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For a reaction `2Ararr3B,` if the rate of formation of B is `x"mol"//L,` the rate of consumption of A isA. xB. `(3x)/(2)`C. 3xD. `(2x)/(3)` |
Answer» Correct Answer - D For reaction, `2Ararr3B` `-(1)/(2) (d[A])/(dt)=(1)/(3)(d[B])/(dt)` `(d[B])/(dt)=xx"mol"//L` `(d[A])/(dt)=-(2)/(3)rArr(d[B])/(dt)=-(2)/(3)x` Hence, the rate of consumption of `A=[(-d[A])/(dt)=(2)/(3)x]` |
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