<u>Answer:</u>
<u>For a:</u> The concentration of NO at equilibrium is ![5.27\times 10^{-7}M](https://tex.z-dn.net/?f=5.27%5Ctimes%2010%5E%7B-7%7DM)
<u>For b:</u> The value of
is ![6.66\times 10^{10}](https://tex.z-dn.net/?f=6.66%5Ctimes%2010%5E%7B10%7D)
<u>For c:</u> The correct answer is product favored.
<u>Explanation:</u>
The given chemical reaction follows:
![N_2(g)+O_2(g)\rightleftharpoons 2NO(g)](https://tex.z-dn.net/?f=N_2%28g%29%2BO_2%28g%29%5Crightleftharpoons%202NO%28g%29)
The expression of
for above equation follows:
![K_c=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)
We are given:
![K_c=1.5\times 10^{-10}](https://tex.z-dn.net/?f=K_c%3D1.5%5Ctimes%2010%5E%7B-10%7D)
![[N_2]_{eq}=0.043M](https://tex.z-dn.net/?f=%5BN_2%5D_%7Beq%7D%3D0.043M)
![[O_2]_{eq}=0.043M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.043M)
Putting values in above equation, we get:
![1.5\times 10^{-10}=\frac{[NO]^2}{0.043\times 0.043}](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-10%7D%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B0.043%5Ctimes%200.043%7D)
![[NO]=\sqrt{(1.5\times 10^{-15}\times 0.043\times 0.043)}=5.27\times 10^{-7}M](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Csqrt%7B%281.5%5Ctimes%2010%5E%7B-15%7D%5Ctimes%200.043%5Ctimes%200.043%29%7D%3D5.27%5Ctimes%2010%5E%7B-7%7DM)
Hence, the concentration of NO at equilibrium is ![5.27\times 10^{-7}M](https://tex.z-dn.net/?f=5.27%5Ctimes%2010%5E%7B-7%7DM)
The given chemical reaction follows:
![2NO(g)\rightleftharpoons N_2(g)+O_2(g)](https://tex.z-dn.net/?f=2NO%28g%29%5Crightleftharpoons%20N_2%28g%29%2BO_2%28g%29)
The expression of
for above equation follows:
![K_c'=\frac{[N_2][O_2]}{[NO]^2}](https://tex.z-dn.net/?f=K_c%27%3D%5Cfrac%7B%5BN_2%5D%5BO_2%5D%7D%7B%5BNO%5D%5E2%7D)
As, the above reaction is the reverse of equation in part a. So, the value of
will be inverse of ![K_c](https://tex.z-dn.net/?f=K_c)
![K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.5\times 10^{-10}}=6.66\times 10^{10}](https://tex.z-dn.net/?f=K_c%27%3D%5Cfrac%7B1%7D%7BK_c%7D%5C%5C%5C%5CK_c%27%3D%5Cfrac%7B1%7D%7B1.5%5Ctimes%2010%5E%7B-10%7D%7D%3D6.66%5Ctimes%2010%5E%7B10%7D)
Hence, the value of
is ![6.66\times 10^{10}](https://tex.z-dn.net/?f=6.66%5Ctimes%2010%5E%7B10%7D)
There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium.
For the reaction in part 'b', the value of
is ![6.66\times 10^{10}](https://tex.z-dn.net/?f=6.66%5Ctimes%2010%5E%7B10%7D)
The value of
is very high than 1. So, the equilibrium in part 'b' is product favored.
Hence, the correct answer is product favored.