Question

Carbon tetrachloride can be produced by this reaction: Suppose 1.1 mol and 3.3 mol are placed in a 1.00-L flask and the flask is sealed. After equilibrium has been achieved, the mixture contains 0.82 mol . Calculate .

Answer 1

The question is incomplete, complete question is:

Carbon tetrachloride can be produced by this reaction:

`CS_2(g) + 3Cl_2(g)` ⇌
`S_2Cl_2(g) + CCl_4 (g)`

Suppose 1.1 mol and 3.3 mol are placed in a 1.00-L flask and the flask is sealed. After equilibrium has been achieved, the mixture contains 0.82 mol .

Calculate
`K_c`.

Answer:

The value of the
`K_c` of the reaction is 4.05.

Step-by-step explanation:

`Concentration=\frac{\text{Moles of solute}}{\text{Volume od solution}(L)}`

Initial concentration of
`CS_2`:

`[CS_2]=(1.1 mol)/(1 L)=1.1 M`

Initial concentration of
`Cl_2`:

`[Cl_2]=(3.3mol)/(1 L)=3.3M`

Equilibrium concentration of
`CCl_4`:

`[CCl_4]=(0.82 mol)/(1 L)=0.82 M`

`CS_2(g) + 3Cl_2(g)` ⇌
`S_2Cl_2(g) + CCl_4 (g)`

initially :

1.1 M 3.3 M 0 0

At equilibrium

(1.1-0.82) M (3.3-3 × 0.82) M 0.82 M 0.82 M

0.28 M 0.84 0.82 0.82

The expression of equilibrium constant
`K_c` is given by :

`K_c=([S_2Cl_2][CCl_4])/([CS_2][Cl_2]^3)`

`=(0.82 M* 0.82 M)/(0.28M* (0.84 M)^3)=4.05`

The value of the
`K_c` of the reaction is 4.05.