Question

Consider the following equilibrium process at ${700}^{\circ }\mathrm{C}$ : $$2{\mathrm{H}}_2(g)+{\mathrm{S}}_2(g)\rightleftarrows 2{\mathrm{H}}_2\mathrm{S}(g)$$ Analysis shows that there are 2.50 moles of ${\mathrm{H}}_2,1.35\times {10}^{-5}$ mole of ${\mathrm{S}}_2$, and 8.70 moles of ${\mathrm{H}}_2\mathrm{S}$ present in a 12.0 -L flask. Calculate the equilibrium constant $K_{\mathrm{c}}$ for the reaction.

Short answer

The equilibrium constant $K_c$ is approximately $1.08\times {10}^7$.

Step-by-step solution

Write the expression for Kc

The expression for the equilibrium constant, $K_c$, involves the concentrations of products and reactants at equilibrium. For the given reaction: $$2{\mathrm{H}}_2(g)+{\mathrm{S}}_2(g)\rightleftharpoons 2{\mathrm{H}}_2\mathrm{S}(g)$$ The equilibrium constant $K_c$ is given by: $$K_c=\frac{[{\mathrm{H}}_2\mathrm{S}{]}^2}{[{\mathrm{H}}_2{]}^2[{\mathrm{S}}_2]}$$ where the brackets denote the concentration (in moles per liter) of each species at equilibrium.

Calculate concentrations

To calculate $K_c$, first determine the concentration of each species in the flask by dividing the number of moles by the volume of the flask (12.0 L):$$[{\mathrm{H}}_2]=\frac{2.50\text{ moles}}{12.0\text{ L}}=0.208\text{ M}$$ $$[{\mathrm{S}}_2]=\frac{1.35\times {10}^{-5}\text{ moles}}{12.0\text{ L}}=1.125\times {10}^{-6}\text{ M}$$ $$[{\mathrm{H}}_2\mathrm{S}]=\frac{8.70\text{ moles}}{12.0\text{ L}}=0.725\text{ M}$$

Substitute into the Kc expression

Substitute the calculated concentrations into the $K_c$ expression:$$K_c=\frac{(0.725{)}^2}{(0.208{)}^2\times 1.125\times {10}^{-6}}$$

Calculate Kc value

Perform the calculation:$$(0.725{)}^2=0.525625$$ $$(0.208{)}^2=0.043264$$ Substitute these values:$$K_c=\frac{0.525625}{0.043264\times 1.125\times {10}^{-6}}=\frac{0.525625}{4.8658\times {10}^{-8}}$$ Finally, calculate $K_c$:$$K_c\approx 1.08\times {10}^7$$

Key concepts

Equilibrium Concentrations

In a chemical reaction, equilibrium concentrations refer to the amounts of reactants and products present in a system when it reaches equilibrium. Achieving equilibrium means that the rates of the forward and reverse reactions are equal, leading to constant concentrations, even though reactions continue to occur. In our example reaction:2 H₂(g) + S₂(g) $\rightleftharpoons$ 2 H₂S(g),we determine equilibrium concentrations by dividing the number of moles of each substance by the volume of the container. For the given problem in a 12.0 L flask, we calculated:

• [H₂] = 0.208 M
• [S₂] = 1.125 x 10⁻⁶ M
• [H₂S] = 0.725 M

These values give us the concentrations in molarity (M), which indicates moles of substance per liter of solution.

Chemical Equilibrium

Chemical equilibrium is a state where the forward and reverse reactions occur at the same rate, and the concentrations of reactants and products remain unchanged. It signifies balance and stability in a reaction, but it doesn't mean the reactants and products are equal in concentration. Instead, they settle in a ratio defined by the equilibrium constant, $K_c$.
At equilibrium, the whole reaction appears static because the changes in concentration happen simultaneously and perpetually, maintaining a consistent composition. In our reaction:2 H₂(g) + S₂(g) $\rightleftharpoons$ 2 H₂S(g),we find balance when the concentrations no longer change. This state allows us to analyze the system mathematically using $K_c$, which helps predict how the system will behave under different conditions.

Reaction Quotient

The reaction quotient, $Q$, is an expression that indicates the ratio of the concentrations of products to reactants at any point during a reaction—not just at equilibrium. It has the same mathematical form as $K_c$, but differs because it might not be at equilibrium.
To use $Q$, you substitute the current concentrations of substances into the equilibrium expression:$$Q=\frac{[H_{2}S{]}^2}{[H_2{]}^2[S_2]}$$Comparing $Q$ to $K_c$ tells us about the direction the reaction will proceed:

• If $Q<K_c$, the reaction moves forward to produce more products.
• If $Q=K_c$, the system is at equilibrium.
• If $Q>K_c$, the reaction goes in reverse to produce more reactants.

Understanding $Q$ provides insight into the dynamics of a chemical reaction, enabling predictions and adjustments to achieve desired outcomes.