Question

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 1

Answer : The partial pressure of $O_2$ 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 :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{168.6min}$

$k=4.11\times 10^{-3}min^{-1}$

Now we have to calculate the partial pressure of $O_2$

The balanced chemical reaction is:

                           $2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)$

Initial pressure   760                0             0

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

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

$t=\frac{2.303}{k}\log\frac{P_o}{P_t}$

where,

k = rate constant

t = time passed by the sample  = 215 min

a = initial pressure of $N_2O_5$ = 760 torr

a - x = pressure of $N_2O_5$ at equilibrium = (760-2x) torr

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

$215=\frac{2.303}{4.11\times 10^{-3}}\log\frac{760}{760-2x}$

$x=222.93torr$

The partial pressure of $O_2$ = x = 222.93 torr