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The rate constant for the first-order decompositio

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Answer #1

2N2O5(g)----------> 4NO2(g) + O2(g)      K = 3.38*10-5 sec-1

   t1/2   = 0.693/K

            = 0.693/3.38*10-5

             = 2.05*104 sec

K   = 2.303/t log[P0]/[P]

3.38*10-5   = 2.303/50 log500/P

log500/P    = 3.38*10-5 *50/2.303

log500/P    = 0.000734

500/P          = 100.000734

500/P         = 1

P               = 500torr

ii)

t = 20min = 20*60 = 1200sec

3.38*10-5   = 2.303/1200 log500/P

log500/P    = 3.38*10-5 *1200/2.303

log500/P    = 0.01761

   500/P       = 100.01761

500/P       = 1.0414

P              = 500/1.0414   = 480.13torr >>>answer

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Answer #2

2N2O5(g)----------> 4NO2(g) + O2(g)      K = 3.38*10-5 sec-1

   t1/2   = 0.693/K

            = 0.693/3.38*10-5

             = 2.05*104 sec

K   = 2.303/t log[P0]/[P]

3.38*10-5   = 2.303/50 log500/P

log500/P    = 3.38*10-5 *50/2.303

log500/P    = 0.000734

500/P          = 100.000734

500/P         = 1

P               = 500torr

ii)

t = 20min = 20*60 = 1200sec

3.38*10-5   = 2.303/1200 log500/P

log500/P    = 3.38*10-5 *1200/2.303

log500/P    = 0.01761

   500/P       = 100.01761

500/P       = 1.0414

P              = 500/1.0414   = 480.13torr >>>answer

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