Reaction Reversibility Worksheet With Answers Page 3
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1
2
3
4forward versus reverse reactions, and how concentration affects reaction rate).
As NO
molecules are produced, they also collide with each other. Occasionally they can
2
combine to form N
O
. Molecules never stop colliding. There is always the possibility that
2
4
any two molecules will react.
8.Assume the spectrophotometer cannot directly measure [N
O
], but can measure [NO
] because of its brown color. Explain how
2
4
2
the [N
O
] values would be calculated.
2
4
Two molecules of NO
are made from one molecule of N
O
. Take the number of NO
2
2
4
2
molecules present, and it is known that đ as many N
O
molecules broke apart. Subtract
2
4
the number reacted from the starting amount of N
O
and that gives the current amount;
2
4
for example:
[NO
] = 0.8 moles; 0.4 moles of N
O
reacted. [N
O
] = 1 mole – 0.4 = 0.6 moles
2
2
4
2
4
9.Define reaction reversibility.
A chemical reaction is reversible if the products can combine/react to recreate reactants.
10.Can the same equilibrium be reached starting with pure NO
? If so, how much NO
would be required?
2
2
NO
molecules can combine to give N
O
(that’s why this is a reversible reaction, after all)
2
2
4
2 NO
molecules can combine to make 1 N
O
.
2
2
4
If you started with 2.0 moles of pure NO
, then after 0.4 moles combined, you would get
2
0.2 moles of N
O
and you would have 1.6 moles of NO
remaining.
2
4
2
→
11. HI can form via the equation: H
(g) + I
(g)
2HI(g). 1 mol H
and 1 mol I
were placed in a
2
2
2
2
1 L vessel (no HI was present initially). At equilibrium [HI] was measured to be 1.56 mol/L. Sketch a graph for the equilibrium
reaction, clearly indicating the starting and equilibrium concentrations of all chemicals. (Be careful to draw the general shape of
the lines correctly: straight vs. curving up vs. curving down). N
Notice that two molecules of HI are
made for each one molecule of H
&
2
2
I
reacted.
1.8
2
[HI]
If 1.56 moles of HI were made, then
1.6
half that much H
and I
reacted =
2
2
1.4
0.78 moles of H
reacted and 0.78
2
1.2
moles of I
(2 •0.78 = 1.56).
2
1
Since you started with 1 mole of
each, they end up at a
0.8
concentration of 1–0.78 = 0.22
0.6
[H
& [I
]
]
moles.
0.4
2
2
0.2
A table shows the amounts in the
0
problem and how the amounts
Time
reacted lead to the amount at
equilibrium.
H
I
HI
2
2
Mole Ratio
1 :
1 :
2
1
1
0
Initial
Amount
Change in
0.78
0.78
1.56
amount
1.56
Equilibrium
1–
1-0.78=0.22
amount
0.78=0.22
12. Ammonia is produced via the reaction: N
(g) + 3H
g)
→
2NH
(g). 1 mol N
and 3 mol H
were placed in an empty 2 L
2
2(
3
2
2
container. The equilibrium [ ] of N
was 0.3 mol/L. Sketch a graph for this reaction.
2
N
H
NH
2
2
3
Mole Ratio
1
3
2
Initial
1
3
0
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