Kinetics - Part 2 - Solutions

Chem Worksheet - Week 6 Kinetics - Part 2 - Solutions Key Questions 1. Consider the reaction N2(g) + 3 H2(g) ÷ 2 NH3(g). At a certain point in the course of an experiment the rate of disappearance of N2(g) is 3.56 x 10–3 mol@L–1@s–1. How would you define the Rate on the basis of the disappearance of H2(g) such that you would obtain the same numerical value at this point in the reaction? How would you define it on the basis of the appearance of NH3(g)? All of the following expressions are equivalent: Rate = –Δ[N2]/Δt = –aΔ[H2]/Δt = +½Δ[NH3]/Δt 2. For each of the following experimentally determined rate law expressions, give the units for k, assuming that all species concentrations are in units of mol@L–1 and all rates are in units of mol@L–1@s–1. 2 N2O5(g) 6 4 NO2(g) + O2(g) 2 NO2(g) + F2(g) 6 2 NO2F(g) 2 NO(g) + O2(g) 6 2 NO2(g) CH3CHO(g) 6 CH4(g) + CO(g) 3. Rate = k[N2O5] Rate = k[NO2][F2] Rate = k[NO]2[O2] Rate = k[CH3CHO]3/2 s–1 (mol@L–1)–1@s–1 (mol@L–1)–2@s–1 (mol@L–1)–½@s–1 Consider the reaction: 2 NO(g) + 2 H2(g) ÷ N2(g) + 2 H2O(g) a. The rate law for this reaction is second order in NO and first order in H2. Write the rate law. Rate = k[NO]2[H2] b. If the rate constant for this reaction at 1000 K is 6.0 x 104 M–2@s–1, what is the reaction rate when [NO] = 0.035 M and [H2] = 0.015 M? Rate = (6.0 x 104 M–2@s–1)(0.035 M)2(0.015 M) = 1.1025 M@s–1 c. What is the reaction rate at 1000 K when [NO] = 0.070 M and [H2] = 0.015 M? Compared to conditions in part b, [NO] has doubled while [H2] is the same. Rate % [NO]2, so the rate will increase by a factor of 4. Rate = 4 × 1.1025 M@s–1 = 4.41 M@s–1 d. What is the reaction rate at 1000 K when [NO] = 0.035 M and [H2] = 0.030 M? Compared to conditions in part b, [H2] has doubled while [NO] is the same. Rate % [H2], so the rate will increase by a factor of 2. Rate = 2 × 1.1025 M@s–1 = 2.205 M@s–1 e. What is the reaction rate at 1000 K when [NO] = 0.070 M and [H2] = 0.030 M? Compared to conditions in part b, both reactant concentrations have doubled. The reaction is third order overall, so the rate will increase by a factor of (2)3 = 8. Rate = 8 × 1.1025 M@s–1 = 8.82 M@s–1 4. Consider the following kinetic data for the reaction 2 NO(g) + O2(g) ÷ 2 NO2(g). [NO] (M) [O2] (M) Initial Rate (M/s) 1 0.0126 0.0125 1.41 × 10–2 2 0.0252 0.0250 1.13 × 10–1 3 0.0252 0.0125 5.64 × 10–2 Exp. a. What is the rate law for the reaction? Compare Exp. 1 with Exp. 3, where [O2] is the same. The concentration of NO was doubled, causing a four-fold increase in the rate. Therefore, the reaction is second order in [NO]; i.e., Rate % [NO]2. Compare Exp. 3 with Exp. 2, where [NO] is the same. The concentration of O2 was doubled, causing a doubling of the rate. Therefore the reaction is first order in [O2]; i.e., Rate % [O2]. The overall rate law expression is Rate = k[NO]2[O2], which is third order overall. b. What is the value of the rate constant with the proper units? Solving the rate expression for k with any line of specific data will yield a value. Ideally, we would solve for k for each set of data and then take the average. Using just the first line of data (Exp. 1), we have 5. Consider a first order reaction, A ÷ products. Sketch a plot of ln [A] vs. t for such a reaction. Indicate on your sketch where the value of the logarithm of initial concentration of A, ln [A]o, is located. Is the slope positive or negative? How is the slope determined and how is it related to k? slope = ln[A]/t = -k ln[A] ln[A]o ln[A] t t1 t2 t the natural logarithm of [A]o is the y intercept. The slope, determined by measuring Δln[A] over Δt from some segment of the straight line, is negative and is equal to –k. 6. We have been calculating the values of the rate constant, k, from tables of kinetic data using the value of the rate for a specific set of concentrations in any one experiment. If the reaction were first order, why would it be better to use a plot of the logarithm of reactant concentration versus time to obtain the value of k? In a real experimental situation all of the data contain experimental errors. By choosing one line from a table of concentration and rate data, as we have been doing, we obtain a single value of k that is subject to all of the errors that this particular experiment might have. For data from a first order reaction, a plot of ln [A] vs. time would yield a series of data points that (owing to experimental error) would fall above or below the ideal straight line for such a plot. By using statistical methods (linear least-squares regression) we could draw the best straight line through all these experimental points, thus averaging out their individual errors. The k value obtained from this plot would reflect all of the data, and by using methods of statistics we could assess the degree of variation in its error. In other words, we would obtain a value of k that represents all of the data, not just a single experiment.