Respuesta :
Answer:
a) [H2O] = 0.05093 M
[Cl2O] = 0.01843 M
[HClO] = 0.00914 M
b) [H2O] = 0.2174M
[Cl2O] = 0.2174M
[HClO] = 0.0652 M
Explanation:
Step 1: Data given
Temperature = 25 °C
K = 0.090
Step 2: The balanced equation
H2O(g) + Cl2O(g) ⇆ 2HOCl(g)
(a) 1.0g H2O and 2.0g Cl2O are mixed in a1.0-L flask.
moles H20 = 1.00 / 18.02 g/mol = 0.0555 moles
Moles Cl2O = 2.00 grams / 86.91 g/mol = 0.0230 moles
Step 3: The initial concentration
[H2O] = 0.0555 M
[Cl2O]= 0.0230 M
[HOCl] = 0M
Step 4: Calculate concentration at equilibrium
For 1 mol H2O we need 1 mol Cl2O to produce 2 moles of HOCl
There will react X of H2O and X of Cl2O, there will be produce 2X of HOCl
The concentration at the equilibrium is:
[H2O] = (0.0555-X) M
[Cl2O]= (0.0230-X)M
[HClO] = 2X
Kc = 0.090
Kc = [HOCl]² / [H2O][Cl2O]
0.090 = 4X² / (0.0555-X)(0.0230-X)
K = 0.090 = 4X² / (0.072-X)(0.018-X)
X = 0.0045 7
[H2O] = 0.0555 - 0.00457 = 0.05093 M
[Cl2O] = 0.0230 - 0.00457 = 0.01843 M
[HClO] = 2*0.00457 = 0.00914 M
b) 1.0 mol pure HOCl is placed in a 2.0-L flask.
H2O(g) + Cl2O(g) ⇆ 2HOCl(g)
Step 1: The initial concentration
[H2O] = 0 M
[Cl2O]= 0M
[HOCl] = 1.00 mol / 2L = 0.500 M
Step 2: The concentration at the equilibrium
For 1 mol H2O we need 1 mol Cl2O to produce 2 moles of HOCl
There will be produced X of H2O and X of Cl2O, while there will be consumed 2X of HOCl
[H2O] = XM
[Cl2O]= XM
[HOCl] = (0.500 -2X) M
Kc = 0.090 = (0.50 - 2X)² / X²
0.090 = (0.0025 -2 + 4X²) / X²
X =0.2174
[H2O] = 0.2174M
[Cl2O] = 0.2174M
[HClO] = 0.500 - 2*0.2174 = 0.0652 M
The concentration are as follows
(a) [H2O] = 0.05093 M
[Cl2O] = 0.01843 M
[HClO] = 0.00914 M
(b) [H2O] = 0.2174M
[Cl2O] = 0.2174M
[HClO] = 0.0652 M
Given that Temperature = 25 °C and K = 0.09
(a) 1.0g H2O and 2.0g Cl2O are mixed in a 1.0-L flask.
moles H20 = 1.00 / 18.02 g/mol = 0.0555 moles
Moles Cl2O = 2.00 grams / 86.91 g/mol = 0.0230 moles
[H2O] = 0.0555 M
[Cl2O]= 0.0230 M
[HOCl] = 0M
The concentration at equilibrium
For 1 mol H2O, 1 mol Cl2O is needed to produce 2 moles of HOCl
The amount of reactant is X moles of H2O and X moles of Cl2O, to produce 2X of HOCl
The concentration at the equilibrium is:
[H2O] = (0.0555-X) M
[Cl2O]= (0.0230-X)M
[HClO] = 2X
Kc = 0.090
Kc = [HOCl]² / [H2O][Cl2O]
0.090 = 4X² / (0.0555-X)(0.0230-X)
K = 0.090 = 4X² / (0.072-X)(0.018-X)
X = 0.0045 7
[H2O] = 0.0555 - 0.00457 = 0.05093 M
[Cl2O] = 0.0230 - 0.00457 = 0.01843 M
[HClO] = 2*0.00457 = 0.00914 M
(b) 1.0 mol pure HOCl is placed in a 2.0-L flask.
H2O(g) + Cl2O(g) ⇆ 2HOCl(g)
The initial concentration
[H2O] = 0 M
[Cl2O]= 0M
[HOCl] = 1.00 mol / 2L = 0.500 M
The concentration at the equilibrium
For 1 mol H2O we need 1 mol Cl2O to produce 2 moles of HOCl
The production of X moles of H2O and X moles of Cl2O, 2X of moles HOCl will be consumed
[H2O] = XM
[Cl2O]= XM
[HOCl] = (0.500 -2X) M
Kc = 0.090 = (0.50 - 2X)² / X²
0.090 = (0.0025 -2 + 4X²) / X²
X =0.2174
[H2O] = 0.2174M
[Cl2O] = 0.2174M
[HClO] = 0.500 - 2*0.2174 = 0.0652 M
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