Question

In: Chemistry

The specific rate constant for the first-order decomposition of N2O5(g) to NO2(g) and O2(g) is 7.48×10−3s−1...

The specific rate constant for the first-order decomposition of N2O5(g) to NO2(g) and O2(g) is 7.48×10−3s−1 at a given temperature.

A. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.100 atm to rise to 0.150 atm .

B. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.100 atm to rise to 0.200 atm .

C. Find the total pressure after 110 s of reaction.

Solutions

Expert Solution

Answer – We are given, first order reaction –

N2O5(g) -----> NO2(g) + O2(g) , rate constant , k = 7.48*10-3 s-1

A) In this one there is given, Ao = 0.100 atm , At = 0.150 atm, t = ?

We know first order integrated law –

ln At / Ao = -k*t

ln 0.150 /0.100 = -7.48*10-3 s-1 * t

0.405 = -7.48*10-3 s-1 * t

So, t = 54.21 s

B) In this one there is given, Ao = 0.100 atm , At = 0.200 atm, t = ?

We know first order integrated law –

ln At / Ao = -k*t

ln 0.200 /0.100 = -7.48*10-3 s-1 * t

0.405 = -7.48*10-3 s-1 * t

So, t = 92.67 s

C) We know Ao = 0.100 atm , t = 110 s, so

ln At / Ao = -k*t

ln At /0.100 = -7.48*10-3 s-1 * 110 s

                   = -0.823

Taking antiln from both side

At /0.100 = 0.439

At = 0.100 atm * 0.439

     = 0.0439 atm

So total pressure after 110 s of reaction = 0.100+0.0439

                                                                = 0.144 atm

queen_honey_blossom answered 2 years ago
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