Question

The equilibrium constant for the following reaction is 1.2 * 10 ^ - 2 at 500K PCl 5 (g) rightleftharpoons PCl 3 (g)+Cl 2 (g) If an equilibrium mixture of the three gases at 500K contains 1.19 x10 ^ - 2 M PCls(g) and 2.59 x 10 ^ - 2 M PCl3 what is the equilibrium concentration of Cl2? ________ m

Answer 1

The equilibrium constant expression for the reaction is:

Kc = [PCl3][Cl2] / [PCl5]

We are given the value of Kc = 1.2 × 10^−2 and the initial concentrations of PCl5 and PCl3. Let x be the change in concentration of PCl5 and PCl3 at equilibrium and let y be the equilibrium concentration of Cl2. Then we can write:

PCl5 = 1.19 × 10^−2 − x

PCl3 = 2.59 × 10^−2 − x

Cl2 = y

Substituting these expressions into the equilibrium constant expression and solving for y, we get:

Kc = [PCl3][Cl2] / [PCl5]

1.2 × 10^−2 = [(2.59 × 10^−2 − x)][y] / (1.19 × 10^−2 − x)

Assuming x is small compared to the initial concentrations of PCl5 and PCl3, we can make the approximation that the change in concentration is negligible compared to the initial concentrations. Therefore, we can simplify the expression as:

Kc = [PCl3][Cl2] / [PCl5]

1.2 × 10^−2 = (2.59 × 10^−2)[y] / (1.19 × 10^−2)

y = (1.2 × 10^−2)(1.19 × 10^−2) / (2.59 × 10^−2)

y = 0.552 M

Therefore, the equilibrium concentration of Cl2 is 0.552 M.

Answer 2

Equilibrium expression for the reaction system-

`\:\star\:\sf\:PCl_5\:(g)\:⇌\:PCl_3\:(g)\:+Cl_2\:(g)\\`

• The equilibrium constant expression for the given reaction is:

`\:\:\:\:\:\:\star\longrightarrow \sf \underline{Kc = (\bigg[PCl_3\bigg]\:\bigg[Cl_2\bigg] )/( \bigg[PCl_5\bigg]\:)}\\`

• As per question, following concentrations and equilibrium constant present in the system are given -

• 
  `\sf{[PCl_5] = 1.19×10^(−2)\:M}`
• 
  `\sf{ [PCl_3] =2.59×10^(−2 )\:M}`

• 
  `\sf{K_(c)=1.2×10^(-2)}`

Now that, we have all the required molar equilibrium concentrations, so we can substitute the molar equilibrium concentrations into the expression and calculate the concentration of
`\sf Cl_2`:-

`\:\:\:\:\:\:\star\longrightarrow \sf \underline{Kc = (\bigg[PCl_3\bigg]\:\bigg[Cl_2\bigg] )/( \bigg[PCl_5\bigg]\:)}\\`

`\:\:\:\:\:\:\longrightarrow \sf 1.2* 10^(-2) = (2.59* 10^(-2)* \bigg[Cl_2\bigg])/(1.19* 10^(-2))\\`

`\:\:\:\:\:\:\longrightarrow \sf 1.2* 10^(-2) = \frac{2.59* \cancel{10^(-2)}* \bigg[Cl_2\bigg]}{1.19* \cancel{10^(-2)}}\\`

`\:\:\:\:\:\:\longrightarrow \sf 1.2* 10^(-2) = (2.59)/(1.19)* \bigg[Cl_2\bigg]\\`

`\:\:\:\:\:\:\longrightarrow \sf\bigg[Cl_2\bigg] = 1.2* 10^(-2)* (1.19)/(2.59)\\`

`\:\:\:\:\:\:\longrightarrow \sf\bigg[Cl_2\bigg] = 1.2* 10^(-2)* 0.459......\\`

`\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf\bigg[Cl_2\bigg] = 1.2* 0.46* 10^(-2)\\`

`\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf\bigg[Cl_2\bigg] = 0.552* 10^(-2)\\`

`\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf\bigg[Cl_2\bigg] = \underline{5.52* 10^(-3)\:M}\\`

• Therefore,the equilibrium concentration of
  `\sf Cl_2` at 500 K is
  `\bf\underline{ 5.52* 10^(-3) \:M.}\\`