When the compound A2X(g) is placed in a flask at 1200 degrees C it dissociates i

Question

When the compound A2X(g) is placed in a flask at 1200 degrees C it dissociates into A2(g) and X2(g) as shown in the following equation.

       2A2X(g)<----> 2A2(g) + X2 (g)

If the inital concentration of A2X is 2.0 M and 15.0% of the A2X vapor is dissociated when equilibrium is reached, what is the value of the equilibrium constant for the reaction?

Hint 1: Use an equilibrium chart to organize the information.

Hint 2: What does "15.0% of A2X is dissociated" imply about the concentrations of the 3 gases?

Explanation / Answer

Let there be 2 moles of A2X in a 1L container (2M) to start with. 15% or 0.3 moles dissociate. Since 1 moles of 2A2X give rise to1 moles of 2A2(g) and 0.5 moles of X2 (g), we now have 0.3 moles of A2(g) and 0.15 moles of X2 (g). Since the volume is 1L, we have 0.3moles/L or 0.3M 2A2(g) and only 0.15M X2.

Equilibrium constant = [products]/[reactans] = [0.3]^2 * [0.15] / [2-0.3]^2 = 4.67*10^-3

Hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.