please help within five minutes 5
please help within five minutes 5
NH_{4}^{+} (aq) + NO_{2}^{–} (aq) â†’ N_{2} (g) + H_{2}O (l)
Experiment  [NH_{4}^{+}]_{i}  [NO_{2}^{–}]_{i}  Initial rate (M/s) 
1  0.24  0.10  7.2 x 10^{4} 
2  0.12  0.10  3.6 x 10^{4} 
3  0.12  0.15  5.4 x 10^{4} 
4  0.12  0.12  4.3 x 10^{4} 
First determine the rate law and rate constant.
Under the same initial conditions as in
Experiment 4, calculate [NH
_{4}
^{+}] at 284 seconds after the start of the reaction. In this experiment, both reactants are present at the same initial concentration.
The units should be M, and should be calculated to three significant figures.
2.
NO_{2}(g) + CO(g)> NO(g) + CO_{2}(g) 
The rate of the above reaction depends only on the concentration of nitrogen dioxide at temperatures below 225Â°C.
At a temperature below 225Â°C the following data were obtained:
Time (s) 
[NO_{2}] (mol/L) 
0  0.520 
1.10Ã—10^{3}  0.481 
2.75Ã—10^{3}  0.433 
4.12Ã—10^{3}  0.400 
8.25Ã—10^{3}  0.325 
1.65Ã—10^{4}  0.236 
Which of the following expressions for the rate law (either differential or integrated) are completely consistent with the above experimental data.
k[NO_{2}]^{3} = d[CO]/dt
ln([NO_{2}]/[NO_{2}]_{0}) = kt
k[NO_{2}]^{2} = d[NO]/dt
d[CO]/dt = k[NO_{2}]^{2}
1/[NO_{2}] – 1/[NO_{2}]_{0} = kt
3.The data below was collected for the reaction: 2N
_{2}O
_{5}(g) â†’ 4NO
_{2}(g) + O
_{2}(g) at a temperature of 25Â°C:
time (hr)  [N_{2}O_{5}](M) 

0.0  0.374 
0.9  0.337 
1.8  0.299 
2.9  0.262 
4.2  0.224 
5.7  0.187 
7.5  0.150 
9.9  0.112 
13.2  0.075 
18.9  0.037 
Using the data, Find t
_{Â½}.
Plot [N
_{2}O
_{5}], ln[N
_{2}O
_{5}] and 1/[N
_{2}O
_{5}] as a function of time. You can cut and paste the data from the table into an Excell worksheet. From the graphs, determine the reaction order and the reaction rate constant.
Reaction order:
Rate constant:
The effect of temperature on the rate of a reaction was studied and the following data obtained:
k (s^{1})  T (Â°C) 
1.91Ã—10^{4}  5 
2.25Ã—10^{4}  7 
3.07Ã—10^{4}  11 
3.58Ã—10^{4}  13 
4.84Ã—10^{4}  17 
7.49Ã—10^{4}  23 
1.06Ã—10^{3}  28 
1.22Ã—10^{3}  30 
It is known that the variation of the rate constant k with the absolute temperature T is described by the Arrhenius equation:

where E_{a} is the activation energy, R is the universal gas constant and A is the preexponential factor (units of the rate constant). Taking the natural logarithm of both sides affords:

For a plot of y = ln k versus x = 1/T, calculate the slope of the best straight line using linear regression.
Calculate the activation energy E_{a}.