1.

For a reaction 1//2 A rarr 2B rate of disappearance of A is related to rate of appearance of B by the expression

Answer»

`(-d[A])/(dt)=4(d[B])/(dt)`
`(-d[A])/(dt) = (1)/(4)(d[B])/(dt)`
`(-d[A])/(dt)=1/2(d[B])/(dt)`
`(-d[A])/(dt)=(d[B])/(dt)`

Solution :For the given CHEMICAL EQUATION, we have
`(1)/(V_A)(d[A])/(dt) = (1)/(V_B)(d[B])/(dt)`
i.e `-(1)/(1//2)(d[A])/(dt) = (1)/(2)(d[B])/(dt)`
HENCE, `-(d[A])/(dt) = 1/4 (d[B])/(dt)`


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