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Chapter 15: Problem 81

Nitrogen dioxide obtained as a cylinder gas is always a mixture of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) .\) A \(5.00 \mathrm{g}\) sample obtained from such a cylinder is sealed in a \(0.500 \mathrm{L}\) flask at \(298 \mathrm{K}\). What is the mole fraction of \(\mathrm{NO}_{2}\) in this mixture? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3}$$

Short Answer

Expert verified
The mole fraction of \(NO_2\) in this mixture will be [(5 - 92.02PV/RT) / (46.01 - 92.02)]/(PV/RT), where P is the pressure of the mixture, V is the volume, R is the ideal gas constant and T is the temperature in Kelvin.

Step by step solution

01

Calculate the Total moles

Let's first calculate the total moles of the cylinder gas. We know that the total mass of the cylinder gas sample is 5.00 g. We know that \( NO_2 \) has a molar mass of 46.01 g/mol and \( N_2O_4 \) has a molar mass of 92.02 g/mol. We can state that x moles are \( NO_2 \) and the rest (n - x) are \( N_2O_4 \), where n is the total moles. Using the molar mass of each gas and the given mass, we can write the following equation: 5 = 46.01x + 92.02(n - x)
02

Apply the Ideal Gas Law

We then apply the Ideal Gas Law to the problem, which states PV = nRT; where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature in Kelvin. Here, Pressure is not given, we assume it to be 1 atm, V is 0.500 L and T is 298 K. R can be taken as 0.0821 L.atm/mol.K for our calculation purposes. We use T in Kelvin in all Ideal Gas Law related calculations. Solving the Ideal Gas Law for n, we find that n = PV/RT.
03

Solve the equation from Step 1 for x

Substituting the equation for n from step 2 into our equation from step 1 and solve for x, we get x = (5 - 92.02PV/RT) / (46.01 - 92.02)
04

Calculate the Mole fraction of \(NO_2\)

Molecule fraction of any component in a mixture is the ratio of the number of moles of that component to the total number of moles of all components in the mixture, therefore the mole fraction of \(NO_2\) will be x/n. Substitute x and n with the equations from previous steps and we get the mole fraction of \(NO_2\) = [(5 - 92.02PV/RT) / (46.01 - 92.02)] / (PV/RT).

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Most popular questions from this chapter

Explain how each of the following affects the amount of \(\mathrm{H}_{2}\) present in an equilibrium mixture in the reaction \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) $$ \Delta H^{\circ}=-150 \mathrm{kJ} $$ (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst.

For the following reaction, \(K_{\mathrm{c}}=2.00\) at \(1000^{\circ} \mathrm{C}\) $$2 \operatorname{COF}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{CF}_{4}(\mathrm{g})$$ If a \(5.00 \mathrm{L}\) mixture contains \(0.145 \mathrm{mol} \mathrm{COF}_{2}, 0.262 \mathrm{mol}\) \(\mathrm{CO}_{2},\) and \(0.074 \mathrm{mol} \mathrm{CF}_{4}\) at a temperature of \(1000^{\circ} \mathrm{C}\) (a) Will the mixture be at equilibrium? (b) If the gases are not at equilibrium, in what direction will a net change occur? (c) How many moles of each gas will be present at equilibrium?

Determine values of \(K_{c}\) from the \(K_{p}\) values given. (a) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=2.9 \times 10^{-2} \mathrm{at} 303 \mathrm{K}\) (b) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=1.48 \times 10^{4} \mathrm{at} 184^{\circ} \mathrm{C}\) (c) \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(K_{\mathrm{p}}=0.429\) at \(713 \mathrm{K}\)

At \(500 \mathrm{K}\), a 10.0 L equilibrium mixture contains 0.424 \(\mathrm{mol} \mathrm{N}_{2}, 1.272 \mathrm{mol} \mathrm{H}_{2},\) and \(1.152 \mathrm{mol} \mathrm{NH}_{3} .\) The mixture is quickly chilled to a temperature at which the \(\mathrm{NH}_{3}\) liquefies, and the \(\mathrm{NH}_{3}(1)\) is completely removed. The 10.0 L gaseous mixture is then returned to \(500 \mathrm{K}\), and equilibrium is re-established. How many moles of \(\mathrm{NH}_{3}(\mathrm{g})\) will be present in the new equilibrium mixture? $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3} \quad K_{\mathrm{c}}=152 \text { at } 500 \mathrm{K}$$

Refer to Example \(15-4 . \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) at \(747.6 \mathrm{mmHg}\) pressure and a \(1.85 \mathrm{g}\) sample of \(\mathrm{I}_{2}(\mathrm{s})\) are introduced into a \(725 \mathrm{mL}\) flask at \(60^{\circ} \mathrm{C} .\) What will be the total pressure in the flask at equilibrium? $$\begin{aligned} \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{S}(\mathrm{s}) & \\ K_{\mathrm{p}}=& 1.34 \times 10^{-5} \mathrm{at} 60^{\circ} \mathrm{C} \end{aligned}$$
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