Question

For a homogeneous gaseous reaction , A + 2B gives C at 300 K, the value of Kc = 0.1. When two moles each of A and B are mixed, then what will be the approximate equilibrium pressure if 30% of A is converted to C? The options are : 100 atm, 90atm., 178 atm and 72 atm....it's urgent ​

Answer 1

The approximate equilibrium pressure if 30 % of A is converted to C is 178 atm.

Given-

• Temperature of the reaction = 300 K
• Value of Kc = 0.1
• Moles of A and B = 2 moles
• Percentage of A converted into B = 30 %

The reaction is

                        A     +       2B       ↔           C

Initial                2                2

At equilibrium 2- α          2-2α                   α

Here α is the degree of dissociation

30 % of A means = 30/100 × 2 = 0.6 moles

So, Concentration of A at equilibrium is = 2 - 0.6 = 1.4

Concentration of B at equilibrium = 2 - 2 (0.6) = 0.8

Concentration of C at equilibrium = 0.6

Mole fraction of A = 1.4 / 1.4+0.8+0.6 = 1.4 /2.8 = 1/2 = 0.5

Mole fraction of B = 0.8 / 2.8 = 0.286

Mole fraction of C = 0.6/2.8 = 0.214

We know that

Kp = Kc (RT)^Δn

By substituting the value we get

Kp = 0.1 (0.0821 × 300)⁻²

Kp = 0.000165

0.000165 = Pc/Pa× Pb²

Where Pa, Pb and Pc is the partial pressure of A, B and C respectively at equilibrium.

Let total pressure at equilibrium is P. So

Pc = 0.214 × P

Pa = 0.5 × P

Pb = 0.286 × P

By substituting the values we get

0.000165 = 0.214 P / 0.5 P × (0.286 P)² = 5.23 /P²

P² = 5.23/0.000165 = 31712.3

P = 178 atm