An amount of 6.0 moles of 'A', 'B' and 'D' is added to a 1.0 L container.
A(g) + B(g) ⇌ C(g); K_C = 9 × 10¹¹ M^-1
A(g) + D(g) ⇌ E(g); K_C = 10¹¹ M^-1
If at equilibrium, the moles of 'B' is 'x', then the value of '200x' is

Question

An amount of 6.0 moles of 'A', 'B' and 'D' is added to a 1.0 L container.

A(g) + B(g) ⇌ C(g); K_C = 9 × 10¹¹ M^-1

A(g) + D(g) ⇌ E(g); K_C = 10¹¹ M^-1

If at equilibrium, the moles of 'B' is 'x', then the value of '200x' is

TestHub · Accepted Answer

Both constants are very large, so A is almost completely used: [A] ≈ 0 Mass balance: [B] + [C] = 6, [D] + [E] = 6, [A] + [C] + [E] = 6

With [A] ≈ 0: [C] + [E] = 6, so [B] = [E] = x and [D] = [C]

K₁/K₂ = ([C]/[B])/([E]/[D]) = 9

([C]/x)/(x/[C]) = [C]²/x² = 9, so [C] = 3x

3x + x = 6, so x = 1.5

200x = 300, written as 0300

Chemistry - Chemical Equilibrium Question with Solution | TestHub