8. in the reaction 2 so₂(g) + o₂(g) ⇌ 2 so₃(g), 0.455 mol so₂, 0.183 mol o₂, and 0.568 mol so₃ are…

8. in the reaction 2 so₂(g) + o₂(g) ⇌ 2 so₃(g), 0.455 mol so₂, 0.183 mol o₂, and 0.568 mol so₃ are introduced simultaneously into a 1.90 l vessel at 1000k. if kc = 2.8 x 10², is the mixture at equilibrium? if not, in which direction will the reaction proceed?

8. in the reaction 2 so₂(g) + o₂(g) ⇌ 2 so₃(g), 0.455 mol so₂, 0.183 mol o₂, and 0.568 mol so₃ are introduced simultaneously into a 1.90 l vessel at 1000k. if kc = 2.8 x 10², is the mixture at equilibrium? if not, in which direction will the reaction proceed?

Answer

Explanation:

Step1: Calculate the initial concentrations

The formula for concentration $c=\frac{n}{V}$. For $SO_2$: $c_{SO_2}=\frac{0.455\ mol}{1.90\ L}= 0.239\ M$ For $O_2$: $c_{O_2}=\frac{0.183\ mol}{1.90\ L}=0.0963\ M$ For $SO_3$: $c_{SO_3}=\frac{0.568\ mol}{1.90\ L}=0.299\ M$

Step2: Write the reaction - quotient expression

The reaction is $2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$. The reaction - quotient $Q_c$ is given by the expression $Q_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}$. Substitute the initial concentrations into the $Q_c$ expression: $Q_c=\frac{(0.299)^2}{(0.239)^2\times0.0963}$ $Q_c=\frac{0.0894}{0.0571\times0.0963}$ $Q_c=\frac{0.0894}{0.00550}$ $Q_c = 162.5$

Step3: Compare $Q_c$ and $K_c$

Given $K_c = 2.8\times10^{2}=280$. Since $Q_c(162.5)<K_c(280)$, the reaction is not at equilibrium. The reaction will proceed in the forward direction to increase the concentration of products and decrease the concentration of reactants until $Q_c = K_c$.

Answer:

The mixture is not at equilibrium. The reaction will proceed in the forward direction.