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The rate of change in concentration of C...

The rate of change in concentration of `C` in the reaction, `2A+Brarr 2C+3D`, was reported as `1.0` mol `litre^(-1) sec^(-1)`. Calculate the reaction rate as well as rate of change of concentration of A, B and D.

Text Solution

Verified by Experts

We have,
`-(1)/(2) (d[A])/(dt) =(d[B])/(dt)`
`=(1)/(2)(d[C])/(dt)=(1)/(3) (d[D])/(dt)=` rate of reaction
`therefore" "(d[C])/(dt) =1.0" mol litre"^(-1) sec^(-1)`
`therefore" "-(d[A])/(dt) =(d[C])/(dt)=1.0" mol L"^(-1) sec^(-1)`
`-(d[B])/(dt) =(1)/(2) (d[C])/(dt) =(1)/(2) =0.5" mol L"^(-1) sec^(-1)`
`(d[D])/(dt) =(3)/(2) (d[C])/(dt) =(3)/(2)xx1=1.5" mol L"^(-1) sec^(-1)`
Also,
`?" Rate "=(1)/(2) (d[C])/(dt)`
`therefore" Rate "=(1)/(2)xx1=0.5" mol L"^(-1) sec^(-1)`
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Knowledge Check

  • For the reaction, 2A+Brarr3C , the reaction rate is equal to

    A
    `1/2 (d[A])/(dt)`
    B
    `1/3 (d[C])/(dt)`
    C
    `(d[B])/(dt)`
    D
    `(-d[A]^(2))/(dt)`
  • For the hypothetical reaction 2A rarr3C , the reaction rate r in terms of the rate of change of the concentration is given by

    A
    `r=-(d[A])/(dt)`
    B
    `r=-1/2(d[A])/(dt)`
    C
    `r=1/3(d[A])/(dt)`
    D
    `r=(d[A])/(dt)`
  • For the reaction, N_2 + 3H_2

    A
    `1.25 xx 10^(-3) mol L^(-1)s^(-1)`
    B
    `2.5 xx 10^(-3) mol L^(-1)s^(-1)`
    C
    `3.75 xx 10^(-3) mol L^(-1)s^(-1)`
    D
    `6.25 xx 10^(-5) mol L^(-1)s^(-1)`
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