Question

Consider this reaction:

2Cl2O5 (g) ------> 2Cl2(g) +5O2 (g)
At a certain temperature it obeys this rate law.
rate = (17.4 M-1*s -1 ) [Cl2O5]2
1. Suppose a vessel contains Cl2O5 at a concentration of 1.46M. Calculate the concentration of Cl2O5 in the vessel .400 seconds later. You may assume no other reaction is imporant. Round your answer to 2 significant digits.

Answer 1

[Cl2O5] = 1.60 M

Step-by-step explanation:

• 2Cl2O5(g) → 2Cl2(g) + 5O2(g)

at T ⇒ r = (17.4/M.s)*[Cl2O5]²

1) [Cl2O5] = 1.46 M

⇒ t = 0.400 s ⇒ [Cl2O5] = ?

rate law gnral:

• - ra = K(Ca)∧α = - δCa/δt

∴ K = 17.4 /M.s

∴ α = 2

⇒ - ∫δCa/Ca² = K∫δt

⇒ [ 1/Ca - 1/Cao ] = K*t

⇒ 1/Ca = K*t + 1/Cao

⇒ Ca = 1/K*t + Cao = [Cl2O5]

∴ Cao = 1.46 M

∴ t = 0.400 s

⇒ [Cl2O5] = 1/[(17.4/M.s)*(0.400s)] + 1.46 M

⇒ [Cl2O5] = 0.1436 M + 1.46 M

⇒ [Cl2O5] = 1.604 M