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:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
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B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.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.
Answer:
depending on what school you go to and what classes you are allowed to take in which grades, you should take forensics.
Answer:
The value of the equilibrium constant = 5.213
Explanation:
Here
(equilibrium constant) is referred to as the partial pressure of product divided by the partial pressure of reactant with each pressure term raised to power that is equal to its stoichiometric coefficient in balanced equation
.
As such only gas appear in
expression as solids takes a value of 1;
SO ; in the given equation from the question:
2 A (g) + B (s) ----> 2 C(s) + D (g)
![K_p = \dfrac{[D]}{[A]^2}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BD%5D%7D%7B%5BA%5D%5E2%7D)


The value of the equilibrium constant = 5.213
No, it does not. I believe it's false.