(D) (E) r=k r=k- IA] 14( If the rate law is: r = k[N2][H2], halving ration of N2

Question

(D) (E) r=k r=k- IA] 14( If the rate law is: r = k[N2][H2], halving ration of N2 and doubling the the concent concentration of H2 will have what effect on the reaction rate? (A) The rate is same (B) Double the rate (C) Halve the rate (D) Increase the rate by four times (E) Decrease the rate by a factor of ¼ 15: Which energy diagram best matches an endothermic reaction with the following mechanism? Step 1: E + S Es Step 2: ES EP Step 3: EP E+ P 90 o (B) 80 70 70 550 so 50 40 20 10 30 20 10 Reaction Progress Reaction progress 90 s (D) 70 70 30 10 10 Reaction Progress Reaction Progress

Explanation / Answer

14 : Since N2 is halved , H2 is doubled :

rate = k* [N2 /2 ]*[2H]2

New rate = 1/2 * 22 = 2 time the old rate

15: C
Firstly , since the reaction is endothermic, the final product should be at a lower energy level . (Delta energy is neative). So only third and fourth options are possible.
now: E+S ES and EP E+P are equilibrium reaction whereas ES -> EP is a completely forward reaction . So, energy barrier for E+S ES and EP E+P is lower than ES -> EP as complete transformation is not raking place in the reaction . Due to equlibrium taking place, the energy barrier peaks for first and third step is lesser.

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