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Rate of formation `SO_(3)` in this reaction is `1.6xx10^(-3)kg//min` `2SO_(2)+O_(2)to2SO_(3)` then rate at which ` SO_(2)` reacts is :-A. `1.6xx10^(-3) kg//mi n^(-1)`B. `8xx10^(-4) kg//mi n^(-1)`C. `3.2xx10^(-3)kg mi n^(-1)`D. `1.28xx10^(-3)kg mi n^(-1)` |
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Answer» Correct Answer - D `(dSO_(3))/(dt)=1.6xx10^(-3) Kg//min` `=(1.6xx10^(-3)1000)/(80)mol//min` `(dSO_(2))/(2dt)-(-dO_(2))/(dt)=(dSO_(3))/(2dt)` `(-dSO_(2))/(dt)=(1.6)/(80)mol//min` `(1.6)/(80)xx64xx10^(-3)kg//mol` `=1.28xx10^(-3)mol//min` |
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