Solution 7 2024 PDF

17A The rates of chemical 1. rate expressed as a change in concentration divided by time. The berate rearranged Forequation example,can if the law is theas: one shown in eqn 17A.4, with −3 0 =concentrations -2 NOBr(g) + 2NO (g) + 1 Br 2 (g) expressed in mol dm , then the units of kr will And thus3 m...

17A The rates of chemical 1. rate expressed as a change in concentration divided by time. The berate rearranged Forequation example,can if the law is theas: one shown in eqn 17A.4, with −3 0 =concentrations -2 NOBr(g) + 2NO (g) + 1 Br 2 (g) expressed in mol dm , then the units of kr will And thus3 mol−1 s−1 because be dm ν (NO) = +2 ν (Br2) = 1 kr [A] [B] !# #= "-2 ## $ !#"# $ !#"# $ ν (NOBr) dm3 mol −1 s −1 × moldm −3 × moldm −3 = moldm −3 s −1 Because ν (NO) = +2, the rate of the reaction is reported as v =which d[NO]/dt / ν (NO) = 0.080 mmol dm −3s−1. are the units of v. If the concentrations are expressed The power to which the concentration of a sp or a reactant) is raised in a rate law of this kin the reaction with respect to that species. A re rate law in eqn 17A.4 is first order in A and The overall order of a reaction with a rate law 17A.7 is the sum of the individual orders, a + b all order of the rate law in eqn 17A.4 is 1 + 1 = therefore said to be second-order overall. A reaction need not have an integral or gas-phase reactions do not. For example, a re rate law in molecules cm−3, and the rate in molecules cm−3 s−1, then the Rate of consumption for NOBr: -d[NOBr]/dt rate constant is expressed in cm3 molecule−1 s−1. The approach v = k [A]1/2[B] just developed can be used to determine the units of the rate v = d[NOBr]/dt / ν (NOBr) = d[NOBr]/dt / -2 = 0.080 mmol dm −3s−1. r constant from rate laws of any form. Therefore, -d[NOBr]/dt = 0.080 mmol dm −3s−1 x 2 = 0.160 mmol dm −3s−1 is half order in A, first order in B, and thr overall. Brief illustration 17A.2 Similarly, rate constant the reaction rateThe formation for Brfor ]/dt O(g) + O3(g) → 2 O2(g) is 2: d[Br −15 3 −1 −12 8.0 × 10 cm molecule s at 298/ K.1 To expressmmol this rate −3 −1 v = d[Br2]/dt / 3 ν (Br ) = d[Br2]/dt = 0.080 dm−1cons. −1 2−1 stant in dm mol s , make use of the relation 1 cm = 10 dm to convert the volume: 2. (10−1 dm)3 ! kr = 8.0 × 10−15 cm3 molecule−1 s −1 = 8.0 × 10−18 dm3 molecule−1 s −1 Now note that the number of molecules can be expressed as an amount in moles by division by Avogadro’s constant expressed as molecules per mole: kr = 8.0 × 10−18 dm3 molecule−1 s −1 −1 1 molecule ⎛ ⎞ −1 = 8.0 × 10−18 dm3 × ⎜ 23 −1 ⎟ s ⎝ 6.022 × 10 molecules mol ⎠ = 8.0 × 10−18 × 6.022 × 1023 dm3 mol −1 s −1 = 4.8 × 106 dm3 mol −1 s −1 3. A practical application of a rate law is that once the law and the value of the rate constant are known, it is possible to predict the rate of reaction from the composition of the mixture. Moreover, as demonstrated in Topic 17B, by knowing the rate law, it is also possible to predict the composition of the reaction mixture at a later stage of the reaction. A rate law also provides evidence used to assess the plausibility of a proposed mechanism of the reaction. This application is developed in Brief illustration 17A.3 The experimentally determined rate law for reaction H2(g) + Br2(g) → 2 HBr(g) is given In the rate law the concentration of H2 appea power +1, so the reaction is first order in H concentrations of Br2 and HBr do not appear raised to a power, so the reaction has an indefi respect to both Br2 and HBr, and an indefinite Some reactions obey a zeroth-order rate la have a rate that is independent of the conc reactant (so long as some is present). Thu decomposition of phosphine (PH3) on hot t pressures has the rate law v = kr This law means that PH3 decomposes at a cons has entirely disappeared. As seen in Brief illustration 17A.3, when a r the form in eqn 17A.7, the reaction does not order and might not even have definite orders each participant. These remarks point to three important task To identify the rate law and obtain the ra the experimental data. This aspect is d Topic. 4. 5. 6. 7. Comment: note that k is replaced by 2k in the formulas compared to those we derived in class. 8. 9. 10.