Question

For the following reaction at a certain temperature $${\mathrm{H}}_2(g)+{\mathrm{F}}_2(g)\rightleftharpoons 2\mathrm{HF}(g)$$.it is found that the equilibrium concentrations in a 5.00 -L rigid container are $\left[{\mathrm{H}}_2\right]=0.0500\mathrm{M},\left[{\mathrm{F}}_2\right]=0.0100\mathrm{M},$ and $[\mathrm{HF}]=$ $0.400\mathrm{M}.$ If 0.200 mole of ${\mathrm{F}}_2$ is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Short answer

The new equilibrium concentrations after adding 0.200 moles of F₂ and reestablishing equilibrium are $[{\mathrm{H}}_2]=0.0725\mathrm{M}$, $[{\mathrm{F}}_2]=0.0275\mathrm{M}$, and $[\mathrm{HF}]=0.355\mathrm{M}$.

Step-by-step solution

Determine the equilibrium constant (Kc)

Given the equilibrium concentrations: $\left[{\mathrm{H}}_2\right]=0.0500\mathrm{M},\left[{\mathrm{F}}_2\right]=0.0100\mathrm{M}$, and $\left[\mathrm{HF}\right]=0.400\mathrm{M}$, we can find Kc using the following expression for the given reaction: $$K_c=\frac{[\mathrm{HF}{]}^2}{[{\mathrm{H}}_2][{\mathrm{F}}_2]}$$ Now, we can use the given concentrations to calculate Kc: $$K_c=\frac{(0.400{)}^2}{(0.0500)(0.0100)}$$ Calculating the Kc value: $$K_c=320$$

Set up an ICE table

After adding 0.200 moles of F₂, we can set up an ICE table to find the new equilibrium concentrations: $$\text{Unknown environment 'tabular'}$$ Note that we find the change in F₂ concentration by dividing the added moles (0.200 moles) by the volume of the container (5.00 L), which gives us a change of 0.0400 M.

Solve for the new equilibrium concentrations

We know that the equilibrium constant doesn't change, so we can write the equilibrium expression with the new concentrations as follows: $$K_c=\frac{[\mathrm{HF}{]}^2}{[{\mathrm{H}}_2][{\mathrm{F}}_2]}=\frac{(0.400-2x{)}^2}{(0.0500+x)(0.0500-x)}$$ Now, we can solve for x: $$320=\frac{(0.400-2x{)}^2}{(0.0500+x)(0.0500-x)}$$ At this point, we can approximate by assuming that x is small compared to 0.0500, i.e., x << 0.0500. Therefore, we can simplify the expression: $$320=\frac{(0.400-2x{)}^2}{(0.0500)(0.0500)}$$ Now, we can solve for x: $$x=0.0225$$

Calculate the new equilibrium concentrations

Lastly, we can find the new equilibrium concentrations by plugging x back into the Equilibrium row of the ICE table: $[{\mathrm{H}}_2]=0.0500+x=0.0500+0.0225=0.0725\mathrm{M}$ $[{\mathrm{F}}_2]=0.0500-x=0.0500-0.0225=0.0275\mathrm{M}$ $[\mathrm{HF}]=0.400-2x=0.400-2(0.0225)=0.355\mathrm{M}$ So, the equilibrium concentrations after adding 0.200 moles of F₂ and reestablishing equilibrium are as follows: $[{\mathrm{H}}_2]=0.0725\mathrm{M}$, $[{\mathrm{F}}_2]=0.0275\mathrm{M}$, and $[\mathrm{HF}]=0.355\mathrm{M}$

Key concepts

Chemical Reaction

Chemical reactions are processes in which substances, known as reactants, transform into different substances called products. These transformations occur due to the breaking and forming of chemical bonds, and they often involve changes in energy, such as the release or absorption of heat. When the rates of the forward and reverse reactions in a chemical process are equal, the system is said to be at chemical equilibrium. This means the concentrations of the reactants and products become constant over time, not because the reactions have stopped, but because they are occurring at the same rate.

Let's consider the reaction from the exercise: ${\mathrm{H}}_2(g)+{\mathrm{F}}_2(g)\rightleftharpoons 2\mathrm{HF}(g)$. Here, hydrogen gas (${\mathrm{H}}_2$) and fluorine gas (${\mathrm{F}}_2$) react to form hydrogen fluoride ($\mathrm{HF}$). At equilibrium, these gases are present at specific concentrations that do not change unless the system is disturbed, such as by adding more ${\mathrm{F}}_2$, as described in the original exercise.

Equilibrium Constant

The equilibrium constant, denoted by $K_c$, is a number that expresses the ratio of the concentrations of products to reactants for a reversible reaction at equilibrium, each raised to the power of their stoichiometric coefficients. The expression for $K_c$ depends on the balanced chemical equation and is defined by the law of mass action. For the reaction given, the equilibrium constant expression would be $K_c=\frac{[\mathrm{HF}{]}^2}{[{\mathrm{H}}_2][{\mathrm{F}}_2]}$.

An important aspect to note is that the value of $K_c$ is constant at a given temperature. It provides vital information about the position of equilibrium; a larger $K_c$ indicates a greater concentration of products, while a smaller $K_c$ suggests a reaction that favors the reactants. The calculation of $K_c$ from the given equilibrium concentrations aids in predicting the extent of the reaction and the concentrations of substances when equilibrium is re-established after a change.

ICE Table

An ICE table, which stands for Initial, Change, and Equilibrium, is a convenient method to organize the initial concentrations, the changes that occur, and the new equilibrium concentrations of each species in a chemical reaction. This approach is particularly useful for solving equilibrium problems, as it helps visualize the shifts in the system when disturbances are applied.

The steps in setting up an ICE table include:

• Writing down the balanced chemical equation
• Filling in the initial concentrations of reactants and products
• Determining the changes in concentrations when the system is perturbed
• Expressing the equilibrium concentrations in terms of the change

Through the ICE table, the impact of adding additional ${\mathrm{F}}_2$ on the system can be systematically determined, and the new equilibrium concentrations—$[{\mathrm{H}}_2]=0.0725\mathrm{M}$, $[{\mathrm{F}}_2]=0.0275\mathrm{M}$, and $[\mathrm{HF}]=0.355\mathrm{M}$—are calculated after accounting for this change and re-establishing equilibrium.