Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0654 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C.

Answer 1

To calculate the equilibrium concentrations of N2O4 and NO2 at 25 °C in a vessel with an initial N2O4 concentration of 0.0654 M, we can use the equilibrium constant expression. The equilibrium concentrations will be approximately 0.0151 M for N2O4 and 0.0303 M for NO2.

Step-by-step explanation:

To calculate the equilibrium concentrations of N2O4 and NO2, we can use the equilibrium constant expression: Kc = [NO2]^2 / [N2O4]. Given an initial concentration of N2O4 = 0.0654 M and Kc = 4.64 × 10^-3, we can substitute these values into the expression to solve for the equilibrium concentrations.

Assuming the equilibrium concentrations of N2O4 and NO2 are x and 2x respectively (based on the stoichiometry of the reaction), we can write the equation: 4.64 × 10^-3 = (2x)^2 / x. Solving this equation will give us the equilibrium concentrations of N2O4 and NO2.

Therefore, the equilibrium concentrations of N2O4 and NO2 at 25 °C in the vessel will be approximately x = 0.0151 M and 2x = 0.0303 M, respectively.

Answer 2

Answer: The equilibrium concentration of
`N_2O_4\text{ and }NO_2` are 0.0164 M and 0.0572 M

Step-by-step explanation:

We are given:

Initial concentration of
`N_2O_4` = 0.0654 M

The given chemical equation follows:

`N_2O_4(g)\rightleftharpoons 2NO_2(g)`

Initial: 0.0654

At eqllm: 0.0654 - x 2x

The expression of
`K_c` for above reaction follows:

`K_c=([NO_2]_(eq)^2)/([N_2O_4]_(eq))`

We are given:

`K_c=4.64* 10^(-3)`

`[NO_2]_(eq)=2x`

`[N_2O_4]_(eq)=0.0654-x`

Putting values in above equation, we get:

`4.64* 10^(-3)=((2x)^2)/((0.0654-x))\\\\4x^2+(4.64* 10^(-3))x-(0.303* 10^(-3))=0`

Solving for the value of 'x', we get:

`x=0.0082,-0.0093`

Neglecting the negative value of 'x', as concentration cannot be negative.

Now,

`[NO_2]_(eq)=2x=2(0.0082)=0.0164M`

`[N_2O_4]_(eq)=0.0654-x=(0.0654-0.0082)=0.0572M`

Hence, the equilibrium concentration of
`N_2O_4\text{ and }NO_2` are 0.0164 M and 0.0572 M