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Chapter 15: Problem 94

In one of Fritz Haber's experiments to establish the conditions required for the ammonia synthesis reaction, pure \(\mathrm{NH}_{3}(\mathrm{g})\) was passed over an iron catalyst at \(901^{\circ} \mathrm{C}\) and 30.0 atm. The gas leaving the reactor was bubbled through 20.00 mL of a HCl(aq) solution. In this way, the \(\mathrm{NH}_{3}(\mathrm{g})\) present was removed by reaction with HCl. The remaining gas occupied a volume of 1.82 L at STP. The \(20.00 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) through which the gas had been bubbled required \(15.42 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Another \(20.00 \mathrm{mL}\) sample of the same HCl(aq) through which no gas had been bubbled required \(18.72 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Use these data to obtain a value of \(K_{\mathrm{p}}\) at \(901^{\circ} \mathrm{C}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})\)

Short Answer

Expert verified
After performing all the necessary calculations, the value of the equilibrium constant (Kp) for the reaction of nitrogen and hydrogen to produce ammonia at 901 degrees Celsius can be obtained.

Step by step solution

01

Calculate the Amount of Reacted Ammonia

First, determine the difference in titration volumes of HCI solution with and without gas bubbling through it. Subtract the volume of KOH solution needed to titrate the HCI when gas was bubbled through from when no gas was bubbled through (18.72 ml - 15.42 ml = 3.3 ml). This difference is due to the reaction of ammonia with HCI. When converted to liters (0.0033 L) and multiplied by the molarity of the KOH (0.0523 mol/L), this gives the amount of reacted ammonia in moles (0.0033 L * 0.0523 mol/L = 0.00017259 mol).
02

Calculate the Pressure of Gases

Next, calculate the partial pressures of nitrogen, hydrogen, and ammonia. The total pressure of the gaseous reaction mixture after ammonia was removed is given as 30 atm. According to the reaction stoichiometry, the molar ratio of nitrogen to hydrogen to ammonia is 1:3:2. So for every mole of nitrogen, there is 3 moles of hydrogen while ammonia is produced as 2 moles. The moles of each gas can be calculated from the measured volume of gas, which is 1.82 L at STP. Using the ideal gas law, at STP 1 mole of any gas has a volume of 22.4 L, hence the quantity of gas left after removal of the ammonia can be calculated (1.82 L / 22.4 L = 0.08125 mol). The pressures of nitrogen, hydrogen, and ammonia can be calculated by multiplying the mole fraction of each (1/4, 3/4, and 0) in the equilibrium reaction mixture times the total pressure (30 atm).
03

Calculate the value of Kp

The final step is to calculate the equilibrium constant, Kp, using the calculated pressures. The expression for the equilibrium constant for the reaction is: Kp = [NH3]^2 / ([N2] * [H2]^3). Given the mole fractions, the pressure of each should be inserted into the equation (Kp = (2 * 0.00017259 * 30)^2 / ((1/4 * 30) * (3/4 * 30)^3). Proceding with these calculation, the value of Kp at 901 degrees Celsius will be found.

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Most popular questions from this chapter

In the reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}), K=1.0 \times\) \(10^{5}\) at \(25^{\circ} \mathrm{C} .\) Would you expect a greater amount of product or reactant?

For the reaction $$ \mathrm{A}(\mathrm{s}) \rightleftharpoons \mathrm{B}(\mathrm{s})+2 \mathrm{C}(\mathrm{g})+\frac{1}{2} \mathrm{D}(\mathrm{g}) \quad \Delta H^{\circ}=0 $$ (a) Will \(K_{p}\) increase, decrease, or remain constant with temperature? Explain. (b) If a constant-volume mixture at equilibrium at 298 K is heated to 400 K and equilibrium re-established, will the number of moles of \(\mathrm{D}(\mathrm{g})\) increase, decrease, or remain constant? Explain.

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

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 \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

The formation of nitrosyl chloride is given by the following equation: \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) \(K_{\mathrm{c}}=4.6 \times 10^{4}\) at \(298 \mathrm{K} .\) In a \(1.50 \mathrm{L}\) flask, there are \(4.125 \mathrm{mol}\) of \(\mathrm{NOCl}\) and \(0.1125 \mathrm{mol}\) of \(\mathrm{Cl}_{2}\) present at equilibrium (298 K). (a) Determine the partial pressure of \(\mathrm{NO}\) at equilibrium. (b) What is the total pressure of the system at equilibrium?
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