Respuesta :
Answer:
For a: The concentration of NO at equilibrium is [tex]5.27\times 10^{-7}M[/tex]
For b: The value of [tex]K_c'[/tex] is [tex]6.66\times 10^{10}[/tex]
For c: The correct answer is product favored.
Explanation:
- For a:
The given chemical reaction follows:
[tex]N_2(g)+O_2(g)\rightleftharpoons 2NO(g)[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[NO]^2}{[N_2][O_2]}[/tex]
We are given:
[tex]K_c=1.5\times 10^{-10}[/tex]
[tex][N_2]_{eq}=0.043M[/tex]
[tex][O_2]_{eq}=0.043M[/tex]
Putting values in above equation, we get:
[tex]1.5\times 10^{-10}=\frac{[NO]^2}{0.043\times 0.043}[/tex]
[tex][NO]=\sqrt{(1.5\times 10^{-15}\times 0.043\times 0.043)}=5.27\times 10^{-7}M[/tex]
Hence, the concentration of NO at equilibrium is [tex]5.27\times 10^{-7}M[/tex]
- For b:
The given chemical reaction follows:
[tex]2NO(g)\rightleftharpoons N_2(g)+O_2(g)[/tex]
The expression of [tex]K_c'[/tex] for above equation follows:
[tex]K_c'=\frac{[N_2][O_2]}{[NO]^2}[/tex]
As, the above reaction is the reverse of equation in part a. So, the value of [tex]K_c'[/tex] will be inverse of [tex]K_c[/tex]
[tex]K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.5\times 10^{-10}}=6.66\times 10^{10}[/tex]
Hence, the value of [tex]K_c'[/tex] is [tex]6.66\times 10^{10}[/tex]
- For c:
There are 3 conditions:
- When [tex]K_{c}>1[/tex]; the reaction is product favored.
- When [tex]K_{c}<1[/tex]; the reaction is reactant favored.
- When [tex]K_{c}=1[/tex]; the reaction is in equilibrium.
For the reaction in part 'b', the value of [tex]K_c'[/tex] is [tex]6.66\times 10^{10}[/tex]
The value of [tex]K_c'[/tex] is very high than 1. So, the equilibrium in part 'b' is product favored.
Hence, the correct answer is product favored.
