Question

Answer the questions according to the following equation: Data for the temperature dependence of the reaction, NO2(g) O2(g) NO(g) O3(g) + + → are illustrated in the table below. k (M/s) 1.08 x 109 T(°C) 78.15 -43.15 2.95 x 10o9 -13.15 5.42 x 10 24.85 12.0x 10 95.85 35.5 x 10 Graph the data according to the Arrhenius equation and the appropriate axes. (Using graph paper will give more accurate results more points) (10 pts) a. b. What is the equation of the line for this graph? (2 pts) b. What is the activation energy for this reaction in kJ/mol? (2 pts) C. Calculate the k at 300 ℃ (2 pts)

Answer 1

Ans. #a.
T(OC) k (M/s) T(K) 1/ In k 78.15 1.08E+09 195.00 0.00513 20.8002 24.5 -43.15 2.95E+09 230.00 0.00435 21.8051 24 13.15 5.42E+09 260.00 0.00385 22.4134 24.85 1.20E+10 298.00 0.00336 23.2082 95.85 3.55E+10 369.00 0.00271 24.2928 23.5 23 22.5 21.5 21 y =-1435-7x + 28.071 R2 0.9941 20.5 0 0.001 0.002 0.003 0.004 0.005 0.006 1/T

#b. Equation of line: y = -1435.7x + 28.071

#c. Using        slope = (-Ea) / R

            Or, -1435.7 K = (-Ea) / (0.008314 kJ mol^-1 K^-1)

            Or, Ea = 1435.7K x (0.008314 kJ mol^-1 K^-1)

            Hence, Ea = 11.9364 kJ mol^-1

#d. The trendline equation of Arrhenius equation y = -1435.7x + 28.071 is in form of y = mx + C as            ln k = ln A + (-Ea / R) (1/ T)

            Where,

                        Slope = m = (-Ea / R) = 1435.7

                        X = 1 /T

                        C = y-intercept = ln A = 28.071

Given, temperature, T = 300.0⁰C = 573.15 K

Now, Putting the values in trendline equation-

            ln k = 28.071 + (-1435.7)K x (1 / 573.15 K)          

            Or, 2.303 log k = 28.071 + (-2.5049) = 25.5661

            Or, log k = 25.5661 / 2.303 = 11.101

            Or, k = 10^11.101

            Hence, k = 1.262 x 10¹¹