At 900.0 K,the equilibrium constant (K_p)for the following reaction is 0.345 bar^-1: $2 \mathrm { SO } _ { 2 }$ + $\mathrm { O } _ { 2 }$ (g)? $2 \mathrm { SO } _ { 3 }$ (g) At equilibrium,the partial pressure of $\mathrm { SO } _ { 2 }$ is 35.0 bar and that of $\mathrm { O } _ { 2 }$ is 15.9 bar.The partial pressure of $\mathrm { SO } _ { 3 }$ is ________ bar.

Question

Quizplus · Accepted Answer

The equilibrium constant expression is:$$K_p = \frac{p_{SO_3}^2}{p_{SO_2}^2 \cdot p_{O_2}}$$Substituting the given values:$$0.345 = \frac{p_{SO_3}^2}{(35.0)^2 \cdot (15.9)}$$Solving for p_{SO_3}:$$p_{SO_3} = 82.0 \ bar$$

At 9000 K,the Equilibrium Constant (Kp)for the Following Reaction Is 0

At 9000 K,the Equilibrium Constant (K_p)for the Following Reaction Is 0