Question

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant, Kc. 3 A(g) + 2 B(g) = 4C(g) Kc = 2.93 x 1015 If, at this temperature, 1.40 mol of A and 3.90 mol of B are placed in a 1.00 L container, what are the concentrations of A, B, and C at equilibrium? [A] = 0 B) = 1.8667 [C] = | 2.967

Answer 1

Kc is very large

So, reaction will proceed to equilibrium

Balanced chemical equation is:

3 A + 2 B ---> 4 C

3 mol of A reacts with 2 mol of B

for 1.4 mol of A, 0.9333 mol of B is required

But we have 3.9 mol of B

so, A is limiting reagent

So, remaining amount of A would be 0

According to balanced equation

mol of C formed = (4/3)* moles of A

= (4/3)*1.4

= 1.867 mol

According to balanced equation

mol of B reacted = (2/3)* moles of A

= (2/3)*1.4

= 0.9333 mol

mol of B remaining = mol initially present - mol reacted

mol of B remaining = 3.9 - 0.9333

mol of B remaining = 2.967 mol

Since volume in 1 L

[A] = 0 M

[B] = 2.967 M

[C] = 1.867 M

Answer:

[A] = 0 M

[B] = 2.97 M

[C] = 1.87 M