Answer :

2 N_{2}O_{5}(g)2 N_{2}O_{4} + O_{2(G)}

GIVEN,

At t = 0 0. 5 atm 0 atm 0 atm

At time t 0. 5 - 2x atm 2x atm x atm

__To find total pressure__:

P_{t} = sum of partial pressures of all reactants and product_{s}

P_{t} = 0. 5 - 2x + 2x + x

x = p_{t} - 0. 5

p = 0. 5 - 2x

= 0. 5 - 2(p_{t} - 0. 5)

= 1. 5 - 2p_{t}

At t = 100 s ;p_{t} = 0. 512 atm

p = 1. 5 - 20. 512 = 0. 476 atm

__Formula to find rate constant of a first order reaction__

k = log

k = log

k = 0. 0216 = 4. 9810 ^{- 4} s ^{– 1}

__conclusion__

the rate constant for the following reaction is 4. 9810 ^{- 4} s ^{- 1}

**OR**

Given:

Difference between activation energies = 24. 9 kJ/mol

__The Arrhenius equation:__

take log on both sides, log k = log (Ae^{-Ea})

log k = log A -

LETS ASSUME,

For reaction 1 log k_{1} = log A -

For reaction 2 log k_{2} = log A -

Subtract 1 from 2

__Conclusion:__

Ratio between rate constants = 2. 198 ^{4}

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