The reaction
`2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)`
is found to be firt order with respect to `N_(2)O_(5)(g)`. Which of the following is correct ?

Options

A. Graph between log `P_(N_(2)O_(5))` and time t is a straight line with zero slope and zero intercept
B. Graph between lig `P_(N_(2)O_(5))` and time t is a straight line with `-ve` slope and non zero intercept
C. Graph between log `P_(N_(2)O_(5))` and time t is a straight line with `+ve` slope and nonzero intercept
D. Graph between loh `P_(N_(2)O_(5))` and time t is a straight line with `-ve` slope and zero intercept

Answer

B

Explanation

Since it is a first order reaction, we can write the integrated rate law as
In`(P_(N_(2)O_(3)))_(t)=-Kt+In(P_(N_(2)O_(s)))_(0)`
where `(P_(N_(2)O_(3))_(t)` is the pressure of `N_(2)O_(5)(g)` at any time `t` and
This equation is the from `y=mx+b` , which says that `In(P_(N_(2)O_(5))_(t)` is a linear function of time `t` :
`underset(y)underset(uarr)(ln(P_(N_(2)O_(5)))_(t))=underset(m)underset(uarr)((-k))underset(x)underset(uarr)(t)+lnunderset(b)underset(uarr)((P_(N_(2)O_(5)))_(0))`
If we graph `In(P_(N_(2)O_(5))_(0)` . versus time `(t)` , we obtain a staight line having a negative slope `(m=-K)` and an intercept `b=In(P_(N_(2)O_(5)))_(0)` .