Five mol of calcium carbide are combined with 10 mol of water in a closed, rigid

Question

Five mol of calcium carbide are combined with 10 mol of water in a closed, rigid, high-pressure vessel of 1800cm^3 internal empty volume. Acetylene gas is produced by the reaction: CaC2(s) + 2H_2O(l) rightarrow C_2H_2(g) + Ca(OH)_2(s) The vessel contains packing with a porosity of 40% to prevent explosive decomposition of the acetylene. Initial conditions are 25 degree C and 1 hour and the reaction goes to completion. The reaction is exothermic, but owing to heat transfer, the final temperature is only 125 degree C. Determine the final pressure in the vessel. Note: At 125 degree C, the molar volume of Ca(OH)_2 is 33.0 cm^3 mol^-1. Ignore the effects of any gases (e.g., air) initially present in the vessel.

Explanation / Answer

Answer is not 197.8 bar it is 297.8 bar may be typo error

I didn't follow that whole porosity bit - I'll take it to mean that the total volume of gas and solid is only
V = 40% (1800 mL) = 0.72 L

The gas volume is less by the volume of calcium hydroxide
Vsolid = (5 mol CaC2) (1 mol Ca(OH)2 / 1 mol CaC2) (33 mL Ca(OH)2 / 1 mol Ca(OH)2) = 0.165 L
Vgas = V - Vsolid = 0.555 L

The amount of acetylene produced is
n = (5 mol CaC2) (1 mol C2H2 / 1 mol CaC2) = 5 mol C2H2
If it behaves like an ideal gas (probably an absurd assumption, since you're reacting in a high-pressure container), pVgas = nRT, then its partial pressure is
p = nRT / Vgas = (5 mol) (0.08206 L atm / K mol) (398.15 K) / (0.555 L) = 294 atm

since 1 atm = 1.01325 bar

Hence 294 atm = 297.896 bar

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