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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 10.00-L vessel is found to contain \(0.050 \mathrm{~mol}\) $\mathrm{CH}_{3} \mathrm{OH}, 0.850 \mathrm{~mol} \mathrm{CO},\( and \)0.750 \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 given reaction at 500 K can be calculated by first determining the concentrations of each species and then plugging them into the expression \(K_c = \frac{[\mathrm{CH_3OH}]}{[\mathrm{CO}][\mathrm{H_2}]^2}\). The concentrations are found to be: \([\mathrm{CH_3OH}] = 0.005 \text{ M}\), \([\mathrm{CO}] = 0.085 \text{ M}\), and \([\mathrm{H_2}] = 0.075 \text{ M}\). Substituting these values into the equilibrium constant expression, we get \(K_c = \frac{0.005}{(0.085)(0.075)^2} = 12.35\). Thus, the equilibrium constant at this temperature is 12.35.

Step by step solution

01

Write the equilibrium constant expression for the given reaction.

The general expression for an equilibrium constant is given by the formula: \[K_c = \frac{\text{Product Concentrations}^{a}}{\text{Reactant Concentrations}^{b}}\] where 'a' and 'b' are the stoichiometric coefficients of the balanced chemical reaction. For the given reaction, the equilibrium constant expression is: \[K_c = \frac{[\mathrm{CH_3OH}]}{[\mathrm{CO}][\mathrm{H_2}]^2}\]
02

Calculate the concentration of each species.

To find the concentration, we will use the following formula: \[\text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}}\] Calculate the concentration of each species in the mixture: \[[\mathrm{CH_3OH}] = \frac{0.050 \text{ mol}}{10.00 \text{ L}} = 0.005 \text{ M}\] \[[\mathrm{CO}] = \frac{0.850 \text{ mol}}{10.00 \text{ L}} = 0.085 \text{ M}\] \[[\mathrm{H_2}] = \frac{0.750 \text{ mol}}{10.00 \text{ L}} = 0.075 \text{ M}\]
03

Substitute the concentrations into the equilibrium constant expression.

Plug the calculated concentrations of the species from Step 2 into the expression for \(K_c\): \[K_c = \frac{0.005}{(0.085)(0.075)^2}\]
04

Calculate \(K_c\).

Calculate the equilibrium constant \(K_c\) using the values obtained in the previous step: \[K_c = \frac{0.005}{(0.085)(0.075)^2} = 12.35\] The equilibrium constant \(K_c\) for this reaction at 500 K is 12.35.

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Most popular questions from this chapter

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for $2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\( if \)K_{p}=0.0572$ at this temperature.

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction $4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\( in a fetus, compared to \)K_{c}$ for the same reaction in an adult.

For $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{2}\( at \)700 \mathrm{~K}$. In a 2.00-L vessel, the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (a) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)
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