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Chapter 15: Problem 86

The equilibrium constant \(K_{c}\) for \(C(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) 2 \(\mathrm{CO}(g)\) is 1.9 at 1000 \(\mathrm{K}\) and 0.133 at 298 \(\mathrm{K}\) . (a) If excess\(\mathrm{C}\) is allowed to react with 25.0 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) in a 3.00 -L vessel at \(1000 \mathrm{K},\) how many grams of CO are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

Short Answer

Expert verified
(a) 15.74 g of CO are produced. (b) 3.373 g of C are consumed. (c) The yield of CO will be smaller if a smaller vessel is used for the reaction. (d) The reaction is endothermic.

Step by step solution

01

Calculate the initial moles of CO2

Since we are given the initial mass of CO2 and the volume of the vessel, we can calculate the moles of CO2 before the reaction takes place: \(n_{CO_2(initial)} = \frac{25.0 \, g}{44.01 \, g/mol} = 0.568 \, mol \)
02

Relate the changes in moles of CO2, CO, and C

Let the change in the number of moles of CO2, CO, and C be represented by x. At equilibrium: \(n_{CO_2(eq)} = 0.568 - x \, mol\) \(n_{CO(eq)} = 2x \, mol\)
03

Find moles of CO at equilibrium

We can use the equilibrium constant expression to find the moles of CO at equilibrium for 1000 K: \(K_{c} = 1.9 = \frac{[CO]^2}{[CO_{2}]} = \frac{(2x)^2}{(0.568 - x)}\) Solve for x: \(x \approx 0.281 \, mol\)
04

Calculate the mass of CO produced and C consumed

Now we can calculate the mass of CO produced and C consumed at equilibrium using their respective molar masses: Mass of CO produced: \(mass_{CO} = n_{CO(eq)} \times \frac{28.01 \, g}{1 \, mol} = 2(0.281) \times 28.01 = 15.74 \, g \) Mass of C consumed: \(mass_{C} = x \times \frac{12.01 \, g}{1 \, mol} = 0.281 \times 12.01 = 3.373 \, g \)
05

Discuss the effect of vessel size on the yield of CO

Using a smaller vessel will result in an increase in pressure, which will favor the side of the reaction with fewer moles of gas. In this case, since the reaction is producing more moles of CO on the right side, a smaller volume will result in a smaller yield of CO.
06

Determine if the reaction is endothermic or exothermic

From the given data, we can see that \(K_c\) at 1000 K is greater than \(K_c\) at 298 K: \(K_{c(1000 \, K)} > K_{c(298 \, K)} \) This indicates that the reaction is favored at higher temperatures, which means that the reaction is endothermic, as an increase in temperature favors the endothermic side of the equilibrium. Answers: a) The mass of CO produced at equilibrium at 1000 K is 15.74 g. b) The mass of C consumed at equilibrium at 1000 K is 3.373 g. c) The yield of CO will be smaller if a smaller vessel is used for the reaction. d) The reaction is endothermic.

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Most popular questions from this chapter

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g) . \mathrm{A} 7.5-\mathrm{L}\) gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{atm}, P_{\mathrm{Cl}_{2}}=0.157 \mathrm{atm},\) and \(P_{\mathrm{PCl}_{\mathrm{s}}}=1.30 \mathrm{atm}\) (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at 450 \(\mathrm{K}\)

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