Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

13) Determine the rate law and the value of k for the following reaction using t

ID: 635589 • Letter: 1

Question

13) Determine the rate law and the value of k for the following reaction using the data provided. 2NO(g) + O2(g)-2NO2(g) [Noli (M) [O2li (M) Initial Rate (Mpls-1) 0.030 0.0055 8.55 10-3 Lo2 0.030 0.060 0.0110 1.71 x 10-2 0.0055 3.42 x 10-2 [No 1 o) A Rate $7 M-1s1 [NOJ[O21 B) Rate 1.7 x 103 M-2s [No2102 C) Rate 3.8 M-12s1[NOJIO212 D) Rate 3.1 x 10 M-3 NO21022 E) Rate 94 x 103 M-2s-1 [Nolo212 s C 14) Nickel has a face-centered cubic structure and has a density of 8.90 g/cm3. What is its atomic radius? (face- centered cubic structure has an edge length of 2Rr, 1 pm?012 m; N-602210) 58. 3 A) 249 pm B) 353 pm 27to-23 25 pm D) 997 pm u-4.37 24.37 15) Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reaction: 78 9 PCI3(g) + Cl2(g) PCls(g) 153)(94.2 An equilibrium mixture at 450 K contains PPC 94.2 mm Hg, Pc2- 119.3 mm Hg, and 938 s-988 mm Hg. What is the value of Kp at this temperature?(4.2)19.3 A)).50 x 10-2 B) 2.53 x 10-2 C) 66.7 D) 1.02 E) 4.63 16) Determine the [OH ] concentration in a 0.344 M Ba(OH)2 solution. A) 0.344 M B) 0.463 C) 0.334 . D) 13.666 0.688 M 344 2 344-

Explanation / Answer

Ans 13 : B) 1.7 M-2s-1[NO]2[O2]

When the concentration of O2 is kept constant and that of NO is doubled , the rate of reaction becomes four times , indicating the order of reaction to be 2 with respect to NO.

Now when the concentration of NO is kept constant and that of O2 is doubled , the rate of reaction also doubles , indicating the order of reaction to be 1 with respect to O2.

So the rate law will be given as :

Rate = k[NO]2[O2]

Putting the values from trial 1 :

8.55 x 10-3 = k (0.030)2 (0.0055)

k = 1.7 x 103 M-2s-1

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote