In which of the following cases will the LEAST time be required to arrive at equilibrium? K below refers to the equilibrium cons
1 answer:
Answer: Option (4) is the correct answer.
Explanation:
It is known that equilibrium constant is represented as follows for any general reaction.
K =
As equilibrium constant is directly proportional to the concentration of products so more is the value of equilibrium constant more will be the number of products formed.
As a result, more is the time taken by the reaction to reach towards equilibrium. Whereas smaller is the value of equilibrium constant more rapidly it will reach towards the equilibrium.
Thus, we can conclude that cases where K is a very small number will require the LEAST time to arrive at equilibrium.
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Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.