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Chapter 15: Problem 18

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.06 \mathrm{g} \mathrm{H}_{2} \mathrm{S}\) in a 0.500 Lflask comes to equilibrium at \(1670 \mathrm{K}: 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{S}_{2}(\mathrm{g})\) found is \(8.00 \times 10^{-6}\) mol. Determine the value of \(K_{p}\) at 1670 K.

Short Answer

Expert verified
By converting mass of gases into moles, then determining their pressures using the ideal gas law, and finally including those pressures in the expression for the equilibrium constant Kp, the value of Kp at 1670 K is found.

Step by step solution

01

Calculate moles of reactants

The first step is to calculate the moles of H2 and H2S, using their respective molar masses. For H2, it's 2 g/mol, so there are 1 g / 2 g/mol = 0.5 moles of H2. For H2S, it's 34.08 g/mol, so there are 1.06 g / 34.08 g/mol = 0.0311 moles of H2S.
02

Chemical Equations

Using the stoichiometry of the balanced chemical reaction, \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\)we can see that 2 moles of hydrogen reacts with 1 mole of sulfur to form 2 moles of hydrogen sulfide.However, we were given that the equilibrium amount of S2 is \(8.00 \times 10^{-6}\)mol ,which would have reacted with \(2*8.00 \times 10^{-6}\)mol= \(1.6 \times 10^{-5} \)mol of H2. Since the initial moles of H2 and H2S were 0.5 mol and 0.0311 mol respectively, at equilibrium, moles of H2 will be 0.5-\(1.6 \times 10^{-5} \)mol=0.499984 mol and moles of H2S will be 0.0311 +\(1.6 \times 10^{-5} \)=0.031116 mol.
03

Ideal Gas Law

Use the Ideal Gas Law PV=nRT to calculate pressures at equilibrium. Remember that the pressure for an individual gas is the partial pressure. Rearranging the formula for pressure we get P=nRT/V. For \( R=0.0821 L atm / K mol \), \( T=1670 K \), and \( V=0.500L \), calculate the pressure for each gas.
04

Equilibrium Constant Kp

Plug the pressures obtained in step 3 into the equilibrium constant Kp expression which from the balanced chemical reaction is : \(Kp=(P_{H_{2}S})^2/(P_{H_{2}})^2*P_{S_{2}}\) . This will give the equilibrium constant Kp at 1670 K.

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Most popular questions from this chapter

An important environmental and physiological reaction is the formation of carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) from carbon dioxide and water. Write the equilibrium constant expression for this reaction in terms of activities. Convert that expression into an equilibrium constant expression containing concentrations and pressures.

What effect does increasing the volume of the system have on the equilibrium condition in each of the following reactions? (a) \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

A mixture of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) and \(\mathrm{CH}_{4}(\mathrm{g})\) in the mole ratio 2: 1 was brought to equilibrium at \(700^{\circ} \mathrm{C}\) and a total pressure of 1 atm. On analysis, the equilibrium mixture was found to contain \(9.54 \times 10^{-3} \mathrm{mol} \mathrm{H}_{2} \mathrm{S} .\) The \(\mathrm{CS}_{2}\) pre- sent at equilibrium was converted successively to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and then to \(\mathrm{BaSO}_{4} ; 1.42 \times 10^{-3} \mathrm{mol} \mathrm{BaSO}_{4}\) was obtained. Use these data to determine \(K_{\mathrm{p}}\) at \(700^{\circ} \mathrm{C}\) for the reaction $$\begin{aligned} 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+& 4 \mathrm{H}_{2}(\mathrm{g}) \\\ & K_{\mathrm{p}} \text { at } 700^{\circ} \mathrm{C}=? \end{aligned}$$

A sample of \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s})\) is placed in a \(2.58 \mathrm{L}\) flask containing 0.100 mol \(\mathrm{NH}_{3}(\mathrm{g}) .\) What will be the total gas pressure when equilibrium is established at \(25^{\circ} \mathrm{C} ?\) $$\begin{aligned} \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) & \\ K_{\mathrm{p}} &=0.108 \text { at } 25^{\circ} \mathrm{C} \end{aligned}$$

What is the apparent molar mass of the gaseous mixture that results when \(\mathrm{COCl}_{2}(\mathrm{g})\) is allowed to dissociate at \(395^{\circ} \mathrm{C}\) and a total pressure of 3.00 atm? $$\begin{aligned} \operatorname{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+& \mathrm{Cl}_{2}(\mathrm{g}) \\ & K_{\mathrm{p}}=4.44 \times 10^{-2} \mathrm{at} 395^{\circ} \mathrm{C} \end{aligned}$$ Think of the apparent molar mass as the molar mass of a hypothetical single gas that is equivalent to the gaseous mixture.
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