is B at equilibrium? is C at equilibrium? is D at equilibrium? B proceeds to the

Question

is B at equilibrium?

is C at equilibrium?

is D at equilibrium?

B proceeds to the right= yes/no

B proceeds to the left=yes/no

C proceeds to the right=yes/no.

C proceeds to the left=yes/no

D proceeds to the right=yes/no

D proceeds to the left=yes/no

Be sure to answer all parts.
The beakers shown contain 0.300 L of aqueous solutions of a moderately weak acid HY. Each particle represents 0.018 mol; solvent molecules are omitted for clarity. Protons are smaller and light blue. Oxygen atoms are red. Therefore, hydronium ions are red with three light blue hydrogen atoms.

(a) The reaction in beaker A is at equilibrium. Calculate Q for B, C, and D to determine which, if any, are also at equilibrium.
QB = ?
QC = ?
QD = ?   

is B at equilibrium?

is C at equilibrium?

  

is D at equilibrium?

(b) For any not at equilibrium, in which direction does the reaction proceed?   

B proceeds to the right= yes/no

  

B proceeds to the left=yes/no

  

C proceeds to the right=yes/no.

  

C proceeds to the left=yes/no

  

D proceeds to the right=yes/no

  

D proceeds to the left=yes/no

Explanation / Answer

a)

There are 8 molecules of HY, 4 of H3O+ and 4 of Y-.

mol HY= 8 x0.018mol=0.144mol ---> molarity= 0.144mol/0.3L= 0.48M

mol Y- =mol H3O+= 4 x 0.018mol = 0.072mol ---> molarity= 0.072mol/0.3L= 0.24M

Keq= [H3O+][Y-]/[HY]= 0.24 x0.24/0.48 = 0.12

B:

There are 6 molecules of HY, 2 of H3O+ and 2 of Y-.

mol HY= 6 x0.018mol= 0.108mol ----> molarity= 0.108mol/0.3L= 0.36M

mol H3O+= mol Y- = 2 x 0.018mol= 0.036mol -----> molarity= 0.036mol/0.3L= 0.12

QB= 0.12 x 0.12/0.36= 0.04

QB <Keq ---> B in not at equilibrium

C:

There are 4 molecules of HY, 2 of H3O+ and 2 of Y-.

mol HY= 4 x0.018mol= 0.072mol ----> molarity= 0.072mol/0.3L= 0.24M

mol H3O+= mol Y- = 2x0.018mol= 0.036mol ----> molarity= 0.036mol/0.3L= 0.12M

QC= 0.12x0.12/0.24= 0.06

QC<Keq ---> C is not at equilibrium

D:

There are 2 molecules of HY, 2 of Y- and 2 of H3O+

mol HY= mol Y- = mol H3O+= 2 x 0.018= 0.036mol ----> molarity= 0.036/0.3L= 0.12M

QD= 0.12 x 0.12/0.12= 0.12

QD=Keq ---> D is at equilibrium

b) B and C are not at equilibrium and in both cases Q is <Keq. So, in order to increase the value of Q the concentration of products must increase and the concentration of reagents decrease. This means that in both cases the reaction proceeds to the right.

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