`2SO_(3)hArr2SO_(2) + O_(2)`. If `K_(c) = 100`, `alpha = 1`, half of the reaction is completed, the concentration of `SO_(3)` and `SO_(2)` are equal, the concentration of `O_(2)` is

2SO_(2)+O_(2)hArr2SO_(3) , if the volume of the reaction vessel is doubled, the rate of forward reaction will be

SO_(2)+(1)/(2)O_(2)hArrSO_(3) , for the above reaction, if 'a' and 'b' mole/1 are the initial concentrations of SO_(2) and O_(2) respectively and x mole SO_(3) is formed at equlibrium. The equilibrium concentration of O_(2) will be

SO_(2)+(1)/(2)O_(2)hArrSO_(3) , for the above reaction, if 'a' and 'b' mole/1 are the initial concentrations of SO_(2) and O_(2) respectively and x mole SO_(3) is formed at equlibrium. The equilibrium concentration of O_(2) will be

K_(C) for 2 SO_(2) + O_(2) hArr 2 SO_(3) in a 10 lit flask at certain T is 100 lit - mol^(-1) . Now , if equilibrium pressures of SO_2 and SO_3 are equal , then mass of O_(2) present at equilibrium is

Write the expressions of K_(c) and K_(p) for the reaction: 2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g) .

The relation between K_(p) and K_(c) for the following reaction : 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) is -

For a reaction, 2SO_(2(g))+O_(2(g))hArr2SO_(3(g)) , 1.5 moles of SO_(2) and 1 mole of O_(2) are taken in a 2 L vessel. At equilibrium the concentration of SO_(3) was found to be 0.35 mol L^(-1) The K_(c) for the reaction would be

For a reaction, 2SO_(2(g))+O_(2(g))hArr2SO_(3(g)) , 1.5 moles of SO_(2) and 1 mole of O_(2) are taken in a 2 L vessel. At equilibrium the concentration of SO_(3) was found to be 0.35 mol L^(-1) The K_(c) for the reaction would be