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Chapter 14: Problem 28

A 2.50 -mole quantity of \(\mathrm{NOCl}\) was initially in a \(1.50-\mathrm{L}\) reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the NOCl had dissociated: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Short Answer

Expert verified
The equilibrium constant \(K_c\) for the reaction is 0.041.

Step by step solution

01

Determine initial number of moles

Before the reaction occurs, we are told that there are 2.50 moles of NOCl present in a 1.5 L reaction chamber and no NO or Cl2.
02

Calculate initial concentrations

To calculate the initial concentration of NOCl, divide the number of moles of NOCl by the volume of the chamber: \n\[\frac{2.50 \ \mathrm{mol}}{1.50 \ \mathrm{L}} = 1.67 \ \mathrm{M}]\n. Since NO and Cl2 are not initially present, their concentrations are zero.
03

Calculate equilibrium concentrations

At equilibrium, we know that 28% of the NOCl has dissociated. Therefore, the change in concentration for NOCl is given by: \(-0.28 \times 1.67 = -0.47 \ \mathrm{M}\). \nFor each mole of NOCl that decomposes, one mole of NO and half a mole of Cl2 are produced. Therefore, at equilibrium, the changes in concentrations for NO and Cl2 are \(+0.47 \ \mathrm{M}\) and \(+0.47/2 = +0.23 \ \mathrm{M}\), respectively. The equilibrium concentrations will therefore be: \n\[NOCl: 1.67 - 0.47 = 1.20 \ \mathrm{M}\]\n\[NO: 0.47 \ \mathrm{M}\]\n\[Cl2: 0.23 \ \mathrm{M}\]
04

Use the equilibrium concentrations to calculate \(K_c\)

The equilibrium constant \(K_c\) is given by \[K_c= \frac{[NO]^2 \ [Cl_2]}{[NOCl]^2}\] Insert the equilibrium concentrations calculated in Step 3 to obtain \[K_c= \frac{(0.47)^2 \times 0.23}{(1.20)^2} = 0.041\]

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Most popular questions from this chapter

Consider the following reaction at \(1600^{\circ} \mathrm{C}\). $$\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)$$ When 1.05 moles of \(\mathrm{Br}_{2}\) are put in a 0.980 - \(\mathrm{L}\) flask, 1.20 percent of the \(\mathrm{Br}_{2}\) undergoes dissociation. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

The dissociation of molecular iodine into iodine atoms is represented as $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ At \(1000 \mathrm{~K},\) the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(\mathrm{I}_{2}\) in a 2.30 -L flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

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Consider the dissociation of iodine: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ A 1.00-g sample of \(I_{2}\) is heated to \(1200^{\circ} \mathrm{C}\) in a \(500-\mathrm{mL}\) flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in \(14.117(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.]
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