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Chapter 13: Problem 26

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 \(\left[\mathrm{N}_{2}\right]=\) \(0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calcu- late the value of \(K\) for the reaction.

Short Answer

Expert verified
The equilibrium constant, K, for the given reaction at the specified temperature is approximately 6.91.

Step by step solution

01

Write the balanced reaction

Write the balanced chemical reaction for the given process: \[ \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \]
02

Write the equilibrium expression

Based on the balanced reaction, we can write the expression for the equilibrium constant K: \[ K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} \]
03

Insert the equilibrium concentrations

Now we substitute the given equilibrium concentrations of each component: \[ K = \frac{(4.7 \times 10^{-4}\ \mathrm{M})^2}{(0.041\ \mathrm{M})(0.0078\ \mathrm{M})} \]
04

Calculate the value of K

Perform the calculations to find the value of K: \[ K = \frac{(2.209\times 10^{-7}\ \cancel{\mathrm{M^2}})}{(3.198 \times 10^{-4}\ \cancel{\mathrm{M^2}})} \] \[ K = 6.907 \] So the equilibrium constant, K, for the given reaction at the specified temperature is approximately 6.91.

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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.

At a particular temperature, \(8.1\) moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0-L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}\) : $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be \(1.4 \mathrm{~mol} / \mathrm{L}\). Calculate the value of \(K\) for this reaction.

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

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ In an experiment, \(1.00\) mole of \(\mathrm{H}_{2}, 1.00 \mathrm{~mole}\) of \(\mathrm{I}_{2}\), and \(1.00\) mole of HI are introduced into a \(1.00-\mathrm{L}\) container. Calculate the concentrations of all species when equilibrium is reached.

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following cases. a. \(1.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) and \(2.0 \mathrm{~g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a \(1.0-\mathrm{L}\) flask. b. \(1.0\) mole of pure \(\mathrm{HOCl}\) is placed in a \(2.0-\mathrm{L}\) flask.
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