Problem 26 - General Chemistry: Principles and Modern Applications

Question

At \(2000 \mathrm{K}, K_{c}=0.154\) for the reaction \(2 \mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If a \(1.00 \mathrm{L}\) equilibrium mixture at \(2000 \mathrm{K}\) contains \(0.10 \mathrm{mol}\) each of \(\mathrm{CH}_{4}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (a) what is the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) present? (b) Is the conversion of \(\mathrm{CH}_{4}(\mathrm{g})\) to \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) favored at high or low pressures? (c) If the equilibrium mixture at \(2000 \mathrm{K}\) is transferred from a 1.00 L flask to a 2.00 L flask, will the number of moles of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) increase, decrease, or remain unchanged?

Step-by-Step Solution

Answer

(a) The mole fraction of \(C2H2(g)\) present is 0.89. (b) The conversion of \(CH4(g)\) to \(C2H2(g)\) is favored at low pressures. (c) The number of moles of \(C2H2(g)\) will increase when the mixture is transferred from a 1.00L flask to a 2.00L flask.

Step 1: Understanding Given Data & Equilibrium expression

Given that at \(2000K\), \(Kc = 0.154\) for the reaction, \(2CH4(g) ⇌ C2H2(g) + 3H2(g)\) Also, we are given that 1L equilibrium mixture contains \(0.1 mol\) each of \(CH4(g)\) and \(H2(g)\). We can write equilibrium expression as \(K_c = [C_2H2][H2]^3/[CH4]^2\).

Step 2: Calculating Concentration & Mole Fraction for Part (a)

Using the equilibrium expression: \(0.154 = [C_2H2][0.1]^3/[0.1]^2\). Therefore, \(0.154 = [C_2H2][0.1]\), which gives \([C_2H2] = 1.54 M\) which is the concentration of \(C_2H2\). Total moles of the mixture will now be \(0.1+0.1+1.54 = 1.74\), thus the mole fraction of \(C_2H2=g\) will be \(1.54/1.74 = 0.89\).

Step 3: Understanding Pressure Impact for Part (b)

According to Le Chatelier's principle, if a system at equilibrium is disturbed, the system tends to counteract the disturbance and a shift in the position of equilibrium is observed. Increasing the pressure shifts the equilibrium to the side with fewer moles of gas to reduce the pressure. Here, the left side of the equation has 2 moles of gas whereas the right side has 4. So, the conversion of \(CH4\) to \(C2H2\) is favored at low pressures.

Step 4: Understanding Volume Change Impact for Part (c)

Increasing the volume is equivalent to decreasing the pressure. According to Le Chatelier's principle, this will shift the equilibrium toward the side of the reaction that has more moles of gas. Therefore, the number of moles of \(C2H2(g)\) will increase when the mixture is transferred from a 1.00L flask to a 2.00L flask.

Key Concepts

Le Chatelier's Principle

Chemical equilibrium is a dynamic state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products.
If a disturbance, such as a change in concentration, pressure, or temperature, affects a system at equilibrium, Le Chatelier's Principle helps predict the direction in which the equilibrium will shift.
- **Pressure Changes**: When pressure is increased, the system will shift towards the side with fewer moles of gas to reduce pressure, and conversely, it will shift towards the side with more moles if pressure decreases.
- **Volume Changes**: Increasing the volume of the system decreases the pressure, causing the equilibrium to shift towards the side with more gas moles.
In the given reaction, increasing the volume (or decreasing the pressure) shifts the equilibrium towards the products (\(\text{C}_2\text{H}_2\text{(g)} + 3\text{H}_2\text{(g)}\)) which have more moles of gas. This results in an increased formation of acetylene \(\text{C}_2\text{H}_2\text{(g)}\). Conversely, compressing the volume would shift the equilibrium towards the reactant side \(\text{2 CH}_4\text{(g)}\).

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), quantifies the ratio of the concentration of products to the concentration of reactants at equilibrium.
It's crucial as it offers insight into the position of equilibrium and the extent of a reaction.
The formula is represented as:\[ K_c = \frac{[\text{C}_2\text{H}_2][\text{H}_2]^3}{[\text{CH}_4]^2} \]This expression involves concentrations, typically in molarity (M). A higher \(K_c\) value indicates a product-favored reaction, while a lower \(K_c\) suggests it's reactant-favored.
In the given exercise, \(K_c\) is 0.154, which implies at equilibrium, the reactants are prevalent over the products.
Calculating \(K_c\) involves substituting equilibrium concentrations into the expression, allowing us to assess how the reaction mixture shifts in response to changes in conditions.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture.
It's the ratio of the number of moles of one component to the total number of moles of all components in the mixture.
For a specific component \(i\), the formula is:\[ \text{Mole Fraction of } i = \frac{\text{moles of } i}{\text{total moles in the mixture}} \]Mole fraction is dimensionless and ranges from 0 to 1. In the given equilibrium mixture, while calculating the mole fraction of \(\text{C}_2\text{H}_2\), we determine the total moles in the system and then divide by the individual moles of acetylene.
This provides the mole fraction to be approximately 0.89, meaning a significant portion of the mixture consists of \(\text{C}_2\text{H}_2\) at equilibrium.
The mole fraction helps in understanding the composition of the mixture and is often used when dealing with gas mixtures and solutions.