A Steady State Approximation - General Chemistry | CHEM 162, Study notes of Chemistry

Download A Steady State Approximation - General Chemistry | CHEM 162 and more Study notes Chemistry in PDF only on Docsity! 1 The Steady-State Approximation (Zumdahl Section 15.7) In some reactions there is no single step in the mechanism that is much slower than the others. This complicates the methods described in the previous section. In such cases, we resort to a Steady-State Approximation, in which we assume that the concentrations of reactive intermediates remain nearly constant through most of the reaction. Steady-State Example: 2NO(g) + H2(g) --> N2O(g) + H2O(g) Proposed Mechanism: 2NO N2O2 N2O2 + H2 N2O + H2O k1 k-1 k2 The double arrows in the first step indicate that both the forward and reverse of this step are important on the time scale of the overall reaction. Let's make the approximation: [N2O2] is constant 2 Example (cont) If [N2O2] is constant, then d[N2O2]/dt = 0 Rate of production Rate of consumption of N2O2 of N2O2 d[N2O2]/dt = k1[NO]2 -d[N2O2]/dt = k-1[N2O2] + k2[N2O2][H2] 2NO N2O2 k1 k-1 N2O2 + H2 N2O + H2O k2 k1[NO]2 = k-1[N2O2] + k2[N2O2][H2] Steady-State Condition: Example (cont): We now use this information to determine the rate law for the overall reaction: 2NO(g) + H2(g) --> N2O(g) + H2O(g) Rate of reaction = -d[H2]/dt = k2[H2][N2O2] (from 2nd step of mechanism) Solving the steady-state equation for [N2O2] and substituting into this rate equation gives: k1[NO]2 k-1 + k2[H2] Rate = -d[H2]/dt = k2[H2] 5 A clue comes from the strong temperature dependence of many rate constants: Note that this temperature dependence is not linear In 1877, Svante Arrhenius suggested that k varies exponentially with 1/T k = Ae-Ea/RT Ea is a constant with dimensions of energy A is a constant with the same dimensions as k k = Ae-Ea/RT Taking the natural logarithm of this equation gives: ln k = ln A - Ea/RT so that a plot of ln k vs. 1/T should be a straight line with slope of -Ea/R. Many rate constants do follow this behavior. 6 Arrhenius believed that for molecules to react upon collision, they must become "activated," and the parameter Ea has become known as the Activation Energy Only some collisions occur with enough energy to overcome the activation barrier Figure 15.9 Finding the activation energy from 2 data points Instead of a plot of of ln k vs. 1/T and getting Ea from the slope, you can calculate Ea from: ln k1 = ln A - Ea/RT1 ln k2 = ln A - Ea/RT2 Taking the difference of these two equations: ln (k2/k1) = (Ea/R)[(1/T1 - 1/T2)] 7 T1 = 500K k1 = 9.51x10-9M-1s-1 T2 = 600K k2 = 1.10x10-5M-1s-1 ln (k2/k1) = (Ea/R)[(1/T1 - 1/T2)] ln (1.1x10-5/9.51x10-9) = (Ea/8.3145J/molK)*[(1/500K - 1/600K)] rearrange and solve: Ea = 176kJ/mol Example: 2HI(g) H2(g) + I2(g) Rate = k[HI]2 Catalysis It is not always practical or convenient to increase reaction rates by increasing the temperature. A Catalyst is a substance that speeds up a reaction without being consumed during by the reaction Catalysis - The use of catalysts to speed up reactions without changing the temperature Homogeneous catalysts- catalysts that are in the same phase (e.g. solution or gas) as the reacting molecules Heterogeneous catalysts- catalysts that are in a different phase from the reacting molecules