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Chapter 15: Problem 70

Consider the dissociation of iodine: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ A \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is heated at \(1200^{\circ} \mathrm{C}\) in a 500 -mL flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in problem \(15.65(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.

Short Answer

Expert verified
The equilibrium constant \(KP\) for the reaction is approximately 199.27.

Step by step solution

01

Analyze the Provided Information

From the equation, it can be seen that for each mole of \(\mathrm{I}_{2}\) that dissociates, 2 moles of \(\mathrm{I}\) are formed. Also, it's noted that a \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is placed in a 500 mL flask and heated to achieve a total equilibrium pressure of 1.51 atm.
02

Calculate the Initial Moles of \(\mathrm{I}_{2}\)

First, calculate the initial moles of \(\mathrm{I}_{2}\) by using the molar mass. So, the initial moles of \(\mathrm{I}_{2}\) is \(1.00 g / 253.8 g/mol = 0.00394 mol\). Since the volume of the flask is 500 mL or 0.5 L, the initial concentration of \(\mathrm{I}_{2}(g)\) is then given by \(0.00394 mol / 0.5 L = 0.00788 M\).
03

Calculate the Degree of Dissociation (\(\alpha\))

The degree of dissociation can be calculated using the formula \(\alpha = 1-(p_{I_2}/P_t)\), where \(p_{I_2}\) is the partial pressure of \(\mathrm{I}_{2}\) and \(P_t\) is the total pressure at equilibrium. Since we calculated the partial pressure expecting none dissociation to be \(0.00788 atm\) and the total pressure is given as \(1.51 atm\), we can plug in these values to find that \(\alpha = 1 - (0.00788 atm / 1.51 atm) = 0.9948.\)
04

Calculate the Equilibrium Constant (KP)

The equilibrium constant for the reaction can be calculated using the formula for KP which is \(KP = \alpha^2/(1-\alpha)\). Substituting \(\alpha=0.9948\) into formula gives \(KP=0.9948^2/(1-0.9948)\approx 199.27\).

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Most popular questions from this chapter

For the synthesis of ammonia $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ the equilibrium constant \(K_{\mathrm{c}}\) at \(375^{\circ} \mathrm{C}\) is \(1.2 .\) Starting with \(\left[\mathrm{H}_{2}\right]_{0}=0.76 \mathrm{M},\left[\mathrm{N}_{2}\right]_{0}=0.60 \mathrm{M},\) and \(\left[\mathrm{NH}_{3}\right]_{0}=0.48 M,\) when this mixture comes to equilibrium, which gases will have increased in concentration and which will have decreased in concentration?

Define reaction quotient. How does it differ from equilibrium constant?

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

One mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\) are placed in a flask at \(397^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is found to be 0.21. The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}.\)

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.2 at \(1650^{\circ} \mathrm{C}\). Initially \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.80 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are injected into a \(5.0-\mathrm{L}\) flask. Calculate the concentration of each species at equilibrium.
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