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3 years ago

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The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

1. If the initial pressure of N2O is 4.70 atm at 730°C, calculate the total gas pressure after one half-life. Assume that the volume remains constant. slatter

2 answers:

jekas [21]3 years ago

8 0

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

grin007 [14]3 years ago

5 0

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

$2N_2O \rightarrow 2N_2 + O_2$

rate constant $k = 1.94\times 10^{-4} min^{-1}$

Half life $t_{1/2} = \frac{0.693}{k} = 3572 min$

Initial pressure $N_2 O = 4.70 atm$

Pressure after 3572 min = P

According to first order kinematics

$k = \frac{1}{t} ln\frac{4.70}{P}$

$1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}$

solving for P we get

P = 2.35 atm

$2N_2O \rightarrow 2N_2 + O_2$

initial 4.70 0 0

change -2x +2x +x

final 4.70 -2x 2x x

pressure of $O_2$ after first half life = 2.35 = 4.70 - 2x

x = 1.175

pressure of $N_2$ after first half life = 2x = 2(1.175) = 2.35 ATM

Total pressure = 2.35 + 2.35 + 1.175

= 5.875 atm

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Explanation:

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According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

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So, P = MRT

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= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

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