Question

At equilibrium, both the forward reaction and the reverse reaction occur simultaneously and at the same rate. Since the rates are the same, the concentrations of the reactants and products do not change. An equilibrium constant expression can be developed, which is a mathematical expression indicating the relationship between the reactants and the products. The equilibrium concentrations can be substituted into the equilibrium expression to determine the equilibrium constant, K. The equilibrium constant can also be denoted as "K", "Kc" or "Keq". What is the concentration of NO, if the concentration of N2 and O2 were measured to be 152 M, and the equilibrium constant, K, is 2.0 × 10-9?

Answer 1

The concentration of NO at equilibrium is calculated using the equilibrium constant and the concentrations of N2 and O2. By rearranging the equilibrium constant expression and substituting given values, the equilibrium concentration of NO is determined to be approximately 6.79 × 10^-3 M.

Step-by-step explanation:

Calculating the Concentration of NO at Equilibrium

The question involves finding the concentration of nitrogen monoxide (NO) at equilibrium, given the equilibrium constant (K) and the concentrations of nitrogen (N2) and oxygen (O2). The balanced chemical equation for the formation of NO from N2 and O2 is:

N2 + O2 → 2NO

The equilibrium constant expression for this reaction can be written as:

K = [NO]2 / ([N2][O2])

To find the concentration of NO, we rearrange the expression to solve for [NO]:

[NO] = √(K[N2][O2])

Substituting the given values:

[NO] = √(2.0 × 10-9 × 152 M × 152 M)

[NO] ≈ √(4.608 × 10-5 M2)

[NO] ≈ 6.79 × 10-3 M

The concentration of NO at equilibrium is approximately 6.79 × 10-3 M.

Answer 2

Answer : The concentration of NO is,
`6.8* 10^(-3)M`

Solution : Given,

Concentration of
`N_2` and
`O_2` = 152 M

Equilibrium constant,
`K_c` =
`2.0* 10^(-9)`

The given equilibrium reaction is,

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

The expression of
`K_c` will be,

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

Now put all the given values in this expression, we get:

`2.0* 10^(-9)=([NO]^2)/(152* 152)`

`[NO]=6.8* 10^(-3)M`

Therefore, the concentration of NO is,
`6.8* 10^(-3)M`