Nitrosyl chloride is produced from the reaction of nitrogen monoxide and chlorin
ID: 1018733 • Letter: N
Question
Nitrosyl chloride is produced from the reaction of nitrogen monoxide and chlorine: 2NO(g) + CI_2(g) rightarrow 2NOCI(g) The following initial rates at a given temperature were obtained for the concentrations listed below. From the data, what is the experimental rate law? A) Rate = k[Cl_2] B) Rate = k[NO][CI_2]^2 C) Rate = k[NO][CI_2] D) Rate = k[NO] E) Rate = k[NO] [CI_2]^1/2 A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH_2(g) N_2(g) + 3H_2(g) At equilibrium, it was found that the concentration of H_2 was 0.481 M, the concentration of N_2 was 0.0160 M, and the concentration of NH_3 was 0.291 M. What was the initial concentration of ammonia? A) 0.23 M B) 0.355 M C) 0.226 M D) 0.307 M E) 0.160 M For the hypothetical first-order reaction A rightarrow products, k = 0.0635 a6-1. If the initial concentration of A is 0.624 M, how long would it take for A to be 56.9% consumed? A) 15.7 s B) 13.3 s C) 10.9 s D) 25.2 s E0 8.88 s Consider the following equilibrium: 4NH_3(g) + 3O_2(g) 2N_2(g) + 6H_2O(g) A) 0.30 - 2/3x B) 0.40 - 3/2 x C) 0.30 - 3/2 x D) 4/6 x E) 0.30 - x A reaction that is second-order in one reactant has a rate constant of 3.4 times 10^-2 L/(mol middot s), If the initial concentration of the reactant is 0.260 mol/L, how long will it take for the concentration to become 0.130 mol/L? A) 110 s B) 20 s C) 110 s D) 200 s E) 57 s A chemical reaction that is first-order in X is observed to have a rate constant of 2.2 times 10^-2 s^-1. If the initial concentration of X is 1.0 M, what is the concentration of X after 180 s? A) 0.60 M B) 0.019 M C) 0.98 M D) 52 M E) 0.20 MExplanation / Answer
19. from the given data we get, rate of the reaction would be,
From 1 and 2 for NO we get order = 2
From 1 and 3 for Cl2 we get order = 1
C) rate = k[NO2]^2[Cl2]
20. From the given equilibrum data,
initial concentration of [NH3] = 0.016 x 2 + 0.291 = 0.323
A) 0.323
21. for first order reaction,
t = (ln[Ao] - ln[A])/k = [ln(0.624) - ln(0.624 x (1 - 0.569))]/0.0635 = 13.3 s
B) 13.3 s
22. For the reaction with x moles of water at equilibrium, moles of NH3 present would be,
A)
23. For second order reaction,
t = (1/A - 1/Ao)/k = (1/0.13 - 1/0.26)/0.034 = 113.12 s
A) 110 s
24. For first order reaction,
ln[A] = ln(1) -180 x 0.022
[A] = 0.019 M
B) 0.019 M
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