Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:

Options

A. increased mass of `BaO_2`
B. increased mass of `BaO`
C. increased temperature of equilibrium
D. increased mass of `BaO_2` and `BaO` both

Answer

C

Explanation

`BaO_2(s)overset(r_1)underset(r_2)hArrBaO(s)+O_2(g),DeltaH=+ve`
According to law of mass action, the rate of forward reaction `=r_1`
`r_1prop[BaO_2]`
or `r_1=k_1[BaO_2]`
`BaO_2` is solid substence in pure state concentration `=1m`.
then, `r_=k_1`
Similarly the rate of backward reaction `=r_2`
`r_2prop[BaO][O_2]`
or `r_2=k_2[BaO][O_2]`
`because` Concentration of solid `[BaO]=1" "[O_2(g)]`
`therefore r_2=k_2[O_2]`
At reuilibrium ,
`r_1=r_2`
`K_1=K_2[O_2]`
or `K_1=K_2.p_(o.2)`
where `(o ._2)` =partial pressure of `O_2`
or `(K_1)/(K_2)=p_(o ._2)" "("equilibrium constant" )`
`because (K_1)/(K_2)=K` or `K=p_(o.2)`