At high temperatures, `N_(2)O` decompose to `N_(2)` and `O_(2)`. If the change in totqal pressure with time is measured, the following data are obtained from an initial pressure of `N_(2)O` of `0.29` atm at `970 K`
`{:(P_(T)(atm),0.29,0.33,0.36,0.39,0.41),(t(sec),0,300,900,2000,4000):}`
The rate of disapperance of `N_(2)O` for the first 300 sec will be

Options

A. `2.7xx10^(-4)atm sec^(-1)`
B. `1.4xx10^(-4)atm sec^(-1)`
C. `2.0xx10^(-5)atm sec^(-1)`
D. `1.0xx10^(-5)atm sec^(-1)`

Answer

A

Explanation

In many kinetic experiments, 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 gases. A total-pressure measurement usually is a good way to determine the rate of a process that causes a change in the number of moles of gas in a system. We can find the rate of disapperarance of nitrous oxide by usingn the total pressure to find the pressure of nitrous oxide at any given time. The relationship between `P_(T)` and `P_(N_(2)O)` is derived from the equation for the reactionL: `underset(s tart),underset(0.29atm)(2N_(2)O(g))rarrunderset(0atm)(2N_(2)(g))+underset(0atm)(O_(2)(g))`
Let `2P` be the pressure in atmospheres of `N_(2)O` that reacts in the time interval of interest. The pressure of each gas at the end of this interval can be tabulated:
tiem `(0.29-2p)atm2patmpatm`
According to Dalton's law of partial pressures, the measured total pressure `(P_(T))` is the sum of these pressures:
`P_(T)=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 `300` sec 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 find the rate of disappearance of `N_(2)O` for the first `300sec`
`-(DeltaP_(N_(2)O))/(Deltat)=-((0.21atm-0.29atm))/(300sec)`
`=2.7xx10^(-4)atmsec^(-1)`