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: $${\mathrm{CH}}_3\mathrm{OH}(aq)\rightleftharpoons {\mathrm{H}}_2\mathrm{CO}(aq)+{\mathrm{H}}_2(aq)$$. =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 $1.24M$ solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Short answer

Given the reaction $C{H}_{3}OH(aq)\rightleftharpoons H_{2}CO(aq)+H_2(aq)$ with an initial methanol concentration of 1.24 M and an equilibrium constant $K=3.7\times {10}^{-10}$, after solving for the equilibrium concentrations, we find: - The equilibrium concentration of methanol $[C{H}_{3}OH{]}_{eq}\approx 1.24M$. - The equilibrium concentration of formaldehyde $[H_{2}CO{]}_{eq}\approx 7.13\times {10}^{-6}M$. - The equilibrium concentration of hydrogen gas $[H_2{]}_{eq}\approx 7.13\times {10}^{-6}M$. As formaldehyde is further converted to formic acid, the change in methanol concentration will be negligible, since the equilibrium shifted significantly towards the reactant side and the equilibrium constant is very small.

Step-by-step solution

Write the equilibrium expression

Based on the given chemical equation, the equilibrium constant expression is given by: $$K=\frac{[H2CO][H2]}{[CH3OH]}$$ Where [H2CO], [H2], and [CH3OH] denote the equilibrium concentrations of formaldehyde, hydrogen gas, and methanol, respectively.

Set up an ICE table

Let's set up an Initial, Change, Equilibrium (ICE) table to keep track of the changes in concentrations: | Species | Initial (M)| Change (M) | Equilibrium (M) | |-------------|------------|------------|--------------| | CH₃OH | 1.24 | -x | 1.24 - x | | H₂CO | 0 | +x | x | | H₂ | 0 | +x | x | Here, x represents the change in the concentration of each species as the reaction proceeds to equilibrium.

Substitute the equilibrium concentrations into the equilibrium expression

Let's substitute the equilibrium concentrations into the equilibrium expression and solve for x: $$3.7\times {10}^{-10}=\frac{[H2CO][H2]}{[CH3OH]}=\frac{x\cdot x}{1.24-x}$$ Simplify the equation and solve for x: $$3.7\times {10}^{-10}(1.24-x)=x^2$$

Solve for the change in concentrations (x)

As the value of $K$ is very small just like x, we can assume that the change in the concentration of methanol (1.24 - x) is approximately equal to 1.24: $$x^2=3.7\times {10}^{-10}(1.24)$$ Using a calculator to solve for x: $$x\approx 7.13\times {10}^{-6}$$

Find the equilibrium concentrations of each species

Now that we have x, we can find the equilibrium concentrations of each species: $$[C{H}_{3}OH{]}_{eq}=1.24-x=1.24-7.13\times {10}^{-6}\approx 1.24M$$ $$[H_{2}CO{]}_{eq}=x=7.13\times {10}^{-6}M$$ $$[H_2{]}_{eq}=x=7.13\times {10}^{-6}M$$

Determine the change in methanol concentration due to further conversion of formaldehyde

We are asked to find the change in methanol concentration as formaldehyde converts further into formic acid. The formaldehyde will be consumed to make formic acid, and the methanol concentration will remain almost the same since the equilibrium constant for the current reaction is very small. The change in methanol concentration will be negligible since the equilibrium shifted significantly towards the reactant side.

Key concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol 'K', is a fundamental concept in chemical equilibrium that quantifies the ratio of the concentrations of products to the concentrations of reactants at equilibrium. Different reactions have different equilibrium constants, which are determined experimentally and remain constant at a given temperature. It reflects the extent to which a reaction proceeds before reaching a state of balance.

For the methanol oxidation reaction given in the problem, $K=\frac{[H_{2}CO][H_2]}{[C{H}_{3}OH]}$, the small value of $K(3.7\times {10}^{-10})$ suggests that at equilibrium, the concentration of reactants (methanol) is much higher in comparison to the concentrations of the products (formaldehyde and hydrogen gas). This indicates that the reaction favors the formation of reactants over products.

ICE Table Method

The ICE table method is a systematic approach to track the concentrations of species involved in a chemical reaction. The acronym ICE stands for Initial, Change, and Equilibrium.

Initially, we identify the concentrations of reactants and products. For changes, we consider 'x' to represent the amount by which reactants decrease and products increase as the system reaches equilibrium. In this question, the initial concentration of methanol is 1.24 M, while formaldehyde and hydrogen begin at 0 M.

As the reaction progresses, some of the methanol is converted into formaldehyde and hydrogen. This change is represented by increasing the product concentrations by 'x' and decreasing the reactant's concentration by 'x'. Finally, we derive the equilibrium concentrations by combining the initial amounts with the changes. The ICE table method is crucial for visually organizing data and simplifying the process of calculating equilibrium concentrations.

Reaction Quotient

The reaction quotient, labeled as 'Q', is closely related to the equilibrium constant but is used to describe the state of a reaction not necessarily at equilibrium. It has the same form as the equilibrium expression but uses the concentrations of reactants and products at any point in time.

$Q=\frac{[H_{2}CO][H_2]}{[C{H}_{3}OH]}$, where the concentrations are those currently present in the reaction mixture. Comparing 'Q' to 'K' helps determine the direction in which the reaction will proceed to achieve equilibrium. If $Q<K$, the forward reaction is favored, and more products will form. For $Q>K$, the reverse reaction is favored, and more reactants will form. When $Q=K$, the system is at equilibrium, and the concentrations of reactants and products will remain constant. Understanding 'Q' is essential for predicting the shifts in equilibrium in response to changing conditions.