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Chapter 14: Problem 108

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 \(0.05 \mathrm{mg} / \mathrm{m}^{3} ?\) (Ignore the volume of furniture and other objects in the laboratory.)

Short Answer

Expert verified
Here, the answers would depend on the computations, but the steps above walk through how to find \(K_c\) and \(K_P\) for the process and how to calculate the mass and concentration of mercury vaporized as well as checking against the safety limit.

Step by step solution

01

Calculation of \(K_P\)

Vapor pressure is defined as the pressure exerted by a vapor in equilibrium with its condensed phases at a given temperature. Here it is given as \(0.0020 \mathrm{mmHg}\). However, equilibrium constants should be dimensionless, so it's best to convert this to a ratio by dividing by the standard atmospheric pressure (760 mmHg). Hence, \(K_P = 0.0020 \, \mathrm{mmHg} / 760 \, \mathrm{mmHg} = 2.63 \times 10^{-6}\).
02

Calculation of \(K_c\)

Given that \(K_P = K_c(RT)^{\Delta n}\) where \(R\) is the gas constant and \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas, which is 1 in this case, we can rearrange to find \(K_c\) as follows: \(K_c = K_P / (RT)\). We will need to convert temperature to Kelvin (26 °C + 273.15 = 299.15 K). Using the universal gas constant \(R = 0.0821 \, \mathrm{L \cdot atm} / \mathrm{K \cdot mol}\), we can then calculate \(K_c = 2.63 \times 10^{-6} \, atm / (0.0821 \, \mathrm{L \cdot atm} \cdot \mathrm{K^{-1} \cdot mol^{-1}} \cdot 299.15 \, K) = 1.08 \times 10^{-4} \, \mathrm{mol/L}\).
03

Calculation of the mass of mercury

Next, we calculate the mass of mercury that vaporizes. Mercury vapor exhibits ideal gas behavior under these conditions, so we use the ideal gas law \(PV = nRT\) rearranged to \(n = PV/RT\). We know that \(P\) is 0.0020 atm (derived from the given vapor pressure), \(V\) is the volume of the room in liters (6.1 m X 5.3 m X 3.1 m X 1000 L/m^3 = 100430 L), \(R\) is 0.0821 L atm / K mol, and \(T\) is 299.15 K. Plugging in these values gives the amount of mercury in moles, which we can then convert to grams using the molar mass of mercury (200.59 g/mol).
04

Calculation of the concentration of mercury

Next, we calculate the concentration of mercury in mg/m^3 by converting this mass to milligrams and dividing by the volume of the room in m^3.
05

Check against safety limit

Finally, compare the calculated concentration to the safety limit of 0.05 mg/m^3 to see if it exceeds this value.

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Most popular questions from this chapter

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00-L flask. At \(648 \mathrm{~K}\), there is \(0.0345 \mathrm{~mole}\) of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$

When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 M\) fructose solution at \(25^{\circ} \mathrm{C}\). At equilibrium, it was found that its concentration had decreased to \(0.113 M .\) (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose?

What do the symbols \(K_{\mathrm{c}}\) and \(K_{P}\) represent?

Consider the reaction between \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) in a closed container: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ Initially, 1 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is present. At equilibrium, \(\alpha\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to form \(\mathrm{NO}_{2}\). (a) Derive an expression for \(K_{P}\) in terms of \(\alpha\) and \(P\), the total pressure. (b) How does the expression in (a) help you predict the shift in equilibrium due to an increase in \(P ?\) Does your prediction agree with Le Châtelier's principle?

Determine the initial and equilibrium concentrations of HI if the initial concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) are both \(0.16 M\) and their equilibrium concentrations are both \(0.072 M\) at \(430^{\circ} \mathrm{C}\). The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) is 54.2 at \(430^{\circ} \mathrm{C}\)
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