Respuesta :
Answer:
Total pressure 5.875 atm
Explanation:
The equation for above decomposition is
[tex]2N_2O \rightarrow 2N_2 + O_2[/tex]
rate constant [tex] k = 1.94\times 10^{-4} min^{-1}[/tex]
Half life [tex]t_{1/2} = \frac{0.693}{k} = 3572 min[/tex]
Initial pressure [tex]N_2 O = 4.70 atm[/tex]
Pressure after 3572 min = P
According to first order kinematics
[tex]k = \frac{1}{t} ln\frac{4.70}{P}[/tex]
[tex]1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}[/tex]
solving for P we get
P = 2.35 atm
[tex]2N_2O \rightarrow 2N_2 + O_2[/tex]
initial 4.70 0 0
change -2x +2x +x
final 4.70 -2x 2x x
pressure of[tex] O_2[/tex] after first half life = 2.35 = 4.70 - 2x
x = 1.175
pressure of [tex]N_2[/tex] after first half life = 2x = 2(1.175) = 2.35 ATM
Total pressure = 2.35 + 2.35 + 1.175
= 5.875 atm
Answer:
The Total Pressure = 5.875
Explanation:
The Equation: 2N2O(g) -> 2N2(g) + O2(g)
The rate costant is "k" of the reaction : k= 19.4 * 10^-4 min^-1
Half period = 0.693 / k
=0.693/19.4 * 10^-4 min^-1 = 3572 min
The initial pressure of N2O, Po = 4.70 atm
The pressure of N2O after 3572 min = Pt
According to the first-order kinetics:
k= 1/t 1n P0/pt
19.4 * 10^-4 min^-1 = 1/3572 min 1n 4.70atm / Pt
1n 4.70atm / Pt = 0.692968
4.70atm / Pt = e^0.692968 = 2.00
Pt = 4.70atm / 2.00 = 2.35 atm
2N2O(g) -> 2N2(g) + O2(g)
The initial(atm) 4.70 0 0
The change(atm) -2x +2x x
Final(atm) 4.70-2x 2x x
Pressure of N20 after one half-life = Pt = 2.35 = 4.70-2x
Pressure of O2 after one half-life = Po = x = 1.175 atm
Pressure of O2 after one half-life = 2x = 2(1.175) = 2.35 atm
Total Pressure = 2.35 atm + 2.35 atm + 1.175 atm
= 5.875 atm
