When 2-propanol reacts to form an equilibrium mixture with propanone and hydrogen, which one of the following best represents how the rates of the forward and back reactions change over time?
A - VCE - SSCE Chemistry - Question 7 - 2007 - Paper 1
Question 7
When 2-propanol reacts to form an equilibrium mixture with propanone and hydrogen, which one of the following best represents how the rates of the forward and back r... show full transcript
Worked Solution & Example Answer:When 2-propanol reacts to form an equilibrium mixture with propanone and hydrogen, which one of the following best represents how the rates of the forward and back reactions change over time?
A - VCE - SSCE Chemistry - Question 7 - 2007 - Paper 1
Step 1
A. $C_3H_8O_{(g)} + H_2(g) \rightarrow C_3H_6O_{(g)}$
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Answer
This option incorrectly signifies that the forward reaction rate decreases to zero too quickly and the reverse reaction rate does not rise appropriately.
Step 2
B. $C_3H_8O_{(g)} + H_2(g) \rightarrow C_3H_6O_{(g)}$
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Answer
This option incorrectly depicts both reaction rates, failing to show the establishment of equilibrium.
Step 3
C. $C_3H_6O_{(g)} + H_2(g) \rightarrow C_3H_8O_{(g)}$
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Answer
This option incorrectly presents the rate of the reverse reaction starting high and does not show stabilization over time.
Step 4
D. $C_3H_8O_{(g)} + H_2(g) \rightarrow C_3H_6O_{(g)}$
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Answer
This is the correct representation. Initially, the forward reaction is faster, but over time, as the product concentration increases, the back reaction rate rises until both rates equalize at equilibrium.
