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.

Explanation:

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

Initial concentration of $CS_2$:

$[CS_2]=\frac{1.1 mol}{1 L}=1.1 M$

Initial concentration of $Cl_2$:

$[Cl_2]=\frac{3.3mol}{1 L}=3.3M$

Equilibrium concentration of $CCl_4$:

$[CCl_4]=\frac{0.82 mol}{1 L}=0.82 M$

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

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 :

$K_c=\frac{[S_2Cl_2][CCl_4]}{[CS_2][Cl_2]^3}$

$=\frac{0.82 M\times 0.82 M}{0.28M\times (0.84 M)^3}=4.05$

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