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sammy [17]
4 years ago
5

In this reaction, what effect does a change in the concentration of substances have on the reverse reaction rate? CaCO3(s) ⇌ CaO

(g) + CO2(g)
A. The rate of the reverse reaction increases with an increase in the concentration of CaCO3(s).
B. The rate of the reverse reaction increases with an increase in the concentration of CaO(g) and CO2(g).
C. The rate of the reverse reaction decreases with a decrease in the concentration of CaCO3(s).
D. The rate of the reverse reaction decreases with an increase in the concentration of CaO(g) and CO2(g).
E. The rate of the reverse reaction increases with a decrease in the concentration of CaO(g) and CO2(g).
Chemistry
2 answers:
steposvetlana [31]4 years ago
7 0

Answer:

B

Explanation:

B on Edmentum

Oksi-84 [34.3K]4 years ago
5 0

Based on Le Chatelier's principle, when the equilibrium of a given reaction is disturbed due to some external factors like temperature, pressure or concentration, the equilibrium shifts in a direction to undo the effects of the change produced.

The given reaction is :-

CaCO3 (s) ↔ CaO (g) + CO2(g)

The forward reaction involves the formation of CaO and CO2 while the reverse involves formation of CaCO3.

If the concentration of the products CaO and CO2 is increased then the equilibrium will shift to a direction to counteract this increase. Hence, the products will tend to get used up and the reverse reaction will be favored.

Ans: B) the rate of reverse reaction will increase if the concentration of caO and CO2 is increased.


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Thus, we can conclude that the mole percent of A in their mixture is 46.7%.

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