In reference to Q.5, the rate of appearance of `O_(2)` for the last 2000 sec is

Options

A. `2.0xx10^(-5) atm s^(-1)`
B. `3.0xx10^(-5)atm sec^(-1)`
C. `4.0xx10^(-5)atm s^(-1)`
D. `1.0xx10^(-5)atm sec^(-1)`

Answer

D

Explanation

In many kinetic expriments, we do not measure concentration directly. Instead, we measure a property that can be related to the concentration of one substance in the system. In a gas phase system, we may measure the total pressure of the gasese. A total-pressure measurement usually is a good way to determine the rate of a process that causes a chagein the number of moles of gas in a system.
We can find the rate of disapparance of nitrou oxide by using the total pressure to find the pressure of nitrous oxide at any given time. The relationship between `P_(tau)` and `P_(N_(2)O)` is derived from the equation for the reaction:
`underset(0.29atm)(2N_(2)O(g))rarrunderset(0atm)(2N_(2)(g))+underset(0atm)(O_(2)(g))`
Let `2P` be the pressure in atmosphere of `N_(2)O` that reacts in the time interval of interest. The pressure of each gaw at the end of this interval can be tabulated:
time `(0.29-2p)` atm `2p` atm`p` atm
According to Dalton's law of partial presssure, the measured total pressure `(P_(tau))` is the sum of these pressre:
`P_(tau)P_(N_(2)O)+P_(N_(2))+P_(O_(2))`
`=(0.29-2p)+2p+p`
`0.29+p`
The total pressure at the end of `300sec` is given as `0.33atm` Therefore
`0.33atm=(0.29+p)atm`
or `p=(0.33-0.29)atm`
`=0.04atm`
Thus, at time `=300sec`
`P_(N_(2)O)=(0.29-2p)atm`
`=0.29atm-2(0.04atm)`
`0.29atm-0.08atm`
`=0.21atm`
This value can be used to final the rate of disappearance of `N_(2)O` for the first `30osec`
`-(DeltaP_(N_(2)O))/(Deltat)=-((0.21atm-0.29atm))/(300sec)`
`=2.7xx10^(-4)atmsec^(-1)`