Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitrogen pentoxide and has a half-life of 2.81 h at 25 ∘C. If a 1.7 −L reaction vessel initially contains 755 torr of N2O5 at 25 ∘C, what partial pressure of O2 will be present in the vessel after 215 minutes?

Answer : The partial pressure of is, 222.93 torr

Explanation :  

Half-life = 2.81 hr = 168.6 min

First we have to calculate the rate constant, we use the formula :

Now we have to calculate the partial pressure of

The balanced chemical reaction is:

Initial pressure   760                0             0

At eqm.             (760-2x)            4x            x

Expression for rate law for first order kinetics is given by:

where,

k = rate constant

t = time passed by the sample  = 215 min

a = initial pressure of  = 760 torr

a - x = pressure of  at equilibrium = (760-2x) torr

Now put all the given values in above equation, we get:

The partial pressure of = x = 222.93 torr

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :