Question

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

Answer 1

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

Answer 2

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