Kc for the reaction A(g) + B(g) =2C is 3.0 at 400k In an experminr “a” mol is mixed with 3 mol of B In a 1L vessel At equilibrium 3mol of C is formed The value of “a” will be

To determine the value of "a" in the reaction A(g) + B(g) ⇌ 2C, given that the equilibrium constant Kc is 3.0 at 400 K and that 3 moles of C are formed at equilibrium, we can use the concept of equilibrium concentrations and the expression for Kc. Let's break this down step by step.

Understanding the Reaction and Initial Conditions

We start with the balanced chemical equation:

• A(g) + B(g) ⇌ 2C

Initially, we have "a" moles of A and 3 moles of B in a 1 L vessel. Therefore, the initial concentrations are:

• [A] = a mol/L
• [B] = 3 mol/L
• [C] = 0 mol/L

Change in Concentrations at Equilibrium

As the reaction proceeds to equilibrium, let's denote the change in concentration of A that reacts as "x". According to the stoichiometry of the reaction:

• For every 1 mole of A that reacts, 1 mole of B reacts and 2 moles of C are produced.

At equilibrium, the concentrations will be:

• [A] = a - x
• [B] = 3 - x
• [C] = 2x

Setting Up the Equilibrium Expression

The equilibrium constant expression for this reaction is given by:

• Kc = \(\frac{[C]^2}{[A][B]}\)

Substituting the equilibrium concentrations into this expression, we have:

• Kc = \(\frac{(2x)^2}{(a - x)(3 - x)}\)

Given that Kc = 3.0 and we know that at equilibrium, 3 moles of C are formed, we can set 2x = 3. Therefore, x = 1.5.

Calculating the Concentrations at Equilibrium

Now, substituting x back into the equilibrium concentrations:

• [C] = 2(1.5) = 3 mol/L
• [A] = a - 1.5
• [B] = 3 - 1.5 = 1.5 mol/L

Substituting into the Kc Expression

Now we can substitute these values into the Kc expression:

• 3.0 = \(\frac{(3)^2}{(a - 1.5)(1.5)}\)

This simplifies to:

• 3.0 = \(\frac{9}{(a - 1.5)(1.5)}\)

Cross-multiplying gives us:

• 3.0(a - 1.5)(1.5) = 9

Expanding this, we have:

• 4.5a - 6.75 = 9

Solving for "a":

• 4.5a = 15.75
• a = \(\frac{15.75}{4.5} = 3.5\)

Final Result

Thus, the value of "a" is 3.5 moles. This means that initially, there were 3.5 moles of A mixed with 3 moles of B in the 1 L vessel.