Question

For the reaction POCl3(g) ⇀↽ POCl(g) + Cl2(g) Kc = 0.30. An initial 0.3 moles of POCl3 are placed in a 3.1 L container with initial concentrations of POCl and Cl2 equal to zero. What is the final concentration of POCl3? 1. final concentration = 0.281774 M 2. final concentration = 0.077 M 3. final concentration = 0.185 M 4. final concentration = 0.039534 M 5. final concentration = 0.019767 M

Answer 1

Answer: The final concentration of $POCl_3$ is 0.019767 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of the solution}}$

Initial moles of $POCl_3$ = 0.3 moles

Volume of solution = 3.1 L

$\text{Initial concentration of }POCl_3=\frac{0.3}{3.1}=0.097M$

For the given chemical equation:

                        $POCl_3(g)\rightleftharpoons POCl(g)+Cl_2(g)$

Initial:                0.097

At eqllm:         0.097-x             x              x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[POCl][Cl_2]}{[POCl_3]}$

We are given:

$K_c=0.30$

Putting values in above expression, we get:

$0.30=\frac{x\times x}{(0.097-x)}\\\\x=0.077233,-0.377$

Neglecting the negative value of 'x' because concentration cannot be negative.

So, final concentration of $POCl_3$ = (0.097 - x) = (0.097 - 0.077233) = 0.019767 M

Hence, the final concentration of $POCl_3$ is 0.019767 M