The data in the table give the temperature dependence of the rate constant for the reaction N2O5 (g) → 2 NO2 (g) + 1/2 O2 (g) Plot these data in the appropriate way to derive the activation energy for the reaction. T(K) k(s*) 338 4.87 x 10¬8 328 1.50 x 10-8 318 4.98 x 10¬4 308 1.35 x 10¬4 298 3.46 x 10-8 273 7.87 x 10-7 Activation energy = kJ/mol

The data in the table give the temperature dependence of the rate constant for the reaction N2O5 (g) → 2 NO2 (g) + 1/2 O2 (g) Plot these data in the appropriate way to derive the activation energy for the reaction. T(K) k(s*) 338 4.87 x 10¬8 328 1.50 x 10-8 318 4.98 x 10¬4 308 1.35 x 10¬4 298 3.46 x 10-8 273 7.87 x 10-7 Activation energy = kJ/mol

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Given that the reaction 2 H2(g) + O2(g)-> 2 H₂O(g) is exothermic, which of the following is true of the reaction 2 H₂O(g) 2 H2(g) + O2(g)? Its activation energy is lower than that of 2 H2(g) + O2(g) -> 2 H₂O(g) Its activation energy is the same as that of 2 H2(g) + O2(g) -> 2 H₂O(g). Its activation energy is higher than that of 2 H2(g) + O2(g) ->2 H₂O(g)- O There is no relationship between its activation energy and that of 2 H2(g) + O2(g)-> 2 H₂O(g).
For a certain reaction, the activation energy is found to be (1.77x10^1) kJ/mol. If the rate constant is (3.6x10^-5) at (3.78x10^2) K, what is the rate constant at (6.290x10^2) K? R = 8.314 J/Kmol; T(K) = T(°C) + 273 Report your answer in the same units as k1, with 2 sig figs.
Calculate the activation energy for the isomerization of reactant X to product Y using the following data for the variation of rate constant (k) with temperature. T (oC) 477 523 577 623 k 0.00018 0.0027 0.03 0.26
x₁ = What two points should be plotted to graphically determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values. x₂ = A- →→ products rise: T (K) k (S-¹) 225 0.392 625 0.637 slope: Determine the rise, run, and slope of the line formed by these points. Ea = What is the activation energy of this reaction? y₁ = y₂ = run: J/mol
For the reaction A + B C + D, various initial rate measurements were run beginning with A and B only, and beginning with C and D only. From the data below, calculate the rate law for both the forward and backward reactions. Then determine the equilibrium constant for the reaction. Rate (M/s) 1.081 x 10 6.577 x 10 6.568 x 10 [A] 0.660 [B] 1.23 4.01 1.23 2.25 4.01 Rate (M/s) 7.805 x 10 1.290 x 10" 1.300 x 10 [C] 2.88 [D] 0.995 2.88 1.01 1.65 1.65
Consider a possible gas-phase reaction between Ca+ and CH4+ to yield Ca+2 and CH4. What important feature of the reactants is neglected in collision theory which might work to reduce the probability of reactive collisions between them?
Determine the activation energy graphically. K(M¹-s¹) T(°C) 1.87×10 600 650 0.0113 0.0569 0.244 700 750
Determine the rate law and the value of k for the following reaction using the data provided. NO,(9) + O3(g) – NO3(g) + O2(g) [NO,li (M) 0.1 0.1 [03li (M) Initial Rate (ms-1) 1 0.33 1.42 2 0.66 2.84 0.25 0.66 7.10 Rate = 43 M's INO,0,] S. -25c O Rate = 1360 M2s"INO,110,] S. Rate = 227 M25s INO,J[0,125 O Rate = 430 M?s INO,10,
[References) CHEMWORK Suppose K = 4.5 x 103 at a certain temperature for the reaction PCl (9) → PC13 (9) + Cl2 (9) If it is found that the concentration of PCI5 is twice the concentration of PCls, what must be the concentration of Cl2 under these conditions? (Cl2] = M Submit Show Hints Submit Answer 10 item attempts remaining Try Another Version pt pt I pt 1 pt 1 pt 1 pt (Previou (? 中 401 >>A 4. 6. 6. ERT F K B
A student collected conducted 6 kinetics experiments for one reaction by holding the concentrations of ALL the reactants and catalyst constant and varying the temperature. After determining the rate constant for this reaction at the 6 different temperatures, an Arrhenius Plot was made. The graph resulted in a straight line. The equation for the straight line was: y= -4,257.15 x + 14.326 What is the Activation Energy (in Joules/mol) for this reaction at 30 oC? Do not include the units in your answer.
Consider the methanol synthesis reaction represented below. CO (g) + 2 H2 (g) = CH3OH (g) The initial concentrations were: [CO ] = 1.75 M, [H2] = 1.00 M, and [CH3OH] = 0.65 M. %3D %3D If the equilibrium [CO] was found to be 1.60 M, calculate the equilibirium constant (K) for this reaction. Note: Reaction temperature was kept constant throughout the process. 44 2.0 1.0 62
Consider the reaction data. What two points should be plotted to graphically determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values. X1 = x2 = A →→ products T (K) k (S-¹) 275 0.370 875 0.675 rise: 0.00364 slope: 0.00114 Determine the rise, run, and slope of the line formed by these points. Ea = Incorrect 0.00250 0.00415 Incorrect What is the activation energy of this reaction? 0.00149 Y₁ = Incorrect y2 = run: -0.994 -0.393 0.601 Incorrect J/mol