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Chapter 15: Problem 49

At \(800 \mathrm{~K},\) the equilibrium constant for $\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\( is \)K_{c}=3.1 \times 10^{-5}$. If an equilibrium mixture in a 5.00-L vessel contains \(30.5 \mathrm{mg}\) of \(\mathrm{I}(g)\), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

Short Answer

Expert verified
In the equilibrium mixture, there are 0.01624 grams of \(I_2\).

Step by step solution

01

Write the balanced equation

The balanced equation for the dissociation of I2 into I atoms is given as: \(I_2 (g) \rightleftharpoons 2\; I(g)\)
02

Convert the mass of I to moles

We are given that there are 30.5 mg of I(g) in the equilibrium mixture. To convert this into moles, use the molecular weight of I (which is 126.90 g/mol): \(n_{I} = \frac{30.5\; \text{mg}}{126.90\; \frac{\text{g}}{\text{mol}}} \times \frac{1\; \text{g}}{1000\; \text{mg}} = 2.403 \times 10^{-4}\; \text{mol}\)
03

Determine the initial moles of each species

Since the reaction has not yet occurred, we know that there are initially no I atoms. Let's assume there are y moles of I2: Initial moles of I2: \(y\; \text{mol}\)
04

Set up the ICE table

An ICE (Initial, Change, Equilibrium) table can be used to track the changes in the moles of each species throughout the reaction: \[ \begin{array}{c|c|c|c} & \text{[I}_2\text{]} & \text{[I]} \\ \hline \text{Initial} & y & 0 \\ \text{Change} & -x & +2x \\ \text{Equilibrium} & y-x & 2x \\ \end{array} \] Since we know the moles of I at equilibrium (2.403 x 10^{-4} mol), we can express the moles of I2 at equilibrium in terms of x: \(y - x = 2x\) We are given the equilibrium constant Kc: \(K_c = 3.1 \times 10^{-5}\)
05

Use the equilibrium constant and the ICE table

From the balanced equation and the ICE table, we can write the expression for the equilibrium constant Kc: \(K_c = \frac{[\text{I}]^2}{[\text{I}_2]} = \frac{(2x)^2}{y-x}\) Now substitute the given value of Kc and the moles of I at equilibrium: \(3.1 \times 10^{-5} = \frac{(2x)^2}{y-x}\) We can use the expression we derived for the moles of I2 at equilibrium in terms of x: \(3.1 \times 10^{-5} = \frac{(2x)^2}{3x}\) Solve this equation for x: \(x = 6.397 \times 10^{-5}\; \text{mol}\)
06

Convert the moles of I2 to mass

We found that at equilibrium, there are 6.397 x 10^{-5} mol of I2. To convert this to grams, use the molecular weight of I2 (which is 253.8 g/mol): Mass of I2 = \(6.397 \times 10^{-5}\; \text{mol} \times 253.8\; \frac{\text{g}}{\text{mol}} = 0.01624\; \text{g}\) Therefore, there are 0.01624 grams of I2 in the equilibrium mixture.

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Most popular questions from this chapter

Gaseous hydrogen iodide is placed in a closed container at $450^{\circ} \mathrm{C},\( where it partially decomposes to hydrogen and iodine: \)2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .$ At equilibrium it is found that $[\mathrm{HI}]=4.50 \times 10^{3} \mathrm{M},\left[\mathrm{H}_{2}\right]=5.75 \times 10^{4} \mathrm{M}$, and \(\left[\mathrm{I}_{2}\right]=5.75 \times 10^{-4} \mathrm{M}\). What is the value of \(K_{c}\) at this temperature?

Write the expression for \(K_{c}\) for the following reactions. \(\operatorname{In}\) each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) (b) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$ (c) \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$ (e) $\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)$ (f) $\mathrm{Fe}^{2+}(a q)+\mathrm{Zn}(s) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Fe}(s)$ (g) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q)$

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g) .\) A 7.5-L gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g)\), which is allowed to equilibrate at 450 \(\mathrm{K} .\) At equilibrium the partial pressures of the three gases are $P_{\mathrm{PCl}_{3}}=12.56 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=15.91 \mathrm{kPa},\( and \)P_{\mathrm{PCl}_{5}}=131.7 \mathrm{kPa}$ (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at \(450 \mathrm{~K}\).

At \(800 \mathrm{~K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g)\) is $K_{c}=3.1 \times 10^{-4}$. (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r} ?\) (b) If the value of $k_{f}=0.27 \mathrm{~s}^{-1}\(, what is the value of \)k_{r}\( at \)800 \mathrm{~K} ?$ (c) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to $1000 \mathrm{~K}\(, will the reverse rate constant \)k_{r}$ increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f}\) ?

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}(g)\) (b) $\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)$ (c) $\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$ (d) $\mathrm{O}_{2}(g)+2 \mathrm{CO}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$ (e) $\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}^{+}(a q)$ (f) $\mathrm{Fe}^{2+}(a q)+\mathrm{Ce}^{4+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)$ (g) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$
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