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 -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. \(655 \mathrm{g} \mathrm{CaCO}_{3}, 95.0 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=2.55 \mathrm{atm}\) 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. \(715 \mathrm{g} \mathrm{CaCO}_{3}, 813 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{atm}\)

Step by step solution

01

Write down the Kp expression for the given reaction

For the given reaction: \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\), the Kp expression is the pressure of CO2 (since solid concentrations are not accounted for): \[K_p = P_{CO_2}\]

02

Calculate Qp for each mixture

Calculate Qp for each mixture using the given amounts and the reaction chamber size (50.0 L): a. \( P_{CO_2}=2.55 \) atm \(Q_{p(a)} = 2.55\) b. \( P_{CO_2}=1.04 \) atm \(Q_{p(b)} = 1.04\) c. \( P_{CO_2}=1.04 \) atm \(Q_{p(c)} = 1.04\) d. \( P_{CO_2}=0.211 \) atm \(Q_{p(d)} = 0.211\)

03

Compare Qp to Kp for each mixture and determine the direction of the reaction

Compare the calculated Qp values against the given Kp value (1.04) to determine if the reaction will move towards equilibrium: a. \(Q_{p(a)} = 2.55\) > \(K_p = 1.04\) → The reaction needs to move left towards higher calcium carbonate concentration and lower CO2 and calcium oxide concentrations. Therefore, the amount of calcium oxide will decrease. b. \(Q_{p(b)} = 1.04\) = \(K_p = 1.04\) → This mixture is already at equilibrium, so the amount of calcium oxide will remain the same. c. \(Q_{p(c)} = 1.04\) = \(K_p = 1.04\) → This mixture is also already at equilibrium, so the amount of calcium oxide will remain the same. d. \(Q_{p(d)} = 0.211\) < \(K_p = 1.04\) → The reaction needs to move right towards higher CO2 and calcium oxide concentrations and lower calcium carbonate concentration. Therefore, the amount of calcium oxide will increase.

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