Question

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 in a $2.0-\mathrm{L}$ container it is found that $[{\mathrm{H}}_2\mathrm{O}(g)]=1.1\times {10}^{-1}\mathrm{M}$ and $[{\mathrm{H}}_2(g)]=1.9\times {10}^{-2}\mathrm{M}.$ Calculate the moles of ${\mathrm{O}}_2(g)$ present under these conditions.

Short answer

The moles of O2 present under these conditions are 0.16 mol.

Step-by-step solution

Write the expression for the equilibrium constant K

Using the given reaction, we can write the expression for K as follows: $$K=\frac{[{\mathrm{H}}_2{]}^2[{\mathrm{O}}_2]}{[{\mathrm{H}}_2\mathrm{O}{]}^2}$$

Substitute the given values into the expression for K and solve for [O2]

We're given that at equilibrium: $$K=2.4\times {10}^{-3}$$ $$[{\mathrm{H}}_2\mathrm{O}]=1.1\times {10}^{-1}\mathrm{M}$$ $$[{\mathrm{H}}_2]=1.9\times {10}^{-2}\mathrm{M}$$ Plugging these values into the expression for K, we get: $$2.4\times {10}^{-3}=\frac{(1.9\times {10}^{-2}{)}^2[{\mathrm{O}}_2]}{(1.1\times {10}^{-1}{)}^2}$$ Now we can solve for [O2]: $$[{\mathrm{O}}_2]=\frac{2.4\times {10}^{-3}(1.1\times {10}^{-1}{)}^2}{(1.9\times {10}^{-2}{)}^2}$$ $$[{\mathrm{O}}_2]=\frac{2.4\times {10}^{-3}\times 12.1\times {10}^{-2}}{3.61\times {10}^{-4}}$$ $$[{\mathrm{O}}_2]=0.080$$

Calculate the moles of O2 present

Now that we have the concentration of O2, we can calculate the number of moles of O2 present by multiplying the concentration by the volume of the container: Moles of O2 = (Concentration of O2) x (Volume of container) Moles of O2 = (0.080 M) x (2.0 L) Moles of O2 = 0.16 mol The moles of O2 present under these conditions are 0.16 mol.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. It's important to note that even though the concentrations remain constant, the reactions are still occurring at the molecular level.

In the context of the equilibrium constant calculation exercise, the reaction given is:
$$2{\text{H}}_2\text{O}(g)\rightleftharpoons 2{\text{H}}_2(g)+{\text{O}}_2(g)$$.

This implies that water vapor (\text{H}_2\text{O}) can decompose into hydrogen gas (\text{H}_2) and oxygen gas (\text{O}_2), and these gases can also react to form water vapor. At chemical equilibrium, the decomposition of water vapor is occurring at the same rate as the formation of water vapor from hydrogen and oxygen gases.

Understanding chemical equilibrium is crucial because it allows us to predict the concentrations of each species present in a reaction mixture at any given time, provided we know the equilibrium constant, which is characteristic of a reaction at a given temperature.

Reaction Quotient

The reaction quotient, denoted as Q, is a measure used to determine the direction in which a reaction needs to shift to reach equilibrium. It has the same form as the equilibrium constant (K) expression, but it's calculated using the initial concentrations or pressures of reactants and products instead of their equilibrium values.

For the reaction in our exercise:
$$Q=\frac{[{\text{H}}_2{]}^2[{\text{O}}_2]}{[{\text{H}}_2\text{O}{]}^2}$$.

By comparing Q to the equilibrium constant K, we can predict which direction the reaction will proceed:

• If Q < K, the reaction will proceed forward to produce more products.
• If Q > K, the reaction will shift backward to produce more reactants.
• If Q = K, the system is at equilibrium, and no net change will occur.

In practice, when a reaction hasn't yet reached equilibrium, calculating Q can provide useful insights on how to alter conditions to push the reaction toward equilibrium. However, when we are given that the reaction has already reached equilibrium, like in our exercise, then Q actually equals K, and the Q calculation is not needed.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of reactants and products in a chemical system at equilibrium. These concentrations are constant over time, as the forward and reverse reactions occur at equal rates. When dealing with equilibrium concentrations, understanding stoichiometry, and the role of coefficients in a balanced chemical equation is essential.

In our exercise, we used the given equilibrium constant and equilibrium concentrations of \text{H}_2\text{O} and \text{H}_2 to calculate the equilibrium concentration of \text{O}_2, illustrated by the equation:
$$\text{K}=\frac{[{\text{H}}_2{]}^2[{\text{O}}_2]}{[{\text{H}}_2\text{O}{]}^2}$$.

This calculation process involved some key steps, including substituting known concentrations into the equilibrium expression, rearranging the expression to solve for the unknown concentration, and finally, multiplying the concentration by the volume of the system to find the number of moles present at equilibrium.

Understanding how to work with equilibrium concentrations can help students predict how changes in conditions, like concentration or volume, will affect the position of equilibrium. For example, if one were to add additional \text{H}_2\text{O} to the system, Le Chatelier’s principle tells us that the equilibrium would shift to counteract this change, likely resulting in the increased conversion of \text{H}_2\text{O} to \text{H}_2 and \text{O}_2 until a new state of equilibrium is attained.