Question

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q)$$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10},\) what are the equilibrium concentrations of each species if you start with a 1.24\(M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Short answer

The equilibrium concentrations of each species in the reaction are approximately as follows: \([CH_3OH] \approx 1.24 M\), \([H_2CO] = [H_2] \approx 2.28 \times 10^{-5} M\). As formaldehyde is further converted to formic acid, the equilibrium will shift towards the products, leading to a decrease in the concentration of methanol in the solution as it is converted into formaldehyde and hydrogen.

Step-by-step solution

Write down the given information

We are given the following information: - The initial concentration of methanol: \([CH_3OH]_0 = 1.24 M\) - The value of the equilibrium constant \(K = 3.7 \times 10^{-10}\) - The chemical reaction: \(\mathrm{CH_{3} OH(aq) \rightleftharpoons H_{2} CO(aq) + H_{2}(aq)}\)

Set up ICE table

We will use the ICE (Initial, Change, Equilibrium) table to represent the information and keep track of the changes in concentrations at equilibrium. The table should look like this: | | CH3OH | H2CO | H2 | |---------|-------|------|-----| | Initial | 1.24 | 0 | 0 | | Change | -x | +x | +x | | Equilib.| 1.24-x| x | x | In the table, "x" represents the amount of CH3OH that reacts to produce H2CO and H2. At equilibrium, the concentrations are as follows: [CH3OH] = 1.24 - x, [H2CO] = x, and [H2] = x.

Set up equilibrium expression

According to the chemical reaction, the equilibrium constant K can be expressed as follows: \(K = \frac{[H_2CO][H_2]}{[CH_3OH]}\) Plugging the equilibrium concentrations from the ICE table, we get: \(3.7 \times 10^{-10} = \frac{x \cdot x}{1.24-x}\)

Solve for x

To find the value of x, we can solve the equation obtained from step 3. Given that K is very small, we can assume that x << 1.24, so the equation can be simplified to: \(3.7 \times 10^{-10} = \frac{x^2}{1.24}\) Now solve for x: \(x^2 = 3.7 \times 10^{-10} \times 1.24\) \(x = \sqrt{3.7 \times 10^{-10} \times 1.24} \approx 2.28 \times 10^{-5}\)

Calculate equilibrium concentrations

We can now plug x back into the equilibrium concentrations obtained from the ICE table to get: - \([CH_3OH] = 1.24 - x = 1.24 - 2.28 \times 10^{-5} \approx 1.24 M\) - \([H_2CO] = x = 2.28 \times 10^{-5} M\) - \([H_2] = x = 2.28 \times 10^{-5} M\)

Discuss the effect on methanol concentration as formaldehyde is converted into formic acid

As formaldehyde is further converted to formic acid, the concentration of formaldehyde will decrease. This leads to a shift in the equilibrium towards the products according to Le Chatelier's Principle, in an attempt to increase the concentration of formaldehyde. Consequently, more methanol will be converted into formaldehyde and hydrogen, which will decrease the concentration of methanol in the solution.

Key concepts

Equilibrium Concentration

When a reversible reaction reaches equilibrium, it means the rate of the forward reaction equals the rate of the reverse reaction. This does not imply that the concentrations of reactants and products are equal, but rather that they remain constant over time. In the given chemical equilibrium example of methanol and formaldehyde, the equilibrium concentration is determined using an ICE (Initial, Change, Equilibrium) table. This allows us to calculate the concentrations of each substance at equilibrium by assuming some of the methanol reacts to form formaldehyde and hydrogen. The concentration of each component at equilibrium is found by substituting a variable for the change in concentration and solving using the equilibrium expression.

Le Chatelier's Principle

Le Chatelier's Principle provides insights into how a chemical equilibrium responds when subjected to changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In our example, if the concentration of formaldehyde is reduced, perhaps by converting it into formic acid, the equilibrium will shift to the right. This means the system will favor the production of more products (formaldehyde and hydrogen) to replace what was lost. As a consequence, more methanol will be converted, thus decreasing its concentration. This principle is crucial in predicting and understanding how reactions will behave under different environmental and experimental conditions.

Equilibrium Constant

The equilibrium constant, denoted as K, is a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their respective coefficients in the balanced equation. For the reaction involving methanol and formaldehyde, the equilibrium constant expression is written as \(K = \frac{[H_2CO][H_2]}{[CH_3OH]}\). This constant provides valuable information about the extent of a reaction; a small K value, as seen in the methanol/formaldehyde reaction, suggests that at equilibrium, the reactants are favored over the products. It's important to remember that K is temperature-dependent and only changes if the temperature is altered.

ICE Table

An ICE table is a valuable tool used in chemistry to organize information about the concentrations of reactants and products at different stages of a chemical reaction: Initial, Change, and Equilibrium. It helps in visually tracking how the concentrations evolve as the reaction progresses to the equilibrium state. In our example, the ICE table is used with initial concentrations and the variable (x) representing the change in concentration. Using this table, we can write the equilibrium expression in terms of x, simplify it for small values of K by assuming \(x \ll 1.24\), and solve for x. This computation then provides us with the concentrations for all substances at equilibrium, making ICE tables a fundamental tool in solving equilibrium problems effectively.