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The equilibrium constant `(K_(p))` for the reaction, `2SO_(2)+O_(2)hArr2SO_(3)` at `1000 K` is `3.5`. The partial pressure of oxygen gas to give equal mole of `SO_(2)` and `SO_(3)` is :A. `0.29 atm`B. `35 atm`C. `0.53 atm`D. `1.87 atm` |
Answer» `2SO_(2)+O_(2)hArr2SO_(3)`, Let at equilibrium, mole of `SO_(2)` and `SO_(3)` be same, then `P_(SO_(3))=P_(SO_(2))` `K_(p)=((P_(SO_(3)))^(2))/((P_(SO_(2)))^(2)*P_(O_(2)))` or `P_(O_(2))=(1)/(K_(P))=(1)/(3.5)=0.29 atm` |
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