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Chapter 14: Problem 55

The equilibrium concentrations for the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) at \(2000^{\circ} \mathrm{C}\) are \(\left[\mathrm{N}_{2}\right]=\) \(0.25 \mathrm{M} ;\left[\mathrm{O}_{2}\right]=1.2 \mathrm{M} ;[\mathrm{NO}]=0.011 \mathrm{M} .\) What is the value of \(K_{\mathrm{eq}}\) for this reaction?

Short Answer

Expert verified
The value of the equilibrium constant, \(K_{\text{eq}}\), for the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) at \(2000^{\circ}\mathrm{C}\) is \(4.03 \times 10^{-4}\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation is given as: \[\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\]
02

Write the expression for the equilibrium constant \(K_{\text{eq}}\)

The equilibrium constant for the given reaction is the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, each raised to the power of the stoichiometric coefficients in the balanced equation: \[K_{\text{eq}} = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}] \cdot [\mathrm{O}_{2}]}\]
03

Insert the equilibrium concentrations into the formula

We are given the following equilibrium concentrations: \([\mathrm{N}_{2}]=0.25\,\mathrm{M}\), \([\mathrm{O}_{2}]=1.2\,\mathrm{M}\), and \([\mathrm{NO}]=0.011\,\mathrm{M}\). Substitute these values into the equilibrium constant expression: \[K_{\text{eq}} =\frac{(0.011)^2}{(0.25) \cdot (1.2)}\]
04

Calculate the value of \(K_{\text{eq}}\)

Perform the calculation to determine the value of the equilibrium constant: \[K_{\text{eq}} =\frac{(0.011)^2}{(0.25) \cdot (1.2)} =\frac{0.000121}{0.3} = 4.03 \times 10^{-4}\] The value of the equilibrium constant, \(K_{\text{eq}}\), for this reaction at \(2000^{\circ}\mathrm{C}\) is \(4.03 \times 10^{-4}\).

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Most popular questions from this chapter

In theory, all reactions are reversible, but in practice, some are not. Explain why.

If \(k_{\mathrm{f}}>k_{\mathrm{r}}\), will \(K_{\mathrm{eq}}\) be less than 1 or greater than 1? Explain your answer.

What is a heterogeneous chemical reaction? Where does a heterogeneous reaction occur?

Consider the gas-state reaction \(2 \mathrm{~A}(g)+3 \mathrm{~B}(g) \rightleftarrows \mathrm{C}(g)+\mathrm{D}(g)\) (a) Write the equilibrium constant expression for the reaction. (b) Write the reaction in reverse. (c) Write the equilibrium constant expression for the reverse reaction that you wrote for part (b). (d) Compare your answers to \((\mathrm{a})\) and \((\mathrm{c})\). What conclusion can you draw from the comparison? (e) Suppose \(K_{\mathrm{eq}}\) for a reaction \(=10.0 .\) What will the value of \(K_{\text {eq }}\) be for the reverse reaction?

A chemist runs a reaction that is known to proceed very rapidly and keeps isolating product that is contaminated with starting material, even though he is following the stoichiometry of the reaction and giving the reaction enough time to run. How might this be explained?
See all solutions

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