7. For the reaction below, the rate law has been found to be rate kINH INO. In o

Question

7. For the reaction below, the rate law has been found to be rate kINH INO. In one experiment where initial [NO21 was 0.0400 M and initial [NH4'] was 0.200 M, the observed rate was 2.15 x 10-6 M/s. What is the value of k in M 's1? 2.69 x 10-4 b. 3.72x103 c. 2.69810-10 d. 1.72x10-8 e. ¡72x104 8. A first order reaction has a half life of 16.0 hours. How long will it take for the concentration to decrease to 12.5% of its original value? a. 2.0 hrs b. 16.0 hrs c 128 hrs d.) 48 hrs e. 32.0 hrs 9. What is the half-life of a first-order reaction having a rate constant of 60 s-12 0.012 s b. 0.017s c. 0.023 s d. 0.033s e. 30 s 10. What is the half-life of a fist-order reaction having a raie congtent of 23 s t of 25 s-1? 166 s b. 750s c. 1040 s 0.0277 s e. 0.29 s a. on has the reaction constant k-2.61x 105 at 1900 °C and k-3.02x 10-3 at 2500 ? What is the value of the activation energy Ea for the reaction? a. 31.3 kJ 160 kJ c. -160 kJ d. 462 kJ e. 13.9 kJ ct 12. For the following reaction, the rate law is: Rate kNO2lIF2l. If the initial rate of this reaction is s when the initial concentration of NO2 is 0.0060 M and F2 is 0.0040 M, the value of the 9.6 × 104 Mi rate constant, k, is 4.0x 101 M-ls d. 2.0 x 101 M-1s-1 e. 0.025 M-1s1 a. 2.3 x 10-8 M-1s-1 b. 6.7103 M-1s-1 13. A reaction with a rate law of Rate k[Aj, has a rate constant of 1. 2 x 102 s1. If the initial concentration of A is 2.0 M, what is the concentration after 200.0 seconds? a. 6.0x 103M b. 1.0M The reaction C c. 1.7 M d. 0.55 M 0.18 M 3 x 104 s. If the initial concentration o 14. -products is a first-order reaction with t12-1 C is 0.200 M, what will be the concentration of C after 2.6 x 104 s? a 0.025 M b. 0.800 M c. 0.100 M d. 0.200 M Which change (at constant temperature and pressure lowest [O 0.050 M ) would not result in a substantial increase in 15. tro 10.5686 2 CH40) CH3CH3(g) + H20) b. FeF30o) +3 H20u Fe(OH)36) + 3 HF(g c. 2 NaNO3(5)-+ 2 NaNO2(s) + 0ap 10.40 e. 2 NO28) 2NO) 028)

Explanation / Answer

Ans 12 : c) 4.0 x 101 M-1s-1

The rate law is given as :

Rate = k[NO2][F2]

putting all the values given :

9.6 x 10-4 = k (0.0060) (0.0040)

k = (9.6 x 10-4) / (0.0060) (0.0040)

= 4.0 x 101 M-1s-1

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