Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:A. increased mass of `BaO_2`B. increased mass of `BaO`C. increased temperature of equilibriumD. increased mass of `BaO_2` and `BaO` both

Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equili

1.	Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:A. increased mass of `BaO_2`B. increased mass of `BaO`C. increased temperature of equilibriumD. increased mass of `BaO_2` and `BaO` both
Answer» Correct Answer - C `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)`

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Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:A. increased mass of `BaO_2`B. increased mass of `BaO`C. increased temperature of equilibriumD. increased mass of `BaO_2` and `BaO` both

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