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Chapter 12: Problem 44

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?

Short Answer

Expert verified
Answer: The total pressure in the flask at equilibrium is 12.4 atm, and 0.463 moles of ammonium iodide decompose at equilibrium.

Step by step solution

01

Write the balanced chemical equation and the expression for equilibrium constant

The balanced chemical equation for the decomposition of solid ammonium iodide is given as: $$\mathrm{NH}_{4} \mathrm{I}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{HI}(g)$$ The equilibrium constant expression for this reaction can be written as: $$K_c = \frac{\left[\mathrm{NH}_3\right]\left[\mathrm{HI}\right]}{\left[\mathrm{NH}_4 \mathrm{I}\right]}$$ Since the ammonium iodide is a solid, its concentration does not affect the equilibrium constant. We simplify the equilibrium constant expression: $$K_c = \left[\mathrm{NH}_3\right]\left[\mathrm{HI}\right]$$
02

Calculate the initial moles of ammonium iodide

We have 15g of ammonium iodide and need to find the initial moles. The molar mass of ammonium iodide is: $$\mathrm{(1\times14.01)+(1\times1.01)+(1\times126.9)\ \text{g/mol}}=142.92\ \text{g/mol}$$ Find the moles: $$\text{moles of}\ \mathrm{NH_{4}I} = \frac{15\ \text{g}}{142.92\ \text{g/mol}}=0.105\ \text{mol}$$
03

Set up the ICE (Initial, Change, Equilibrium) table

Below is our ICE table representing the initial moles of components, the change of moles at equilibrium and the equilibrium moles of components. NH4I(s) <=> NH3(g) + HI(g) Initial 0.105 0 0 Change -x +x +x Final 0.105-x x x
04

Calculate the molar equilibrium concentrations

We can use the simplified equilibrium constant expression and the equilibrium concentrations calculated from our ICE table to find x: $$K_c = \left[\mathrm{NH}_3\right] \left[\mathrm{HI}\right] = x^2$$ Plug in the value of Kc: $$0.215=x^2$$ Solve for x: $$x=\sqrt{0.215}=0.463$$
05

Calculate the total pressure at equilibrium

At equilibrium, there are x moles of \(\mathrm{NH}_3\) and \(\mathrm{HI}\). We can use the ideal gas law (PV=nRT) to calculate the total pressure at equilibrium: $$P_{total} = \frac{(n_{NH_{3}}+n_{HI})RT}{V}$$ Plug in the values for moles, temperature, and volume: $$P_{total} = \frac{(0.463+0.463)(0.0821)(673)}{5.0}$$ Calculate the total pressure: $$P_{total} = 12.4\ \text{atm}$$ (a): The total pressure in the flask at equilibrium is 12.4 atm.
06

Calculate the amount of ammonium iodide that decomposes

The change in moles of ammonium iodide, x, represents the amount of ammonium iodide that decomposes. Therefore: $$\text{Amount of NH}_{4}\text{I decomposed (mol)} = x = 0.463\ \text{mol}$$ (b): 0.463 moles of ammonium iodide decomposes at equilibrium.

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Most popular questions from this chapter

Consider the decomposition of ammonium hydrogen sulfide: $$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ In a sealed flask at \(25^{\circ} \mathrm{C}\) are \(10.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{HS}\), ammonia with a partial pressure of \(0.692 \mathrm{~atm}\), and \(\mathrm{H}_{2} \mathrm{~S}\) with a partial pressure of \(0.0532 \mathrm{~atm}\). When equilibrium is established, it is found that the partial pressure of ammonia has increased by 12.4\%. Calculate \(K\) for the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) at \(25^{\circ} \mathrm{C}\).

Consider the following reaction at \(250^{\circ} \mathrm{C}\) : $$\mathrm{A}(s)+2 \mathrm{~B}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{D}(g)$$ (a) Write an equilibrium constant expression for the reaction. Call the equilibrium constant \(K_{1}\). (b) Write an equilibrium constant expression for the formation of one mole of \(\mathrm{B}(\mathrm{g})\) and call the equilibrium constant \(K_{2}\). (c) Relate \(K_{1}\) and \(K_{2}\).

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 \(25^{\circ} \mathrm{C}\).

For the system $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ at \(1000 \mathrm{~K}, K=0.45 .\) Sulfur trioxide, originally at \(1.00 \mathrm{~atm}\) pressure, partially dissociates to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) at \(1000 \mathrm{~K}\). What is its partial pressure at equilibrium?

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(g)$$ When equilibrium is established, the partial pressures of the gases are: \(P_{\mathrm{N}_{2}}=\) \(1.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.80 \mathrm{~atm}, P_{\mathrm{NO}}=0.022 \mathrm{~atm} .\) (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure \(1.2\) atm. Calculate the partial pressure of all gases when equilibrium is reestablished.
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