Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e.
Calculations:
The second order rate equation is:
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
Answer:
N₂ = 0.7515atm
O₂ = 0.1715atm
NO = 0.0770atm
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g)
Where Kp is defined as:
Pressures in equilibrium are:
N₂ = 0.790atm - X
O₂ = 0.210atm - X
NO = 2X
Replacing in Kp:
0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]
0.0460 = 4X² / 0.1659 - X + X²
0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²
-3.954X² - 0.0460X + 7.6314x10⁻³ = 0
Solving for X:
X = - 0.050 → False answer. There is no negative concentrations.
X = <em>0.0385 atm</em> → Right answer.
Replacing for pressures in equilibrium:
N₂ = 0.790atm - X = <em>0.7515atm</em>
O₂ = 0.210atm - X = <em>0.1715atm</em>
NO = 2X = <em>0.0770atm</em>
Answer:
20619.4793 years
Explanation:
The half life of carbon-14 = 5730 years
The formula for the half life for a first order kinetic reaction is:
Where,
is the half life
k is the rate constant.
Thus rate constant is:
5730 years=ln(2)/k
k = 1.21×10⁻⁴ years ⁻¹
Using integrated rate law as:
Where,
is the concentration at time t
is the initial concentration
Given that the final concentration contains 8.25 % of the original quantity which means that:
So,
ln(.0825)= -1.21×10⁻⁴×t
<u> t = 20619.4793 years</u>
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