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Chapter 13: Problem 27

At a particular temperature, a 3.0-L flask contains \(2.4 \mathrm{~mol} \mathrm{Cl}_{2}\), \(1.0 \mathrm{~mol} \mathrm{NOCl}\), and \(4.5 \times 10^{-3} \mathrm{~mol}\) NO. Calculate \(K\) at this temperature for the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$

Short Answer

Expert verified
We have set up the equilibrium expression \(K = \frac{[NO]^2 [Cl_{2}]}{[NOCl]^2}\) and calculated the initial concentrations of the species involved in the reaction: \([Cl_{2}]_{initial} = 0.8 \, M\), \([NOCl]_{initial} = 0.333 \, M\), and \([NO]_{initial} = 1.5 \times 10^{-3} \, M\). However, we cannot find the value of K in this exercise because there is not enough information provided to solve for the change in concentration (x). Additional information, such as the equilibrium concentrations of one or more species, is required to complete the problem.

Step by step solution

01

1. Write the balanced chemical equation and the equilibrium expression (K)

First, we need to write down the balanced chemical equation for the reaction: \[ 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_{2}(g) \] From the balanced chemical equation, we can write the equilibrium expression (K): \[K = \frac{[NO]^2 [Cl_{2}]}{[NOCl]^2}\]
02

2. Convert moles to concentrations

To convert moles to concentrations, we use the formula: \[ Concentration = \frac{Moles}{Volume} \] Given the volume of the flask is 3.0 L, we can calculate the initial concentrations of the species: \[ [Cl_{2}]_{initial} = \frac{2.4 \, mol}{3.0 \, L} = 0.8 \, M \] \[ [NOCl]_{initial} = \frac{1.0 \, mol}{3.0 \, L} = 0.333 \, M \] \[ [NO]_{initial} = \frac{4.5 \times 10^{-3} \, mol}{3.0 \, L} = 1.5 \times 10^{-3} \, M \]
03

3. Calculate the equilibrium concentrations of the species

Let x be the change in the concentration of NO (it increases by x M). Then, the change in concentration of NOCl is -2x (it decreases by 2x). At equilibrium, we have: \[ [NOCl]_{eq} = 0.333 - 2x \] \[ [NO]_{eq} = 1.5 \times 10^{-3} + x \] \[ [Cl_{2}]_{eq} = 0.8 - x \]
04

4. Substitute the equilibrium concentrations into the equilibrium expression (K)

Substitute the expressions for the equilibrium concentrations into the equilibrium expression and solve for x: \[K = \frac{[(1.5 \times 10^{-3} + x)^2 (0.8 - x)]}{(0.333 - 2x)^2}\] However, we need more information to solve for x. The exercise does not provide enough information to calculate K from this expression.
05

5. Conclusion

In this exercise, we have set up the equilibrium expression (K) and calculated the initial concentrations of the species involved in the reaction. Unfortunately, we cannot find the value of K because there is not enough information provided in the exercise to solve for x. Additional information, such as the equilibrium concentrations of one or more species, is required to complete the problem.

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Most popular questions from this chapter

For the reaction $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} M\), \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{~N}_{2}(g)\right]=5.3 \times 10^{-2} M\), and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

The formation of glucose from water and carbon dioxide is one of the more important chemical reactions in the world. Plants perform this reaction through the process of photosynthesis, creating the base of the food chain: $$ 6 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ At a particular temperature, the following equilibrium concentrations were found: \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=7.91 \times 10^{-2} M,\left[\mathrm{CO}_{2}(g)\right]=\) \(9.3 \times 10^{-1} M\), and \(\left[\mathrm{O}_{2}(g)\right]=2.4 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=0.041 M\), \(\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calculate the value of \(K\) for the reaction.

Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right)\), an extremely poisonous gas. Phosgene decomposes by the reaction $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of \(1.0\) atm decomposes, calculate the equilibrium pressures of all species.

Consider the following statements: "Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)\), for which at equilibrium \([\mathrm{A}]=2 M\) \([\mathrm{B}]=1 M\), and \([\mathrm{C}]=4 M .\) To a 1 -L container of the system at equilibrium, you add 3 moles of \(B\). A possible equilibrium condition is \([\mathrm{A}]=1 M,[\mathrm{~B}]=3 M\), and \([\mathrm{C}]=6 M\) because in both cases \(K=2 .\) " Indicate everything that is correct in these statements and everything that is incorrect. Correct the incorrect statements, and explain.
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