Answer :
The equilibrium favors NO and
.
(1) The value of equilibrium constant for this reaction is, 76.9
(2) The value of equilibrium constant for this reaction is, 8.77
Explanation:
The given chemical equation is:
![2NO(g)+Br_2(g)\rightarrow 2NOBr(g)](https://tex.z-dn.net/?f=2NO%28g%29%2BBr_2%28g%29%5Crightarrow%202NOBr%28g%29)
The value of equilibrium constant for the above equation is
.
The value of
that means equilibrium lies to the left side. Thus, the equilibrium favors NO and
.
We need to calculate the equilibrium constant for the given equation of above chemical equation, which is:
(1) ![2NOBr(g)\rightarrow 2NO(g)+Br_2(g)](https://tex.z-dn.net/?f=2NOBr%28g%29%5Crightarrow%202NO%28g%29%2BBr_2%28g%29)
The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.
The value of equilibrium constant for reverse reaction is:
![K_{c_1}=\frac{1}{K_c}](https://tex.z-dn.net/?f=K_%7Bc_1%7D%3D%5Cfrac%7B1%7D%7BK_c%7D)
![K_{c_1}=\frac{1}{1.3\times 10^{-2}}=76.9](https://tex.z-dn.net/?f=K_%7Bc_1%7D%3D%5Cfrac%7B1%7D%7B1.3%5Ctimes%2010%5E%7B-2%7D%7D%3D76.9)
Thus, the value of equilibrium constant for this reaction is, 76.9
(2) ![NOBr(g)\rightarrow NO(g)+\frac{1}{2}Br_2(g)](https://tex.z-dn.net/?f=NOBr%28g%29%5Crightarrow%20NO%28g%29%2B%5Cfrac%7B1%7D%7B2%7DBr_2%28g%29)
The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.
If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
![K_{c_2}=(\frac{1}{K_c})^{1/2}](https://tex.z-dn.net/?f=K_%7Bc_2%7D%3D%28%5Cfrac%7B1%7D%7BK_c%7D%29%5E%7B1%2F2%7D)
![K_{c_2}=(\frac{1}{1.3\times 10^{-2}})^{1/2}=8.77](https://tex.z-dn.net/?f=K_%7Bc_2%7D%3D%28%5Cfrac%7B1%7D%7B1.3%5Ctimes%2010%5E%7B-2%7D%7D%29%5E%7B1%2F2%7D%3D8.77)
Thus, the value of equilibrium constant for this reaction is, 8.77