Question

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate the value of Kc at 500. K. For the same reaction, calculate the molar concentration of reactants and products at equilibrium if initially 1.00 mol of (CH3),CCI was placed in a 5.00 L vessel.

Answer 1

Answer: The value of
`K_p` for the reaction is 6.32 and concentrations of
`(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl` is 0.094 M, 0.094 M and 0.106 M respectively.

Step-by-step explanation:

Relation of
`K_p` with
`K_c` is given by the formula:

`K_p=K_c(RT)^(\Delta ng)`

where,

`K_p` = equilibrium constant in terms of partial pressure = 3.45

`K_c` = equilibrium constant in terms of concentration = ?

R = Gas constant =
`0.0821\text{ L atm }mol^(-1)K^(-1)`

T = temperature = 500 K

`\Delta n_g` = change in number of moles of gas particles =
`n_(products)-n_(reactants)=2-1=1`

Putting values in above equation, we get:

`3.45=K_c* (0.0821* 500)^(1)\\\\K_c=(3.45)/(0.0821* 500)=0.084`

The equation used to calculate concentration of a solution is:

`\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}`

Initial moles of
`(CH_3)_3CCl(g)` = 1.00 mol

Volume of the flask = 5.00 L

So,
`\text{Concentration of }(CH_3)_3CCl=(1.00mol)/(5.00L)=0.2M`

For the given chemical reaction:

`(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)`

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of
`K_c` for above reaction follows:

`K_c=([(CH_3)_2C=CH]* [HCl])/([(CH_3)_3CCl])`

Putting values in above equation, we get:

`0.084=(x* x)/(0.2-x)\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178`

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

`[(CH_3)_2C=CH]=x=0.094M`

`[HCl]=x=0.094M`

`[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M`

Hence, the value of
`K_p` for the reaction is 6.32 and concentrations of
`(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl` is 0.094 M, 0.094 M and 0.106 M respectively.