Module 3 EOC Assignment – Part2 Due by 3/3
Chapter 16  15, 23, 25, 27, 29
Chpt16 #15
•
(1) Provide ALL of the given information.
Include constants, density,
conversion factors, etc.
I
2
(g)
2 I(g)
= 3.76 x 10
⁻
³ at 1000K
0.105 mol of I
2
placed in a 12.3L flask at 1000K
•
(1) State the goal of the problem.
What are the concentrations of I
2
and I when the system comes to equilibrium
•
(1) List possible equations that may be used
•
(4) Solve the problem and explain the reason for each step of
multiplication, division, addition, or subtraction.
2 mol of I form for each mol of I
2
that reacts, so
then it’s 2I
and I
2
at equilibrium would be 8.54 a 10
3
– x.
(2I)
2
/ 8.54 x 10
3
= 4I
2
+ 3.76 x 10
3
(8.54 x 10
3
x) this needs to be changed to be
able to use the quadratic equation and I will change the “I” into an x:
4x
2
+ 3.76 x 10
3
x – 3.21 x 10
5
= 0 ; which = 2.40 x 10
3
At equilibrium = I
2
= 8.54 x 10
3
– 2.40 x 10
3
=
6.14 x 10
3
I = 2(2.40 x 10
3
) =
4.79 x 10
3
•
(1) Explain why your answer makes physical and mathematical sense.
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Module 3 EOC Assignment – Part2 Due by 3/3
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 Fall '08
 Thomas
 Chemistry, Addition, Multiplication, Elementary arithmetic, List possible equations

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