Rate constant k?

Question

The following reaction 2NO + H2 -> N2O + H2O goes to completion and is known to follow the rate law as shown below:
dP(N2O)/dt = k[P(NO)]^2[P(NO)]
The data in the following table is obtained:

run P(NO)o P(H2)o Half-life Temperature
(mm Hg) (mm Hg) (sec) (oC)
1 600 10 19.2 820
2 600 20 X 820
3 10 600 830 820
4 20 600 Y 820
5 600 10 10 840

(a) Insert the missing half-lives (X and Y) in the table above.
(b) Calculate k at 820oC in unit of (mm Hg)^-2 s^-1.

Answer 1

Lancenigo di Villorba (TV), Italy

Wow! You are a great source of Kinetic Questions!

This time, the trouble interests a Gas Phase Reaction, the following one :

2 NO(g) + H2(g) <---> N2O(g) + H2O(g)

You gave me a kinetic expression of the Reaction's Rate

Reaction's Rate = k * (p,NO)^2 * p,H2

which will aid in order to correlate the Initial Reactant's Abundances with the Half-Life of the Mixture Preparated.
I may pursue the announced goal since I start arising the Stoichiometric Correspondances

Reaction's Rate = k * (p,NO)^2 * p,H2
d[p,NO] / dt = 0 - 2 * Reaction's Rate
d[p,H2] / dt = 0 - Reaction's Rate

so I rearrange the variables in the Kinetic Expression

d[p,NO] / dt =
= 0 - 2 * k * (p,NO)^2 * (0.5 * p,NO + p°,H2 - 0.5 * p°,NO)

hence I manipulate the Equation in order to estimate Its Integral Forms as the following one

[2 / (p°,H2 - 0.5 * p°,NO)^2] * LN[(p°,H2 - 0.5 * p°,NO +
+ 0.5 * p,NO) / (p°,H2)] - [0.5 / (p°,H2 - 0.5 * p°,NO)^2] *
* LN[(p,NO) / (p°,NO)] - [1 / (p°,H2 - 0.5 * p°,NO)] * [(1 / p,NO) - (1 / p°,NO)] + 2 * k * t = 0

Now, I modify this Integral Expression to introduce the HALF-TIME, that is the time which Main Reactant joins to Half Value than Its Initial One

[2 / (p°,H2 - 0.5 * p°,NO)^2] * LN[(p°,H2 - 0.25 * p°,NO) /
/ (p°,H2)] - [0.5 / (p°,H2 - 0.5 * p°,NO)^2] * LN[0.5] -
- [1 / (p°,H2 - 0.5 * p°,NO)] * [1 / p°,NO] +
+ 2 * k * Half-Time = 0

Since you gave me several experimental data, I related them using the overwritten PARTIAL PRESSURE VS. HALF-TIME EQUATION : in this way, I calculated the requested unknown values

§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§
§§§§§§§§§ This time I refer to RUN n.1 §§§§§§§§
§§§§§§§ k = 1.0E-7 [torr^-2 * s^-1] §§§§§§§§§§§§
§§§§§§ (Kinetic Constant for T = 820°C) §§§§§§§§
§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§§
**********************************************************
**************** This time I refer to RUN n.2 **********
******************** X = 10 [s] ***************************
**** (Half-Time for p°,NO = 600 torr, p°,H2 = 20 torr
and T = 820°C) ****************************************
**********************************************************
#######################################
########This time I refer to RUN n.4 #########
############## Y = 410 [s] ###############
## (Half-Time for p°,NO = 20 torr, p°,H2 = 600 torr
and T = 820°C) ###########################
#######################################

I hope this helps you.

Answer 2

7fc56270e7a7fa81a5935b72eacbe29ctivated complicated 7fc56270e7a7fa81a5935b72eacbe29n activated complicated is a intense potential, risky intermediate which exists very quickly whilst activation potential offered to a reaction is absorbed via the reaction. as quickly as formed it at present day decomposes to type products of decrease potential state or extra stability. 7fc56270e7a7fa81a5935b72eacbe29s cost=ok[A]^3/2 unit of cost consistent= (concentration)^a million-n *(time)^ -a million concentration= mol/litre time= sec n= order 7fc56270e7a7fa81a5935b72eacbe29s in above question n= 3/2 subsequently, unit is= (conc)^a million-3/2 (time)^ -a million = (mol/litre)^ -a million/2 (sec)^ -a million = mol^ -a million/2 litre^a million/2 sec^ -a million

Answer 3

X = 16.4 Y = 740

k = 34.5