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Chapter 15: Problem 91

Recall the formation of methanol from synthesis gas, the reversible reaction at the heart of a process with great potential for the future production of automotive fuels (page 663 ). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 14.5 \mathrm{at} 483 \mathrm{K} \end{aligned}$$ A particular synthesis gas consisting of 35.0 mole percent \(\mathrm{CO}(g)\) and 65.0 mole percent \(\mathrm{H}_{2}(\mathrm{g})\) at a total pressure of 100.0 atm at \(483 \mathrm{K}\) is allowed to come to equilibrium. Determine the partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture.

Short Answer

Expert verified
The partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture is given by the value of \(x\) from Step 4. The exact value will depend on the specific solution approach.

Step by step solution

01

Initial Pressures

Determine the initial pressures of the reactants. The partial pressure of a gas is given by multiplying the total pressure by the mole fraction of the gas. For \(\mathrm{CO}(g)\), this would be \(0.35 \times 100.0\:atm = 35.0\:atm\), and for \(\mathrm{H}_2(g)\), this would be \(0.65 \times 100.0\:atm = 65.0\:atm\). Remember that initially, there is no \(\mathrm{CH}_3\mathrm{OH}(g)\), so its pressure is zero.
02

Equilibrium Pressures

Write the expressions for the equilibrium pressures of all substances. Since the forward reaction consumes 1 mole of CO and 2 moles of H2 to produce 1 mole of CH3OH, the changes in pressures upon achieving equilibrium are as follows: \(\mathrm{CO}(g)\) changes by \(-x\), \(\mathrm{H}_2(g)\) changes by \(-2x\), and \(\mathrm{CH}_3\mathrm{OH}(g)\) changes by \(+x\). Therefore, the pressure of \(\mathrm{CO}(g)\) at equilibrium will be \(35.0 - x\), the pressure of \(\mathrm{H}_2(g)\) at equilibrium will be \(65.0 - 2x\). The pressure of \(\mathrm{CH}_3\mathrm{OH}(g)\) at equilibrium will simply be \(x\).
03

Equilibrium Constant Expression

Using the given equilibrium constant, write down the equilibrium expression. The equilibrium constant \(K_c\) is defined as the ratio of the product of the equilibrium concentrations (or pressures) of the products to that of the reactants, each raised to their respective stoichiometric coefficients. In this case, \(K_c = 14.5 = \frac{x}{(35.0-x)(65.0-2x)^2}\).
04

Solve for x

Solve for \(x\), the equilibrium pressure of \(\mathrm{CH}_3\mathrm{OH}(g)\). This will involve rearranging the equation from Step 3, which is then followed by solving for \(x\) using an appropriate mathematical technique such as root finding or numerical methods.

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Most popular questions from this chapter

A crystal of dinitrogen tetroxide (melting point, \(\left.-9.3^{\circ} \mathrm{C} ; \text { boiling point, } 21.3^{\circ} \mathrm{C}\right)\) is added to an equilibrium mixture of dintrogen tetroxide and nitrogen dioxide that is at \(20.0^{\circ} \mathrm{C} .\) Will the pressure of nitrogen dioxide increase, decrease, or remain the same? Explain.

The volume of the reaction vessel containing an equilibrium mixture in the reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is increased. When equilibrium is re-established, (a) the amount of \(\mathrm{Cl}_{2}\) will have increased; (b) the amount of \(\mathrm{SO}_{2}\) will have decreased; (c) the amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) will have remained the same; (d) the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will have increased.

The decomposition of \(\mathrm{HI}(\mathrm{g})\) is represented by the equation $$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})$$ \(\mathrm{HI}(\mathrm{g})\) is introduced into five identical \(400 \mathrm{cm}^{3}\) glass bulbs, and the five bulbs are maintained at \(623 \mathrm{K}\) Each bulb is opened after a period of time and analyzed for \(I_{2}\) by titration with \(0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq})\) $$\begin{array}{l} \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \end{array}$$ Data for this experiment are provided in the table below. What is the value of \(K_{\mathrm{c}}\) at \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

Write an equilibrium constant, \(K_{c},\) for the formation from its gaseous elements of \((a) 1\) mol \(\mathrm{HF}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g}) ;(\mathrm{c}) 2 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) ;(\mathrm{d}) 1 \mathrm{mol} \mathrm{ClF}_{3}(1)\)

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) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)
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