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)hArrBaO(s) + O_(2)(s)` , `Delta H = +ve`
According to the law of mass action, the rate of forward reaction `=r_(1)`
`r_(1) prop [BaO_(2)]` or `r_(1) = k_(1)[BaO_(2)]`
But concentration of solid = 1, then `r_(1) = k_(1)`
Similarly the rate of backward reaction = `r_(2)`
`r_(2) prop [BaO] [O_(2)]` or `r_(2) = k_(2) [BaO] [O_(2)]`
`.:` conc. of `[BaO] = 1`
or `r_(2) = k_(2)[O_(2)]`
At equilibrium, `r_(1) = r_(2)`
`K_(1) = K_(2)[O_(2)]` or `k_(1) = k_(2)` . `P_(O_(2))`
where, `P_(O_(2))` = partial pressure of `O_(2)`
or `(k_(1))/(k_(2)) = P_(O_(2))` (equilibrium constant)
`.: (k_(1))/(k_(2)) = k` or `k = P_(O_(2))`
So, from the above it is clear that pressure of `O_(2)` does not depend upon the concentration of reactants. The given equilibrium is an enothermic reaction. It the temperature of such reactikon is increased, then dissociation of `BaO_(2)` would increase, and more `O_(2)` is produced.