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Chapter 15: Problem 48

For thereaction, at $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\( at \)700 \mathrm{~K}$. In a 10.0-L flask containing an equilibrium mixture of the three gases, there are $1.30 \mathrm{~g} \mathrm{H}_{2}\( and \)21.0 \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 at equilibrium is 10.6 g.

Step by step solution

01

Convert mass to moles

We will convert the mass of H2 and I2 to moles using their molar mass. Molar mass of H2 = 2.02g/mol Molar mass of I2 = 253.81g/mol Moles of H2 = mass / molar_mass = 1.30g / 2.02 g/mol = 0.6436 mol Moles of I2 = mass / molar_mass = 21.0g / 253.81 g/mol = 0.0827 mol Moles of HI (initially) = 0
02

Use ICE table

Now we will use the ICE table to find equilibrium moles. H2 + I2 ⇌ 2HI Initial(mol) 0.6436 0.0827 0 Change(mol) -x -x +2x Equil(mol) 0.6436-x 0.0827-x 2x Kc = 55.3 = \([HI]^2 / ([H2][I2])\] We will plug the equilibrium concentrations in the Kc expression and solve for x. 55.3 = \((((2x)^2)/(0.6436-x)(0.0827-x))\)
03

Solve for x

After solving the above equation for x, we get: x = 0.0413 mol Now we can find the moles of HI at equilibrium: Moles of HI at equilibrium = 2x = 2(0.0413) = 0.0826 mol
04

Convert moles of HI to mass

Finally, we will convert the moles of HI back to mass using its molar mass. Molar mass of HI = 127.91g/mol (H=1.01g/mol and I=126.9g/mol) Mass of HI = moles × molar_mass = 0.0826 mol × 127.91 g/mol = 10.6 g The mass of HI in the flask at equilibrium is 10.6 g.

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Most popular questions from this chapter

True or false: When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases, which leads to a decrease in the equilibrium constant, \(K_{c}\)

The equilibrium $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\( is established at \)550 \mathrm{~K}$. An equilibrium mixture of the three gases has partial pressures of $10.13 \mathrm{kPa}, 20.27 \mathrm{kPa}\(, and \)35.46 \mathrm{kPa}\( for \)\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.

For $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{2}\( at \)700 \mathrm{~K}$. In a 2.00-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?

Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases, the speed at which a reaction reaches equilibrium increases.

In Section \(11.5,\) we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for $K_{p \cdot}(\mathbf{b})\( By using data in Appendix \)\mathrm{B}$, give the value of \(K_{p}\) for this reaction at \(30^{\circ} \mathrm{C}\). (c) What is the value of \(K_{p}\) for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?
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