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Chapter 15: Problem 31

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 -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 reaction \(\mathrm{CO}(g)+2\,\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) at 500 K is approximately 1.15. This is calculated using the molar concentrations of each substance in the equilibrium mixture: \([CH_3OH] = 0.0203\,\text{M},\ [CO] = 0.0850\,\text{M}\), and \([H_2] = 0.151\,\text{M}\).

Step by step solution

01

Determine the molar concentrations

In order to determine the molar concentrations, we need to divide the molar amounts by the volume of the container (2.00 L): For methanol: \([CH_3OH] = \frac{0.0406\,\text{mol}}{2.00\,\text{L}}\) For carbon monoxide: \([CO] = \frac{0.170\,\text{mol}}{2.00\,\text{L}}\) For hydrogen gas: \([H_2] = \frac{0.302\,\text{mol}}{2.00\,\text{L}}\)
02

Calculate the molar concentrations

Now, let's perform the calculations: \([CH_3OH] = \frac{0.0406\,\text{mol}}{2.00\,\text{L}} = 0.0203\,\text{M}\) \([CO] = \frac{0.170\,\text{mol}}{2.00\,\text{L}} = 0.0850\,\text{M}\) \([H_2] = \frac{0.302\,\text{mol}}{2.00\,\text{L}} = 0.151\,\text{M}\)
03

Write the expression for \(K_c\)

The equilibrium constant for the reaction \(\mathrm{CO}(g)+2\,\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) is equal to the product of the concentrations of the products divided by the product of the concentrations of the reactants, with each concentration raised to the power of its stoichiometric coefficient. In this case, the expression for \(K_c\) is: \(K_c = \frac{[CH_3OH]}{[CO] \times [H_2]^2}\)
04

Calculate \(K_c\)

Now, let's substitute the molar concentrations determined in step 2 into the expression for \(K_c\): \(K_c = \frac{0.0203\,\text{M}}{(0.0850\,\text{M})\times(0.151\,\text{M})^2} \) To find the value of \(K_c\), compute the expression above: \(K_c = \frac{0.0203}{(0.0850)(0.151)^2} \approx 1.15 \) #Conclusion# The equilibrium constant, \(K_c\), for the given reaction at 500 K is approximately 1.15.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rates of the forward reaction and the reverse reaction are equal, resulting in no net change in the amounts of products and reactants. It's important to note that chemical equilibrium refers to the dynamic balance of reactions, meaning that the chemical species are still reacting with each other, but the overall concentrations remain stable over time.

In the context of the exercise, when carbon monoxide and hydrogen react to form methanol, they reach a point where the formation of methanol from the reactants is equal to the decomposition of methanol back into carbon monoxide and hydrogen. At this point, the reaction has achieved equilibrium. The equilibrium is not static; the reactants continue to form products and the products continue to form reactants, but the overall concentration of each species remains constant. Understanding the concept of chemical equilibrium is crucial in calculating the equilibrium constant, which provides a quantitative measure of the reaction's position at equilibrium.
Molar Concentration
Molar concentration, also known as molarity and represented by the symbol 'M,' is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This unit of concentration is commonly used in chemistry because it allows scientists to easily calculate the volumes and moles involved in chemical reactions and solutions.

Calculation of Molar Concentration

As shown in the exercise, calculating molar concentration involves dividing the number of moles of the substance by the volume of the solution in liters. For instance, to find the molar concentration of methanol, we take the moles of methanol (0.0406 mol) and divide it by the volume of the vessel containing the reaction mixture (2.00 L), resulting in a molarity of 0.0203 M.

Understanding molar concentrations is fundamental in determining the quantities of reactants and products at equilibrium, which is crucial when calculating the equilibrium constant for a chemical reaction.
Reaction Quotient
The reaction quotient, denoted as 'Q', is a measure that tells us the direction in which a reaction will proceed to reach equilibrium. It is calculated by taking the concentrations (or pressures for gases) of the products raised to the power of their stoichiometric coefficients and dividing by the concentrations of the reactants, also raised to the power of their stoichiometric coefficients, similar to the equilibrium constant, 'K'.

Comparing Q to K

If 'Q' is less than 'K', the reaction will proceed in the forward direction to form more products. If 'Q' is greater than 'K', the reaction will proceed in the reverse direction to form more reactants. When 'Q' equals 'K', the system is at equilibrium. However, in this exercise, at equilibrium, we find 'K' directly because the concentrations are given for the equilibrium state. The significance of 'Q' lies in its ability to predict the shift of the reaction under non-equilibrium conditions, thereby guiding chemists in adjusting reaction conditions to maximize the yield of desired products.

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Most popular questions from this chapter

Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{B}(g) K_{c}\) and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{B}(g) . \mathrm{A}\) flask is charged with 0.75 atm of pure \(\mathrm{A},\) after which it is allowed to reach equilibrium at \(0^{\circ} \mathrm{C}\) . At equilibrium, the partial pressure of \(\mathrm{A}\) is 0.36 atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?(\mathbf{c})\) What could we do to maximize the yield of B?

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(F_{2}(g) \rightleftharpoons 2 \mathrm{F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{K},\) does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by 100 \(\mathrm{K}\) , does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

At \(25^{\circ} \mathrm{C},\) the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4} .\) What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in a saturated solution of \(\mathrm{CaCrO}_{4} ?\)

The water-gas shift reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) is used industrially to produce hydrogen. The reaction enthalpy is \(\Delta H^{\circ}=-41 \mathrm{kJ}\) . (a) To increase the equilibrium yield of hydrogen would you use high or low temperature? ( b) Could you increase the equilibrium yield of hydrogen by controlling the pressure of this reaction? If so would high or low pressure favor formation of \(\mathrm{H}_{2}(g) ?\)
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