The rate of the reaction?

What is the rate of reaction when 1.0 mole of B has reacted?

• The hypothetical reaction: 2A + B --> Products, is found to follow the rate law: rate = k[A][B]^2 with l= 1.8 x 10^1 M^ -2 sec ^ -1 Suppose 5 moles of A and 2 moles of B are mixed in a 2 liter container and reacted. What is the rate of reaction when 1.0 mole of B has reacted? 2. The reaction: 2N2O5 (g) --> 4NO2 (g) + O2(g), is first order in N2O5 (g). When [N2O5 (g)]=0.025 M, the rate is 7.2 x 10^ -4 M/min. How long will it take for the rate to drop to 3.6 x 10^ -4 M/min? 3. A reaction is first order and it takes 324 minutes for the reaction to be 50.0% complete. How long will it take for the reaction to be 85.0% complete at the same temperature? MUCH THANKS.. I really can't figure out these question. I think the third one is related to half-life, since half-life is in the first order and it is talking about half of it complete. The first one, I thought about using the integrated rate law, but it didn't work out for me... So, I'm pretty confused..

• Answer:

  1.) According to reaction equation 2 moles of A react with 1 mol of B. So when 1 mol of B has reacted 2 moles of A were consumed. So the concentrations of A dn B at this instant are: [A] = (5mol - 2mol) / 2L = 1.5M [B] = (2mol - 1mol) / 2L = 0.5M The rate at this instant is rate = k∙[A]∙[B]² = 1.8×10¹M⁻²s⁻¹ ∙ 1.5M ∙ (0.5M)² = 6.75 Ms⁻¹ 2.) Use integrated first-order rate law ln[N₂O₅] = ln[N₂O₅]₀ - k∙t => [N₂O₅] = e^( ln[N₂O₅]₀ - k∙t) = e^( ln[N₂O₅]₀) ∙ e^(- k∙t) = [N₂O₅]₀∙e^( - k∙t) So the change o rate with time is given by r = k∙[N₂O₅] = k∙[N₂O₅]₀∙e^( - k∙t) with initial rate r₀ = k∙[N₂O₅]₀ follows r = r₀∙e^( - k∙t) <=> ln(r/r₀) = - k∙t So the time t elapsed until rate has dropped from r₀ to r is: t = - ln(r/r₀) / k Rate constant for your rate can be found from initial rate r₀ = k∙[N₂O₅]₀ => k = r₀ / [N₂O₅]₀ = 7.2×10⁻⁴ Mmin⁻¹ / 0.025M = 2.88×10⁻² min⁻¹ Hence t = - ln(r/r₀) / k = - ln(3.6×10⁻⁴ Mmin⁻¹ / 7.2×10⁻⁴ Mmin⁻¹ ) / 2.88×10⁻² min⁻¹ = 24.07 min 3) Half life, i.e. the time elapsed until 50% of initial amount has reacted, and rate constant for a first-order reaction are related as: k = ln(2)/t½ For this reaction k = ln(2) / 324 min = 2.14×10⁻³ min⁻¹ From integrated rate law follows the time elapsed until reactant concentration has dropped to specified level: ln[A] = ln[A]₀ - k∙t <=> t = - (ln[A] - ln[A]₀)/k = - ln( [A]/[A]₀ )/k When reaction is 85% complete, 15% reactant are still there, i.e. [A] = 0.15∙[A]₀ => t = - ln( 0.15 )/2.14×10⁻³ min⁻¹ = 886.5min