At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ At a low temperature, dry ice (solid \(\mathrm{CO}_{2} ),\) calcium oxide, and calcium carbonate are introduced into a \(50.0-\mathrm{L}\) reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C},\) resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2} .\) For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{g} \mathrm{CaCO}_{3}, 95.0 \mathrm{g}\) CaO, \(P_{\mathrm{CO}_{2}}=2.55 \mathrm{atm}\) b. $780 \mathrm{g} \mathrm{CaCO}_{3}, 1.00 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}$ c. $0.14 \mathrm{g} \mathrm{CaCO}_{3}, 5000 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}$ d. $715 \mathrm{g} \mathrm{CaCO}_{3}, 813 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{atm}$

Short Answer

Expert verified
a. The initial amount of CaO will decrease. b. The initial amount of CaO will remain the same. c. The initial amount of CaO will increase. d. The initial amount of CaO will increase.

Step by step solution

01

Case a: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 655g CaCO3, 95.0g CaO, and CO2 pressure (P_CO2) of 2.55 atm to moles.
02

Case b: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 780g CaCO3, 1.00g CaO, and CO2 pressure (P_CO2) of 1.04 atm to moles.
03

Case c: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 0.14g CaCO3, 5000g CaO, and CO2 pressure (P_CO2) of 1.04 atm to moles.
04

Case d: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 715g CaCO3, 813g CaO, and CO2 pressure (P_CO2) of 0.211 atm to moles. #Step 2: Calculate the Reaction Quotient (Qp)# Now we will calculate the reaction quotient (Qp) for each case, using the formula Qp = P_CO2.
05

Case a: Calculate Qp

Calculate Qp for Case a using P_CO2 of 2.55 atm.
06

Case b: Calculate Qp

Calculate Qp for Case b using P_CO2 of 1.04 atm.
07

Case c: Calculate Qp

Calculate Qp for Case c using P_CO2 of 1.04 atm.
08

Case d: Calculate Qp

Calculate Qp for Case d using P_CO2 of 0.211 atm. #Step 3: Compare Qp to Kp for Each Case and Determine System's Behavior# Now, we will compare Qp to the given Kp value (1.04) for each case to determine if the reaction will shift right (forward), left (reverse), or remain unchanged.
09

Case a: Compare Qp to Kp

Compare Qp to Kp for Case a to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.
10

Case b: Compare Qp to Kp

Compare Qp to Kp for Case b to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.
11

Case c: Compare Qp to Kp

Compare Qp to Kp for Case c to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.
12

Case d: Compare Qp to Kp

Compare Qp to Kp for Case d to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.

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Most popular questions from this chapter

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas are placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to \(\mathrm{NO}\) and \(\mathrm{O}_{2} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 \(\mathrm{mol} / \mathrm{L}\) . Calculate the value of \(K\) for this reaction.

In a given experiment, 5.2 moles of pure NOCl were placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

In Section 13.1 of your text, it is mentioned that equilibrium is reached in a "closed system." What is meant by the term "closed system," and why is it necessary to have a closed system in order for a system to reach equilibrium? Explain why equilibrium is not reached in an open system.

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: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q)$$ 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 1.24\(M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Explain the difference between \(K, K_{\mathrm{p}},\) and \(Q\)

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