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Chapter 15: Problem 92

The vapor pressure of mercury is \(0.0020 \mathrm{mmHg}\) at \(26^{\circ} \mathrm{C}\). (a) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the process \(\mathrm{Hg}(l) \rightleftharpoons \mathrm{Hg}(g) .\) (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring \(6.1 \mathrm{~m}\) long, \(5.3 \mathrm{~m}\) wide, and \(3.1 \mathrm{~m}\) high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor in \(\mathrm{mg} / \mathrm{m}^{3}\). Does this concentration exceed the safety limit of \(0.050 \mathrm{mg} / \mathrm{m}^{3} ?\) (Ignore the volume of furniture and other objects in the laboratory.)

Short Answer

Expert verified
The equilibrium constants Kc and Kp for the vaporization of mercury are respectively \(8.74 * 10^{-3}\) mol/L and \(2.63 * 10^{-6}\) atm. The mass of mercury that shall vaporize is calculated to be 52.9 g, causing a mercury vapor concentration of 527 mg/m³, which is significantly above the safety limit of 0.05mg/m³.

Step by step solution

01

Establish the equilibrium constants Kc and Kp for the vaporization of mercury

For the vaporization of mercury, the equilibrium process can be expressed as: Hg(l) <-> Hg(g). \n At equilibrium, all the Hg(l) becomes Hg(g), hence [Hg(g)] = P(Hg)/(RT), where P is the pressure, R the gas constant and T the temperature in Kelvin. \n Kc is hence = [Hg(g)], and utilizing the given pressure of mercury (converted to atm by dividing by 760, to match the units of R and T), R = 0.0821 L·atm/(K·mol), and T = 26+273 = 299 K, Kc = (0.0020/760)/(0.0821*299) = 8.74 * 10^-3 mol/L.\nSimilarly, as the reaction involves the conversion of one mol of liquid to one mol of gas, Kp is the same as the pressure P = 0.0020/760 = 2.63 * 10^-6 atm.
02

Calculate the mass of mercury vaporized and its concentration

For part (b) of the problem, we are given the dimensions of the room, which can be used to calculate the volume V = 6.1 * 5.3 * 3.1 = 100.43 m³ = 100430 L (converted to allow for usage in the ideal gas law). \n Utilizing the ideal gas law, we know that P(Hg) = n/V, where n is the moles of mercury. \n Solving for n, we find n = P(Hg)*V = 2.63 * 10^-6 * 100430 = 0.264 mol. \n The mass is then obtained by multiplying by the molar mass of Hg giving m = n * molar mass of Hg = 0.264 * 200.59 = 52.9 g. \n The concentration C(Hg) is then calculated as the mass / volume = 52.9g / 100430L = 0.53 g/m³ = 527mg/m³.
03

Compare with safety limit and derive conclusions

Finally, we have to compare the derived concentration with the safety limit of 0.05mg/m³ and conclude whether it is hazardous. As 527mg/m³ > 0.05mg/m³, the spill of mercury in such a room would indeed cause hazardous conditions exceeding the safety limit.

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Most popular questions from this chapter

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation $$ \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{CO}(g) $$ At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

At \(1024^{\circ} \mathrm{C}\), the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}\) $$ 4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) $$ (a) What is the \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) decomposed if 0.16 mole of it is placed in a 2.0 - \(L\) flask at \(1024^{\circ} \mathrm{C}\). (c) What would be the fraction if a 1.0 -mole sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

Write the expressions for the equilibrium constants \(K_{P}\) of these thermal decompositions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftharpoons 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\)

Define homogeneous equilibrium and heterogeneous equilibrium. Give two examples of each.

The dissociation of molecular iodine into iodine atoms is represented as $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K}\), the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(I_{2}\) in a 2.30 - \(L\) flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?
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