The rate of the reaction?

Calculating the reaction rate of a chemical reaction?

• "The rate of decomposition of N2O3(g) to NO2(g) and NO(g) is followed by measuring NO2 at different times. The following data are obtained. NO2 (mol/L) 0 ; 0.193 ; 0.316 ; 0.427 ; 0.784 Time 't' in (s) 0 ; 884 ; 1610 ; 2460 ; 50,000 The reaction follows a first-order rate law. Calculate the specific rate constant. Assume that after 50,000 seconds all the N2O3(g) had decomposed." -Do I just find the reaction rate of the reverse reaction (the composing of N2O3) and use the inverse for my rate constant, k? Also, since the decomposition follows first order, would the sum of the two reactants in the reverse reaction also equal 1st order, each being ^1/2? Please help! How do I set this up?

• Answer:

  The reaction can be considered as irreversible, because the reaction goes to completion. A reaction of considerable reversibility would end up in equilibrium with some reactant remaining. I speak of considerable reversibility, because from viewpoint of thermodynamics every reaction is reversible. But if the equilibrium position is far to the product side, like in this reaction, the amount of remaining reactant is beyond level of observability. From viewpoint of chemical kinetics this implies that reverse is rater slow. So the effect of reversibility may ignored for kinetic calculations about such reactions. The reaction rate of this first order reaction is given by: r = - d[N₂O₃]/dt = k∙[N₂O₃] Solving this differential equation with initial value [N₂O₃]₀ at t=0 leads to the integrated rate law: ln[N₂O₃] = -k∙t + ln[N₂O₃]₀ <=> ln( [N₂O₃]₀ / [N₂O₃] ) = k∙t Here you got the product concentration. Reactant and product concentration are linked by the stochiometry of the reaction: N₂O₃ → NO₂ + NO So per mole of dinitrogen trioxide reacted away one mole of nitrogen dioxide is formed. Assuming experiment was carried in a constant volume reactor, these relations hold for the change of concentration: -∆[N₂O₃] = ∆[NO₂] <=> - ( [N₂O₃]₀ - [N₂O₃] ) = [NO₂] - [NO₂]₀ since [NO₂]₀=0 - [N₂O₃]₀ + [N₂O₃] ) = [NO₂] So you can calculate the N₂O₃ at any time from: [N₂O₃] = [N₂O₃]₀ - [NO₂] from the final NO₂ concentration you can calculate the initial N₂O₃ concentration: at t=50000s [N₂O₃] = 0 => 0 = [N₂O₃]₀ - 0.784M => [N₂O₃]₀ = 0.784M Actually the [N₂O₃] drops not to zero for any time, but for sufficiently large times drops to an negligible level. with this you can calculate the LHS in the integrated rate law for each data point y = ln( [N₂O₃]₀ / [N₂O₃] ) = ln( 0.784M/(0.784M - [NO₃] ) except the last point, because you don't know the exact concentration. Next step is a least square fit: fitting y = k∙t leads to k = ∑ y∙t / ∑ t² I get k = 3.20×10⁻⁴ s⁻¹ Note if the reaction is reversible the reverse reaction would be first order with respect to each reactant: r = k_f∙[N₂O₃] - k_r∙ [NO₂]∙[NO] The reason is that the kinetic approach for equilibrium has to be consistent with thermodynamics. From viewpoint of thermodynamics equilibrium concentration are given by mass action law: ( [NO₂]e∙[NO]e ) / [N₂O₃]e = constant From kinetic viewpoint equilibrium is established, when rates equals zero. Hence: k_f∙[N₂O₃]e - k_r∙ [NO₂]e∙[NO]e = 0 <=> ( [NO₂]e∙[NO]e ) / [N₂O₃]e = kf/ke = constant If the order of reverse reaction with respect to NO₂ and to NO is different from one, rate law would violate thermodynamic requirements for equilibrium concentrations.