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In a reaction, `2Ato "products",` the concentration of A decreases from 0.5 mol `L^(-1) to 0.4 mol L^(-1)` in 10 min. The rate during this interval isA. `0.05 "mol"L^(-1)"min"^(-1)`B. `0.42 """mol"" "L^(-1)"min"^(-1)`C. `0.005 """mol"" "L^(-1)"min"^(-1)`D. `0.5 """mol"" "L^(-1)"min"^(-1)` |
Answer» Correct Answer - C Rate of reaction =`-(1)/(2)(d[A])/(dt)` `=-(d[A])/(dt)=-([A]_(2)-[A]_(1))/(t_(2)-t_(1))=-(0.4-0.5)/(10-0)=(0.1)/(10)` `0.01 "mol"L^(-1) "min"^(-1)` `therefore Rate of reaction =(1)/(2)xx0.01` `=0.005 "mol" L^(-1) "min"^(-1)` |
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