The half-life of the substance is 3.106 years.
<h3>What is the formula for exponential decay?</h3>
- The exponential decline, which is a rapid reduction over time, can be calculated with the use of the exponential decay formula.
- The exponential decay formula is used to determine population decay, half-life, radioactivity decay, and other phenomena.
- The general form is F(x) = a.
Here,
a = the initial amount of substance
1-r is the decay rate
x = time span
The equation is given in its correct form as follows:
a =
×
As this is an exponential decay of a first order reaction, t is an exponent of 0.8.
Now let's figure out the half life. Since the amount left is half of the initial amount at time t, that is when:
a = 0.5 a0
<h3>Substituting this into the equation:</h3>
0.5
=
×
0.5 = 
taking log on both sides
t log 0.8 = log 0.5
t = log 0.5/log 0.8
t = 3.106 years
The half-life of the substance is 3.106 years.
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Answer:
The equilibrium concentrations are:
[SO2]=[NO2] = 0.563 M
[SO3]=[NO] = 1.04 M
Explanation:
<u>Given:</u>
Equilibrium constant K = 3.39
[SO2] = [NO2] = [SO3] = [NO] = 0.800 M
<u>To determine:</u>
The equilibrium concentrations of the above gases
Calculation:
Set-up an ICE table for the given reaction

I 0.800 0.800 0.800 0.800
C -x -x +x +x
E (0.800-x) (0.800-x) (0.800+x) (0.800+x)
The equilibrium constant is given as:
![Keq = \frac{[SO3][NO]}{[SO2][NO2]}=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BSO3%5D%5BNO%5D%7D%7B%5BSO2%5D%5BNO2%5D%7D%3D%5Cfrac%7B%280.800%2Bx%29%5E%7B2%7D%7D%7B%280.800-x%29%5E%7B2%7D%7D)

x = 0.2368 M
[SO2]=[NO2] = 0.800 -x = 0.800 - 0.2368 = 0.5632 M
[SO3]=[NO] = 0.800 +x = 0.800 + 0.2368 = 1.037 M
A) 2, 3 , 2 simply valence the rest