Question

Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium concentration (in M to 4 decimal places) of NO?

Answer 1

The equilibrium concentration of NO is 0.02124 M.

Step-by-step explanation:

Given that,

Initial concentration of NOBr = 0.878 M

`k_(c)=3.07*10^(-4)`

Temperature = 24°C

We know that,

The balance equation is

`2NOBr\Rightarrow 2NO+Br_(2)`

Initial concentration is,

`0.878\Rightarrow 0+0`

Concentration is,

`-2x\Rightarrow 2x+x`

Equilibrium concentration

`0.878-2x\Rightarrow 2x+x`

We need to calculate the value of x

Using formula of concentration

`k_(c)=([NO][Br_(2)])/([NOBr]^2)`

Put the value into the formula

`3.07*10^(-4)=([2x][x])/([0.878-2x]^2)`

`2x^2=3.07*10^(-4)*(0.878)^2+3.07*10^(-4)*4x^2-2*2x*0.878*3*10^(-4)`

`2x^2=0.0002367+0.001228x^2-0.0010536x`

`2x^2-0.001228x^2+0.0010536x-0.0002367=0`

`1.998772x^2+0.0010536x-0.0002367=0`

`x=0, 0.01062`

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

`concentration\ of\ NO=2x`

Put the value of x

`concentration\ of\ NO=2*0.01062`

`concentration\ of\ NO=0.02124`

Hence, The equilibrium concentration of NO is 0.02124 M.