In a study of the decomposition of ammonia on a platinum surface at 856^degree C

Question

In a study of the decomposition of ammonia on a platinum surface at 856^degree C NH_3 rightarrow 1/2 N_2 + 3/2 H_2 the following data were obtained: The observed half-life for this reaction when the starting concentration is 1.02 times 10^-2 M is []s and when the starting concentration is 5.10 times 10^-3 M is []s. The average rate of disappearance of NH_3 from t = 0 s to t 997 s is [] M s^-1. The average rate of disappearance of NH_3 from t = 997 s to t = 1.50 times 10^3 s is [] M s^-1 Based on these data, the rate constant for this [] order reaction is [] M s^-1.

Explanation / Answer

Q1.

half life = time required for concentration to drop to 1/2 (50%)

so

for 1.02*10^-2 ---> 1/2*(1.02*10^-2) = 0.0051 = 5.1*10^-3 M

so... t = 997 seconds required

for second point:

5.1*10^-3 --> (5.1*10^-3)/2 = 0.00255 = 2.55*10^-3

time was = 1.5*10^3 = 1500

dt = 1500-997 = 503 seconds

Q2

average rate of NH3

RAte = Change in Concentrations / change in time

RAte = (5.1*10^-3 - 1.02*10^-2) /(997-0) = -0.00000511 M/s

Q3

average rate form 997 to 1500

RAte = (2.55*10^-3 - 5.1*10^-3) / (1500-997) = -0.000005069 M/s

Q4

Based on this data

k = slope...

so

0.00000511 is approx = 0.000005069 so this is constant

k = 0.00000511 1/(Ms)

this must be 2nd order since k is constant

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