole of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$

Short Answer

Expert verified
\(K_{c} = 0.0192\)

Step by step solution

01

Mole Calculations

Calculate the number of moles of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\). Using the molar mass of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) which is \(102.97 \mathrm{~g}/\mathrm{mol}\), the number of moles can be worked out as \(6.75 \mathrm{~g} / 102.97 \mathrm{~g}/\mathrm{mol} = 0.0656 \mathrm{~mol}\)
02

Determine the Initial and Equilibrium Amounts

At the start of the reaction, there are \(0.0656 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) and 0 of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\). At equilibrium, \(0.0345 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) is present. Hence the same quantity of \(\mathrm{SO}_{2}\mathrm{Cl}_{2}\) must have decomposed and the same amount of \(\mathrm{Cl}_{2}\) must have formed. So the mols at equilibrium are: \(\mathrm{SO}_{2}\mathrm{Cl}_{2} = 0.0656 \mathrm{~mol}-0.0345 \mathrm{~mol} = 0.0311 \mathrm{~mol}\), \(\mathrm{SO}_{2} = 0.0345 \mathrm{~mol}\), and \(\mathrm{Cl}_{2} = 0.0345 \mathrm{~mol}\)
03

Find the Equilibrium Concentrations

To convert the number of moles to concentration (in M), divide by the volume of the flask: \([\mathrm{SO}_{2}\mathrm{Cl}_{2}] = 0.0311 \mathrm{~mol} / 2.00 \mathrm{~L} = 0.01555 \mathrm{~M}\), \([\mathrm{SO}_{2}] = 0.0345 \mathrm{~mol} / 2.00 \mathrm{~L} = 0.01725 \mathrm{~M}\) and \([\mathrm{Cl}_{2}] = 0.0345 \mathrm{~mol} / 2.00 \mathrm{~L} = 0.01725 \mathrm{~M}\)
04

Calculate \(K_{c}\)

The equilibrium constant \(K_{c}\) for the reaction is given by the expression: \(K_{c} = [\mathrm{SO}_{2}] [\mathrm{Cl}_{2}] / [\mathrm{SO}_{2}\mathrm{Cl}_{2}] = (0.01725 \times 0.01725) / 0.01555 = 0.0192\)

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Most popular questions from this chapter

At \(1024^{\circ} \mathrm{C}\), the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}\) $$ 4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) $$ (a) What is the \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) decomposed if 0.16 mole of it is placed in a 2.0 - \(L\) flask at \(1024^{\circ} \mathrm{C}\). (c) What would be the fraction if a 1.0 -mole sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

The decomposition of ammonium hydrogen sulfide $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ is an endothermic process. A \(6.1589-\mathrm{g}\) sample of the solid is placed in an evacuated 4.000 - \(L\) vessel at exactly \(24^{\circ} \mathrm{C}\). After equilibrium has been established, the total pressure inside is \(0.709 \mathrm{~atm}\). Some solid \(\mathrm{NH}_{4} \mathrm{HS}\) remains in the vessel. (a) What is the \(K_{P}\) for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?

A quantity of 0.20 mole of carbon dioxide was heated at a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Under this condition, the average molar mass of the gases was found to be \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is the \(K_{P}\) for the equilibrium if the total pressure was 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

One mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\) are placed in a flask at \(397^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is found to be 0.21. The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}.\)

Briefly describe the importance of equilibrium in the study of chemical reactions.
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