Transcript Practice Problem #2

When Compounds do not React
in a 1:1 Molar Ratio
• The molar ratios affect how the
concentrations change during a reaction.
• We must account for these ratios when we
solve ICE box problems.
• Specifically they will affect the CHANGE
line when we are doing these problems.
Practice Problem
Nitrosyl bromide (NOBr) decomposes
according to the reaction below:
2 NOBr  2 NO + Br2
0.500 M of NOBr is used initially and
when the reaction reaches
equilibrium, 0.410 M NOBr remain.
Determine the equilibrium constant!
(Keq)
Start with the Initial Line!
• We know the starting concentration of NOBr.
• What about the concentrations of our products?
• Do we know any other concentrations?
Initial
NOBr
NO
Br2
0.500
0
0
Change
Equilibrium
0.410
Next is the Change Line!
2 NOBr  2 NO + 1 Br2
• We don’t know the exact change amount yet,
but we know how they are going to change. Pay
attention to the coefficients!
NOBr
NO
Br2
Initial
0.500
0
0
Change
-2X
2X
X
Equilibrium
0.410
Solving for X
NOBr
Initial
• Using the info for NOBr, we
can solve for X, the change
in concentration:
0.500 0.500 M – 2x = 0.410 M
X = 0.045 M
•
But
remember
this
is
just
X.
Change
• We are looking for the
equilibrium constant (K) so
we need the rest of the
Equilibrium 0.410
values at equilibrium..
-2X
Solving the Concentrations at Equilibrium
• X = 0.045 M (Previous Slide)
NOBr
NO
Br2
Initial
0.500 M
0
0
Change
-2X
2X
X
Equilibrium
0.410 M 0.090 M
0.045 M
• Note: These numbers are plugged into the equation
2 NOBr  2 NO + Br2
• Remember coefficients K = [NO]2[Br ]
2
when writing the
2
[NOBr]
equilibrium expression.
Equilibrium
NOBr
NO
Br2
0.410
0.090
0.045
• Plug in the values at equilibrium into the
expression and solve for K:
K = [0.090]2[0.045] K = 0.0021 or 2.1x10-3
[0.410]2
Practice Problem – A different kind…
• Hydrogen gas reacts with Iodine gas to produce
hydroiodic acid. Write and balance the
chemical equation:
1 H2 + 1 I2  2 HI
• Calculate all concentrations at equilibrium
when initially 0.200 M of both reactants are
used and the equilibrium constant is equal to
64.0.
• What makes this problem different than
previous ones? – We are given the Keq value
and asked to solve for the concentrations.
Start with the Initial!
• Again, this one is different because we are not
given any concentrations at equilibrium. We are
going to solve for that upcoming.
Initial
H2
l2
Hl
0.200
0.200
0
??
??
??
Change
Equilibrium
Next Comes the Change!
1 H2 + 1 I2  2 HI
• Make sure you’ve balanced the equation first!
H2
l2
Hl
Initial
0.200
0.200
0
Change
-X
-X
2X
Equilibrium
Finally the Equilibrium!
• To fill in the rest of the boxes, it is always:
INITIAL + CHANGE = EQUILIBRIUM
• Great! Now it’s time for us to solve for X !
H2
l2
Hl
Initial
0.200
0.200
0
Change
-X
-X
2X
Equilibrium
0.200 – X 0.200 – X
2X
Wait a second…
• There is no column for us to solve for X!
H2
Equilibrium
l2
0.200 – X 0.200 – X
Hl
2X
• Fear not my downhearted disciples of chemistry!
We were given the Keq intially in the problem.
• We will write equilibrium expression and use the
Keq to solve for X. Then plug that in to get the
concentrations at equilibrium.
1 H2 + 1 I2  2 HI
• Remember coefficients
when writing the
equilibrium expression.
H2
Equilibrium 0.200 – X
K=
[HI]2
[H2][I2]
l2
Hl
0.200 – X
2X
• Plug in the values at equilibrium, and Keq= 64.0
into the expression and solve for X:
64.0 =
[2X]2
[0.200 – X][0.200 – X]
Let’s clean this up a little bit…
• In the bottom, two identical things are
multiplied together. That means we can ???
 64.0 =
[2X]2
2
[0.200
–
X]

• So what can we do to get X by itself?
• Take the square root!
8.0 = 2X
0.200 – X
Let’s clean this up a little bit…
8.0 =
2X
0.200 – X
• Need X, so multiply both sides by???
8.0 (0.200 – X) = 2 X *Distribute Through*
1.6 – 8 X = 2 X
*Combine Like Terms*
1.6 = 10 X
*Solve for X!*
0.16 = X
But remember that’s not the final answer.
That’s just X, now lets plug it into the
expression.
Yes, That is My Final Answer!
• From the last slide, we solved for
X = 0.16. Lets Plug it in!
H2
l2
Hl
0.200
X
0.200
0.040
––0.16
0.200
X
0.200
0.040
––0.16
2X
2(0.16)
0.32
• And that’s it! At equilibrium there will be
[0.040] H2, [0.040] I2, [0.32] HI