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how do i solve this?For this gas phase equilibrium shown below, 6 mol of C is placed in a 2.5 L flask and allowed to equilibrate. If the equilibrium constant for this reaction is 2.50, what are the final concentrations for all the three gases?A + B <-----> C

How Do I Solve ThisFor This Gas Phase Equilibrium Shown Below 6 Mol Of C Is Placed In A 25 L Flask And Allowed To Equilibrate If The Equilibrium Constant For Th class=

Sagot :

For this gas phase equilibrium shown below, 6 mol of C is placed in a 2.5 L flask and allowed to equilibrate. If the equilibrium constant for this reaction is 2.50, what are the final concentrations for all the three gases?

A + B <-----> C

We place 6 mol of C in a 2.5 L flask. So its molarity is:

molarity of C = 6 mol / 2.5 L

molarity of C = 2.4 M

The equilibrium constant for that reaction can be expressed like this:

Keq = [C] / ( [A] * [B] )

To solve the problem we have to set up an ICE table:

A + B <-----> C

I 0 0 2.4 M

C x x - x

E x x 2.4 M - x

Replacing those values in the Keq (Keq = 2.50):

Keq = [C] / ( [A] * [B] )

2.50 = (2.4 - x) / (x * x )

2.50 = (2.4 - x) / (x²)

And we have to solve that equation for X:

2.50 * x² = 2.4 - x

2.50 *x² + x - 2.4 = 0

This equation has 2 roots:

x₁ = -1.2 and x₂= 0.8

We can't have a negative concentration, so -1.2 is not our answer, then we use 0.8 M.

From the ICE table we said that the equilbrium concentrations are:

[A] = x = 0.8 M [B] = x = 0.8 M [C] = 2.4 M - x = 2.4 M - 0.8 M = 1.6 M

Answer: the final concentrations are [A] = 0.8 M [B] = 0.8 M [C] = 1.6 M

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