Question

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol . Calculate at the unknown temperature.

Answer 1

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

`CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)`

Suppose 1.20 mol
`CS_2(g)` of and 3.60 mol of
`Cl_2(g)` were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of
`CCl_4`. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Step-by-step explanation:

Moles of
`CS_2` = 1.20 mole

Moles of
`Cl_2` = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of
`CS_2` =
`(moles)/(volume)=(1.20mol)/(1L)=1.20M`

Initial concentration of
`Cl_2` =
`(moles)/(volume)=(3.60mol)/(1L)=3.60M`

The given balanced equilibrium reaction is,

`CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)`

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

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

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

`K_c=((x)* (x))/((3.60-3x)^3* (1.20-x))`

Given :Equilibrium concentration of
`CCl_4` , x =
`(moles)/(volume)=(0.72mol)/(1L)=0.72M`

`K_c=((0.72)* (0.72))/((3.60-3* 0.72)^3* (1.20-0.72))`

`K_c=0.36`

Thus equilibrium constant at unknown temperature is 0.36