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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

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{ccccccc}\hline \text { Time (s) } & 0 & 30 & 45 & 60 & 75 & 90 \\\ P_{\mathrm{A}} \text { (atm) } & 0.500 & 0.390 & 0.360 & 0.340 & 0.325 & 0.325 \\\ P_{\text {B }} \text { (atm) } & 0.000 & 0.220 & 0.280 & 0.320 & 0.350 & 0.350 \\\\\hline\end{array}$$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after 45 s? After 90 s?

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?

Consider the statement "The equilibrium constant for a reaction at \(400 \mathrm{~K}\) is 792 . It must be a very fast reaction." What is wrong with the statement?

The reversible reaction between hydrogen chloride gas and one mole of oxygen gas produces steam and chlorine gas: $$4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=0.79$$ Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{HCl}}=P_{\mathrm{O}_{2}}=0.20 \mathrm{~atm}\) (b) \(P_{\mathrm{HCl}}=0.30 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.35 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.15 \mathrm{~atm}\)

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