Question

A 100 L reaction container is charged with 0.612 mol of NOBr, which decomposes at a certain temperature** (say between 100 and 150 oC) according to the following reaction:

NOBr(g) ? NO(g) + 0.5Br2(g)

At equilibrium the bromine concentration is 2.12x10-3 M. Calculate Kc (in M0.5)

**Not specifying the temperature allows for a more liberal use of random numbers.

Answer 1

Answer: The value of the equilibrium constant is 0.024

Step-by-step explanation:

Initial moles of
`NOBr` = 0.612 mole

Volume of container = 100 L

Initial concentration of
`NOBr=(moles)/(volume)=(0.612moles)/(100L)=6.12* 10^(-3)M`

equilibrium concentration of
`Br_2=2.12* 10^(-3)M`

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

`NOBr(g)\rightleftharpoons NO(g)+(1)/(2)Br_2(g)`

at t=0
`6.12* 10^(-3)M` 0 0

At eqm. conc.
`(6.12* 10^(-3)-x)M` x x/2

The expression for equilibrium constant for this reaction will be,

`K_c=([NO]^2[Br_2])/([NOBr])`

`K_c=((2x)^2* x/2)/((6.12* 10^(-3)-x))`

we are given : x/2 =
`2.12* 10^(-3)`

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

`K_c=\frac{(2.12* 10^(-3))^{(1)/(2)}* (2.12* 10^(-3))}{(6.12* 10^(-3))-(2.12* 10^(-3))}`

`K_c=0.024`

Thus the value of the equilibrium constant is 0.024