Question

At 2000 ∘C the equilibrium constant for the reaction

2NO(g)←−→N2(g)+O2(g)
is Kc=2.4×103. The initial concentration of NO is 0.250 M .

Part A

What is the equilibrium concentration of NO?

Part B

What is the equilibrium concentration of N2?

Part C

What is the equilibrium concentration of O2?

Please show me how to got your results :)

Thank you!

Answer 1

The given chemical reaction in equilibrium is

2NO(g) < ------- > N2(g) + O2(g), Kc = 2.4x10³

n(g) for the above reaction is

n(g) =(product moles) - (reactant moles) = 1 + 1 - 2 = 0

Since n(g) for the above reaction is 0, the value of Kc is independent of volume.

---------------------2NO(g) < ------- > N2(g) + O2(g), Kc = 2.4x10³

Init. Conc(M): 0.250, ---------------- 0, ------- 0

Eqm.Conc(M):(0.250 - 2x),-------- x --------- x

Kc = [N2(g)] * [O2(g)] / [NO(g)]²

=> 2.4x10³ = x*x / (0.250 - 2x)²  

=> 9599x² - 2400x + 150 = 0

=> x = 0.12374 M

Part - A: Hence equilibrium concentration of NO(g), [NO(g)] = (0.250 - 2x) = 0.00252 M (answer)

Part - B: equilibrium concentration of N2(g), [N2(g)] = x = 0.12374 M (answer)

Part - C: equilibrium concentration of O2(g), [O2(g)] = x = 0.12374 M (answer)