Question

For the reaction

A(g) + B(g) <---> C(g)

At 130.00 Celcius, the equilibrium constant is 0.3

Find the equilibrium conversion and the product composition for this reactor at the following condition.

A) Reaction Mixture at 1 Bar, 1 mol of A, 2 moles of B and 0 moles of inerts gas

B) Reaction Mixture at 3 Bar, 1 mol of A, 3 moles of B and 2 moles of inerts gas

Answer 1

A(g)    + B(g)    <------------------>   C(g)

1 mol        2mol                                     0

1-x            2-x                                        x

Kc = [C] / [A][B]

0.3 = x / (1-x)(2-x)

x = 0.33

equilibrium concentrations :

moles of [A] = 1 - x = 1 - 0.33 = 0.67

moles of [B] = 2 - x = 2 - 0.33 = 1.67

moles of [C] = x = 0.33

equilibrium shifts towards volume increase side that means backward direction .

B)

A + B ----------------------> C

1       3                              0   ------------------------> initial

1+2   3+2                          2 --------------------------------> on addition of inert gas

3       5                              2   -------------------------> equilibrium

here also equlibrium shifts towrads back ward direction where volume increase side