Question

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant, Kc 3 A(g) 2 B(g)= 4C(g) = 1.53 x 1019 К If, at this temperature, 2.10 mol of A and 3.80 mol of B are placed in a 1.00L container, what are the concentrations of A, B, and C at equilibrium? [A] М М [B= [C] М =

Answer 1

For reaction at t = 0 starting moles are given

And at t= t(equilibrium) some mole of C is formed from A&B

3A + 2 B -> 4C

at t=0 2.10 mol 3.80 mol 0.00 mol

at t= t(eq) 2.10 - 3x mol 3.80- 2x mol 4x mol

We have at equilibrium

Kc= [C] ⁴/( [A]³ [B]²)

And also Kc= 1.53 ×10 ¹⁹

As kc is quite high we can take reaction as complete

And solve it by finding limiting reagent

Which here is A .

As A is cosumed fully.

[A] =0

2.10 - 3x= 0

x = 0.7

Now [B] = 3.80 - 2 ×0.7 = 2.40 mol/ L

Now [C] = 4 x = 4 × 0.7 = 2.80 mol/L