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Chapter 16: Problem 50

For the reaction $$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} \mathrm{M}\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} \mathrm{M} .\) What is the concentration of \(\mathrm{O}_{2}(g)\) under these conditions?

Short Answer

Expert verified
The concentration of O鈧�(g) under these conditions is \(8.04 \times 10^{-2}\) M.

Step by step solution

01

Write the expression for the equilibrium constant K

The expression for the equilibrium constant K in terms of the concentrations of reactants and products for the given reaction can be written as: \[K=\frac{[\mathrm{H}_2]^2 \times [\mathrm{O}_2]}{[\mathrm{H}_2\mathrm{O}]^2}\] where [H鈧俔, [O鈧俔, and [H鈧侽] are the molar concentrations of H鈧�(g), O鈧�(g), and H鈧侽(g) at equilibrium, respectively.
02

Plug in the known values

We are given the values of K, [H鈧侽], and [H鈧俔. We can substitute these values into the equation for K: \[2.4 \times 10^{-3}=\frac{(1.9 \times 10^{-2})^2 \times [\mathrm{O}_2]}{(1.1 \times 10^{-1})^2}\]
03

Solve for the concentration of O鈧�(g)

Now, we need to solve the equation for [O鈧俔: \[[\mathrm{O}_2]=\frac{2.4 \times 10^{-3} \times (1.1\times 10^{-1})^2}{(1.9\times 10^{-2})^2}\] First, calculate the numerator: \((2.4 \times 10^{-3})\times (1.1\times 10^{-1})^2\approx 2.904 \times 10^{-5}\) Next, calculate the denominator: \((1.9\times 10^{-2})^2\approx 3.61\times 10^{-4}\) Finally, divide the numerator by the denominator to find the concentration of O鈧�(g): \[[\mathrm{O}_2]\approx \frac{2.904 \times 10^{-5}}{3.61\times 10^{-4}}\approx 8.04\times 10^{-2} \mathrm{M}\] So, the concentration of O鈧�(g) under these conditions is \(8.04 \times 10^{-2}\) M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
When we refer to chemical equilibrium, we're talking about a state in which a chemical reaction and its reverse reaction occur at the same rate. As a result, the concentrations of the reactants and products remain constant over time, not to be confused with them being equal to each other. This balance can be visualized as a scale in perfect harmony, where no side is heavier than the other.

In our example, the equilibrium involves the conversion of water vapor into hydrogen gas and oxygen gas. At equilibrium, these substances are converted back and forth at equivalent rates, so even though their concentrations don't change, the actual molecules are constantly in motion, continually reacting and re-forming. This concept is fundamental to understanding how reactions can be predicted and controlled.
Reaction Quotient
The reaction quotient, denoted as Q, is a snapshot of a reaction that tells us whether it has reached equilibrium and, if not, which direction it will proceed to reach equilibrium. It closely resembles the equilibrium constant expression, but for Q, the concentrations are not necessarily those at equilibrium.

The reaction quotient is calculated exactly like the equilibrium constant, using the same formula, but with the initial concentrations of the reactants and products. If Q is less than the equilibrium constant, K, the forward reaction is favored, and more products will form. If Q is greater than K, the reverse reaction is favored, and reactants will form. This concept helps chemists understand the state of a reaction mixture at any point before equilibrium has been established.
Concentration of Reactants and Products
In the context of chemical reactions, concentration usually refers to molarity, which is the number of moles of a substance per liter of solution (mol/L or M). Knowing the concentration of reactants and products at equilibrium is essential for calculating the equilibrium constant.

Understanding these concentrations allows us to predict the position of equilibrium and the extent to which a reaction will proceed. Higher concentrations of reactants typically push the equilibrium towards the products, while higher concentrations of products push the equilibrium back towards the reactants.
Equilibrium Expression
An equilibrium expression relates the concentrations of the products raised to their stoichiometric coefficients to the reactants raised to their stoichiometric coefficients. For the given reaction, the equilibrium expression in terms of the molarity (M) of the gases involved is written as
\[K=\frac{[\mathrm{H}_2]^2 \times [\mathrm{O}_2]}{[\mathrm{H}_2\mathrm{O}]^2}\]
This expression allows us to plug in known concentrations to determine the unknown concentration of a product or reactant at equilibrium. Importantly, the state of equilibrium is dynamic, and this expression holds true only when the system is at equilibrium.

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Most popular questions from this chapter

Explain why the position of a heterogeneous equilibrium does not depend on the amounts of pure solid or pure liquid reactants or products present.

The solubility product constant, \(K_{\mathrm{sp}}\), for barium carbonate is \(8.2 \times 10^{-9}\) at a particular temperature. Calculate the solubility of \(\mathrm{BaCO}_{3}\) in \(\mathrm{mol} / \mathrm{L}\) at this temperature.

As you learned in Chapter \(7,\) most metal hydroxides are sparingly soluble in water. Write balanced chemical equations describing the dissolving of the following metal hydroxides in water. Write the expression for \(K_{\mathrm{sp}}\) for each process. a. \(\mathrm{Cu}(\mathrm{OH})_{2}(s)\) b. \(\operatorname{Cr}(\mathrm{OH})_{3}(s)\) c. \(\mathrm{Ba}(\mathrm{OH})_{2}(s)\) d. \(\operatorname{Sn}(O H)_{2}(s)\)

Plants synthesize the sugar dextrose according to the following reaction by absorbing radiant energy from the sun (photosynthesis). $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Will an increase in temperature tend to favor or discourage the production of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) ?\)

For a given reaction at a given temperature, the special ratio of products to reactants defined by the equilibrium constant is always equal to the same number. Explain why this is true, no matter what initial concentrations of reactants (or products) may have been taken in setting up an experiment.
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