Question

Consider the following reaction. N2(g) + O2(g) = 2 NO(g) (a) If Kc for this reaction is 1.5 ✕ 10−10, and a reaction system at equilibrium has an [N2] of 0.043 M and an [O2] of 0.043 M, what is the equilibrium concentration of NO? M (b) What is the value of Kc for the reaction 2 NO(g) equilibrium reaction arrow N2(g) + O2(g)? (c) Does the equilibrium in Part (b) favor the reactants or the products? the reactants the products neither reactants nor products

Answer 1

For a: The concentration of NO at equilibrium is
`5.27* 10^(-7)M`

For b: The value of
`K_c'` is
`6.66* 10^(10)`

For c: The correct answer is product favored.

Step-by-step explanation:

• For a:

The given chemical reaction follows:

`N_2(g)+O_2(g)\rightleftharpoons 2NO(g)`

The expression of
`K_c` for above equation follows:

`K_c=([NO]^2)/([N_2][O_2])`

We are given:

`K_c=1.5* 10^(-10)`

`[N_2]_(eq)=0.043M`

`[O_2]_(eq)=0.043M`

Putting values in above equation, we get:

`1.5* 10^(-10)=([NO]^2)/(0.043* 0.043)`

`[NO]=\sqrt{(1.5* 10^(-15)* 0.043* 0.043)}=5.27* 10^(-7)M`

Hence, the concentration of NO at equilibrium is
`5.27* 10^(-7)M`

• For b:

The given chemical reaction follows:

`2NO(g)\rightleftharpoons N_2(g)+O_2(g)`

The expression of
`K_c'` for above equation follows:

`K_c'=([N_2][O_2])/([NO]^2)`

As, the above reaction is the reverse of equation in part a. So, the value of
`K_c'` will be inverse of
`K_c`

`K_c'=(1)/(K_c)\\\\K_c'=(1)/(1.5* 10^(-10))=6.66* 10^(10)`

Hence, the value of
`K_c'` is
`6.66* 10^(10)`

• For c:

There are 3 conditions:

• When
  `K_(c)>1`; the reaction is product favored.
• When
  `K_(c)<1`; the reaction is reactant favored.
• When
  `K_(c)=1`; the reaction is in equilibrium.

For the reaction in part 'b', the value of
`K_c'` is
`6.66* 10^(10)`

The value of
`K_c'` is very high than 1. So, the equilibrium in part 'b' is product favored.

Hence, the correct answer is product favored.