0.010 mol of chloral hydrate, CCl₃CH(OH)₃, is dissolved in a pure organic solvent - VCE - SSCE Chemistry - Question 6 - 2002 - Paper 1
Question 6
0.010 mol of chloral hydrate, CCl₃CH(OH)₃, is dissolved in a pure organic solvent. The resulting solution is made up to one litre exactly. In this solvent, the chlor... show full transcript
Worked Solution & Example Answer:0.010 mol of chloral hydrate, CCl₃CH(OH)₃, is dissolved in a pure organic solvent - VCE - SSCE Chemistry - Question 6 - 2002 - Paper 1
Step 1
Determine the Equilibrium Reaction Expressions
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Answer
For the given reaction, the equilibrium constant K is expressed as:
K=[CCl3CH(OH)3][CCl3CHO][H2O]
At equilibrium, we know the concentration of water, ([H₂O] = 0.0020 , M). The initial concentration of chloral hydrate was 0.010 M.
Step 2
Calculate Initial and Change in Concentrations
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Answer
Let ( x ) be the change in concentration of CCl₃CHO and H₂O at equilibrium. Since water is produced, at equilibrium:
([CCl₃CHO] = x )
([CCl₃CH(OH)₃] = (0.010 - x) )
Given that ( [H₂O] = 0.0020 ), we can substitute ( x = 0.0020 ). Thus, the equilibrium concentrations are:
([CCl₃CHO] = 0.0020 )
([CCl₃CH(OH)₃] = 0.010 - 0.0020 = 0.0080 )
Step 3
Substitute Concentrations into the K Expression
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Answer
Now substituting the equilibrium concentrations into the equilibrium expression:
K=[0.0080][0.0020][0.0020]
Calculating this gives:
K=0.00804.0×10−4=5.0×10−4
Step 4
Select the Correct Answer
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Answer
From the calculations, the equilibrium constant for the reaction at this temperature is:
