For the reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)`, the value of `K_(p)` is `1.7xx10^(3)` at `500K` and `1.7xx10^(4)` at `600K`. Which of the following is/are correct ?

Options

A. The proportions of `NO_(2)` in the equilibrium mixture is increased by decrease in pressure.
B. The standard enthalpy change for the forward reaction is negative
C. Units of `K_(p)` are `atm^(-1)`
D. At `500 K` the degree of dissociation of `N_(2)O_(4)` decreases by `50%` by increasing the pressure by `100%`

Answer

A

Explanation

`Deltan=2-1=1`
a. That is, with the decrease of pressure, reaction shifts towards right i.e. proportions of `NO_(2)` increases. Statement (a) is correct.
b. Value of K increases with increase of temperature and hence reaction is endothermic, i.e., `DeltaH=+ve` Hence statement (b) is incorrect.
c. `K_(p)=([p_(NO_(2))]^(2))/([p_(N_(2)O_(4))])=("atm"^(2))/("atm")="atm"`
Hence statement (c) is incorrect.
d. `{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial",1 "atm",,0),("At eq.",1-alpha,,2alpha):}`
`K_(p)=((2alpha)^(2))/(1-alpha)=1.7xx10^(3)` at `500 K`
`[(1-alpha~~1), "since" alpha "is small"]`
`4alpha^(2)=1.7xx10^(3)`
`alpha_(1)=sqrt((1.7xx10^(3))/4)`
`alpha=0.206xx10^(2)`
When the pressure was `1` atm, now let it be `2` atm `(100% "increase")`.
`{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial",2,,0),("At eq.",2-alpha,,2alpha):}`
`K_(p)=((2alpha)^(2))/(2-alpha)=1.7xx10^(3)` at `500 K`
`[2-alpha~~2, "since" alpha "is small"]`
`(4alpha^(2))/2=1.7xx10^(3)`
`alpha_(2)=sqrt((1.8xx10^(3))/2)`
`alpha_(2)/alpha_(1)=sqrt((1.7xx10^(3)xx4)/(2xx1.7xx10^(3)))=sqrt(2)=1.4`
`:.` When the pressure is increased `100%` the decrease in `alpha` is `1.4` times which is not `50%`. Hence statement is wrong.
e. `K_(p)` at `600 K=1.78xx10^(4)`
`N_(2)O_(4)(g) hArr 2NO_(2)(g)`
`Deltan=2-1=1`
Since, by decrease of pressure reaction goes forward, i.e., more of `N_(2)O_(4)` will dissociate. It means by decreasing pressure dissociation of `N_(2)O_(4)` increases.
Hence, the statement is wrong.