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

Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

$k_{c}=3.07\times10^{-4}$

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

$k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}$

Put the value into the formula

$3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}$

$2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}$

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

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

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

$x=0, 0.01062$

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

$concentration\ of\ NO=2x$

Put the value of x

$concentration\ of\ NO=2\times0.01062$

$concentration\ of\ NO=0.02124$

Hence, The equilibrium concentration of NO is 0.02124 M.