For the following equilibrium system: CaCO_3(s) + H_2O(l) CO_2(g) + Ca(OH)_2(s)
ID: 1005796 • Letter: F
Question
For the following equilibrium system: CaCO_3(s) + H_2O(l) CO_2(g) + Ca(OH)_2(s) Delta H degree = + 113 kJ Will each one of the following changes result in making more or less calcium hydroxide Decrease the temperature at constant pressure (with no phase change) Decrease the volume at constant temperature Add additional calcium hydroxide Increase the partial pressure of carbon dioxide Remove some of the water Increase the pressure by adding some argon A toxicologist studying mustard gas, S(CH_2CH_2Cl)_2, a blistering agent, prepares mixture of 0.875 M SCl_2 and 0.473 M S(CH_2CH_2Cl)_2 in 1.00 L & allows it to react at 80.0 degree C: SCl_2: (g) + 2C_2H_4(g) S(CH_2CH_2Cl)_2 (g) If at equilibrium [S(CH_2CH_2Cl)_2] = 0.350 M, complete the following table: (M) Calculate K_c and K_p for this reaction:Explanation / Answer
Part A ) For the given endothermic reaction, formation of Ca(OH)2 is governed by LeChatellier's principle.
(a) Decreasing Temperature : Less Ca(OH)2 formed
(b) Decrease Volume : Less Ca(OH)2 formed
(c) add Ca(OH)2 : Less Ca(OH)2 formed
(d) Increase partial pressure of CO2 : Less Ca(OH)2 formed
(e) Remove water : : Less Ca(OH)2 formed
(f) Increase P by Ar : Less Ca(OH)2 formed
Part B ) For the reaction,
(a) Table
(M) SCl2 C2H4 S(CH2CH2Cl)2
Initial 0.875 - 0.473
Change 0.123 - 0.123
Eq. 0.752 - 0.350
Kc = 0.350/0.752 = 0.465
Kp = Kc(RT)dn = 0.465 x (0.0821 x (80 + 273))^-2 = 5.53 x 10^-4