In: Chemistry
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!
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)