Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a \(2.00-\mathrm{L}\) vessel is found to contain \(0.0406 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH},\) \(0.170 \mathrm{~mol} \mathrm{CO},\) and \(0.302 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K} .\) Calculate \(K_{c}\) at this temperature.

Short Answer

Expert verified
The equilibrium constant, \(K_c\), for the formation of methanol from carbon monoxide and hydrogen gas at 500 K is 1.96.

Step by step solution

01

(Calculate the molar concentrations of each species)

(To calculate the molar concentrations of each species, divide the number of moles of each species by the volume of the reaction vessel.) [CH₃OH] = \(\frac{0.0406 \ \text{mol}}{2.00 \ \text{L}}\) = 0.0203 M [CO] = \(\frac{0.170 \ \text{mol}}{2.00 \ \text{L}}\) = 0.0850 M [H₂] = \(\frac{0.302 \ \text{mol}}{2.00 \ \text{L}}\) = 0.151 M Step 2: Calculate the equilibrium constant, Kc
02

(Calculate the equilibrium constant, Kc)

(To calculate the equilibrium constant, Kc, for the reaction at 500 K, substitute the equilibrium molar concentrations of each species in the expression for Kc.) \(K_c = \frac{[CH_3OH]}{[CO][H_2]^2} = \frac{0.0203}{(0.0850)(0.151)^2} = 1.96\) So, the value of \(K_c\) for the formation of methanol from carbon monoxide and hydrogen gas at 500 K is 1.96.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Understanding chemical equilibrium is pivotal in grasping how chemical reactions occur under certain conditions. In a chemical reaction, reactants convert into products, and, in a reversible reaction, products can revert back to reactants. Equilibrium is the state at which the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in concentration of reactants and products over time. However, it's critical to note that this does not mean the reactants and products are present in equal amounts, but rather that their rates of formation are equivalent.

Equilibrium can be influenced by changes in concentration, pressure, volume, and temperature. For example, according to Le Chatelier's Principle, if we change the conditions of a reaction at equilibrium, the system will adjust to partially counteract the change and a new equilibrium will be established. In the given problem, methanol is produced in a closed container, which means at the given temperature, the system will reach equilibrium with specific concentrations of CO, H₂, and CH₃OH.
Molar Concentration
The molar concentration, often referred to as molarity and denoted by 'M', indicates the number of moles of a solute per liter of solution. It's given by the formula: \[ M = \frac{moles \text{ of solute}}{volume \text{ of solution in liters}} \]. To visualize this, imagine you have a certain amount of substance dissolved in a container; the molar concentration tells you how packed this container is with the solute's particles.

It's an essential metric in chemistry because it allows chemists to quantify the amount of substances involved in reactions in a precise, reproducible manner. When solving equilibrium problems like the one provided, it is necessary to calculate the molarity of each reactant and product present to determine the position of equilibrium and eventually the equilibrium constant, which provides insight into the extent of the reaction.
Reaction Quotient
The reaction quotient, Q, plays a crucial role in predicting the direction of a reaction. It is calculated using the same form as the equilibrium constant, K, but with the initial concentrations of the reactants and products instead of their concentrations at equilibrium.

Q can be compared to the equilibrium constant K to forecast whether a reaction will proceed forward (to make more products), reverse (to form more reactants), or remain at equilibrium. The mathematical representation is \[ Q = \frac{[products]}{[reactants]} \], where the concentrations are raised to the power of their stoichiometric coefficients. If Q < K, the reaction will tend to produce more products. If Q > K, the reaction will favor the reactants. When Q = K, the reaction is at equilibrium, meaning that the concentrations of products and reactants will remain constant over time, just as demonstrated in the methanol synthesis problem you're studying.

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Most popular questions from this chapter

For the equilibrium $$2 \mathrm{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If 0.025 atm of IBr is placed in a 2.0-L container, what is the partial pressure of all substances after equilibrium is reached?

Two different proteins \(\mathrm{X}\) and \(\mathrm{Y}\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\). The proteins bind in a 1: 1 ratio to form XY. A solution that is initially \(1.00 \mathrm{~m} M\) in each protein is allowed to reach equilibrium. At equilibrium, \(0.20 \mathrm{~m} M\) of free \(\mathrm{X}\) and \(0.20 \mathrm{~m} M\) of free \(\mathrm{Y}\) remain. What is \(K_{c}\) for the reaction?

Can the equilibrium constant ever be a negative number? Explain.

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\) (f) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned}\mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v)\rightleftharpoons & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(s o l v)+\mathrm{H}_{2} \mathrm{O}(so l v)\end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is \(6.68 .\) A pharmaceutical chemist makes up \(15.0 \mathrm{~L}\) of a solution that is initially \(0.275 \mathrm{M}\) in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

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