Question

For a reaction at 273k, it is observed the pressure of NO falls from 700 mmHg to 500 mmHg in 250 secs. Calculate the average rate of reaction​

Answer 1

Answer:

NO(g) + O3(g) --> NO2(g) + O2

Average rate = (change in pressure/change in time)

(∆P)/( ∆t)

=(700 – 500)/(250)

= (200)/ (250) mmHgs-1

1atm = 760 mmHg

Therefore 1 mmHg = (1/760) atm

(200/250) x (1/760)

=1.053 x 10-3 atm sec-1.

Average rate = (∆C)/( ∆t) = (∆n/V)/( ∆t)

Also PV = nRT

Or (n/V) = (P/RT)

∆(n/V) = (∆P/RT)

Average Rate = (∆P/∆t) x (1/RT)

= (1.053 x 10-3) atm sec-1/(0.082 litre atm k-1mol-1 x 273 k)

=4.7 x 10-5 mol-1sec-1