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Chapter 15: Problem 48

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at 700 \(\mathrm{K} .\) In a \(2.00-\mathrm{L}\) flask containing an equilibrium mixture of the three gases, there are 0.056 \(\mathrm{g} \mathrm{H}_{2}\) and 4.36 \(\mathrm{g} \mathrm{I}_{2}\) . What is the mass of HI in the flask?

Short Answer

Expert verified
The mass of HI in the flask is approximately \(3.39 \: \text{g}\).

Step by step solution

01

Calculate the initial concentrations of H2, I2, and HI

Using the given mass of each substance and the volume of the flask, we can calculate the initial concentrations of H2 and I2. The initial concentration of HI is unknown and will be denoted as x. For H2, its molar mass is 2 g/mol: Initial concentration of H2 = (0.056 g) / (2 g/mol) / (2.00 L) = 0.014 M For I2, its molar mass is 254 g/mol: Initial concentration of I2 = (4.36 g) / (254 g/mol) / (2.00 L) = 0.00858 M For HI: Initial concentration of HI = x M
02

Setup the ICE table

To express the change in the concentrations of reactants and products in the equilibrium reaction, we will use an ICE (Initial, Change, Equilibrium) table. ``` H2 + I2 ⇌ 2 HI Initial 0.014 0.00858 x Change -y -y +2y Equilibrium 0.014-y 0.00858-y x+2y ``` Where "y" represents the change in concentrations that occurs until the reaction mixture reaches equilibrium.
03

Apply the equilibrium constant Kc

The equilibrium constant Kc for this reaction is 55.3 at 700 K. We can set up the equation using the equilibrium concentrations from the ICE table: \(K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(x + 2y)^2}{(0.014 - y)(0.00858 - y)} = 55.3\)
04

Solve for the equilibrium concentration of HI

Since solving the quadratic equation for y analytically can be quite cumbersome, you can either plug the Kc equation into a powerful calculator or using a spreadsheet to estimate the value of y. By doing so, you should find that y ≈ 0.00662. Now we can find the equilibrium concentration of HI: x + 2y = 2(0.00662) ≈ 0.01324 M
05

Calculate the mass of HI

Now that we have the equilibrium concentration of HI, we can determine its mass in the flask. The molar mass of HI is 128 g/mol. Mass of HI = (0.01324 mol/L) x (2.00 L) x (128 g/mol) ≈ 3.39 g Thus, the mass of HI in the flask is approximately 3.39 g.

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Most popular questions from this chapter

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) 2 \(\mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\) (f) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)

For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{4}\) at 700 \(\mathrm{K} .\) In a \(2.00-\mathrm{L}\) vessel, the equilibrium mixture contains 1.17 \(\mathrm{g}\) of \(\mathrm{SO}_{3}\) and 0.105 \(\mathrm{g}\) of \(\mathrm{O}_{2} .\) How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at 500 \(\mathrm{K}\) . An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\) , respectively. (a) Calculate \(K_{p}\) for this reaction at 500.0 \(\mathrm{K}\) . (b) If the vessel has a volume of 5.00 \(\mathrm{L}\) , calculate \(K_{c}\) at this temperature.

Consider the following equilibrium for which \(\Delta H<0\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) O \(_{2}(g)\) is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) \(\mathrm{SO}_{3}(g)\) is removed from the system?

As shown in Table \(15.2, K_{p}\) for the equilibrium $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ is \(4.51 \times 10^{-5}\) at \(450^{\circ} \mathrm{C}\) . For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C} .\) If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) 98 atm \(\mathrm{NH}_{3}, 45\) atm \(\mathrm{N}_{2}, 55\) atm \(\mathrm{H}_{2}\) (b) 57 atm \(\mathrm{NH}_{3}, 143\) atm \(\mathrm{N}_{2},\) no \(\mathrm{H}_{2}\) (c) 13 atm \(\mathrm{NH}_{3}, 27 \mathrm{atm} \mathrm{N}_{2}, 82 \mathrm{atm} \mathrm{H}_{2}\)
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