The $K _ { p }$ for the reaction below is 1.49 × 10⁸ bar^-1 at 100.0 °C: CO(g) + $\mathrm { Cl } _ { 2 }$ (g) ? $\mathrm { COCl } _ { 2 }$ (g)
In an equilibrium mixture of the three gases, $\mathrm { PCO }$ = $\mathrm { P } _ { \mathrm { Cl } _ { 2 } }$ = 1.92 × 10^-4 bar.The partial pressure of the product,phosgene ( $\mathrm { COCl } _ { 2 }$ ) ,is ________ bar.

Question

The

K _ { p }

for the reaction below is 1.49 × 10⁸ bar^-1 at 100.0 °C: CO(g) +

\mathrm { Cl } _ { 2 }

(g) ?

\mathrm { COCl } _ { 2 }

(g)
In an equilibrium mixture of the three gases,

\mathrm { PCO }

=

\mathrm { P } _ { \mathrm { Cl } _ { 2 } }

= 1.92 × 10^-4 bar.The partial pressure of the product,phosgene (

\mathrm { COCl } _ { 2 }

) ,is ________ bar.

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is:$$K_p = \frac{P_{COCl_2}}{P_{CO} \cdot P_{Cl_2}}$$Substituting the given values into the expression, we get:$$1.49 \times 10^8 = \frac{P_{COCl_2}}{(1.92 \times 10^{-4})^2}$$Solving for $P_{COCl_2}$, we get:$$P_{COCl_2} = 5.49 \times 10^{-4} \text{ bar}$$

The KpK _ { p }Kp​ For the Reaction Below Is 1

The $K _ { p }$ For the Reaction Below Is 1