Given:A(g) + 3B(g) ⇋ C(g) + 2D(g) 1.0 mole of A and 1.0 mole of B are placed in a 5.0-liter container. After equilibrium has been established, 0.50 mole of D is present in the container. Calculate the equilibrium constant, Kc, for the reaction.

Kc = 26.67

Explanation:

Step 1: Data given

Number of moles of A = 1.0 mol

Number of moles of B = 1.0 mol

Number of moles of D = 0.50 moles

Volume = 5.0 L

Step 2: The balanced equation

A(g) + 3B(g) ⇋ C(g) + 2D(g)

Step 3: Number of initial moles

Moles A = 1.0 mol

Moles B = 1.0 mol

Moles C = 0 moles

Moles D = 0 moles

[A] = 1.0 mol / 5.0 L = 0.2 M

[B] = 1.0 mol / 5.0 L = 0.2 M

[C] = 0 M

[D] = 0 M

Step 4: Moles at the equilibrium

For 1 mol A we have 3 moles B to produce 1 mol C and 2 moles D

There will react X of A, this give (1.0 - X) moles

There will reacto 3X for B this gives (1.0 - 3X) moles

There will be X produced for C this gives X moles

There will be 2X produced for D this gives 2X moles

After equilibrium has been established, 0.50 mole of D

2X = 0.50 moles or X = 0.25 moles this gives 0.25 moles / 5.0 L = 0.05 M

Step 5: Calculate Kc

The concentration at the equilibrium

[A] = 0.2 - 0.05 = 0.15M

[B] = 0.2 - 0.15 = 0.05M

[C] = 0.05M

[D] = 0.1M

Kc = (0.05)(0.1)² / (0.15)(0.05)³

Kc = 26.67

the answer is d.) potassium