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The decomposition of "N"(2)"O"(5)(g), i....

The decomposition of `"N"_(2)"O"_(5)(g), i.e.,"N"_(2)"O"_(5)(g)to4"NO"_(2)(g)+"O"_(2)(g)` is a first order reaction with a rate constant of `5xx10^(-4)"sec"^(-1)" at "45^(@)C`. If intial concentration of `"N"_(2)"O"_(5)` is 0.25 M, calculate its concentration after 2 min. Also calculate half life for the decomposition of `"N"_(2)"O"_(5)(g).`

Text Solution

Verified by Experts

`k=5xx10^(-4)"sec"^(-1),a=0.25" M",(a-x)=?`
`t=2" min "=120" sec"`
For a first order reaction, `k=(2.303)/(t)log""(a)/(a-x):.5xx10^(-4)=(2.303)/(120)log""(0.25)/((a-x))" or "log""(0.25)/((a-x))=0.026`
`:." "(0.25)/(a-x)="Antilog "0.026=1.062" or "(a-x)=(0.25)/(1.062)M=0.235M`
`t_(1//2)=(0.693)/(k)=(0.693)/(5xx10^(-4)"sec"^(-1))=1386" sec"=23" min "6" sec".`
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With rate constant of 5 xx 10^(-4)sec^(-1) at 45^(@) C, If initial concentration of N_(2)O_(5) is 0.25 M, Calcualte the concentration after 2 minutes. Also calculate half life for the decomposition of N_(2)O_(5) ? 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g)

The decomposition of N_(2)O_(5)(g) is a first order reaction with a rate constant of 5xx10^(-4)s^(-1) at 45^(@)C . i.e., 2N_(2)O_(5)(g) to 4NO_(2)(g)+O_(2)(g) . If initial concentration of N_(2)O_(5) is 0.25M, calculate its concentration after 2 min. Also calculate half-life for decomposition of N_(2)O_(5) (g).

Knowledge Check

  • For the first order reaction 2" N"_(2)"O"_(5)(g)to4" NO"_(2)(g)+O_(2)(g)

    A
    the concentration of the reactant decreases exponentially with time
    B
    the half-life of the reaction decreases with increasing temperature
    C
    the half-life of the reaction depends on the initial concentration of the reactant
    D
    the reaction proceeds to `99.6%` completion in eight half-life duration
  • For a reaction, 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) rate of reaction is:

    A
    `1/2 d/(dt) [N_(2)O_(5)]`
    B
    `2d/(dt)[N_(2)O_(5)]`
    C
    `1/4d/(dt)[NO_(2)]`
    D
    `4d/(dt)[NO_(2)]`
  • The decompostion of N_(2)O_(5 (g)) to NO_(3(g)) Proceeds as a first order reaction with a half-life period of 30 seconds at a certain temperature . If the initial concentration [N_(2)O_(5)] = 0.4 M , what is the rate constant of the reaction

    A
    `0.00924 sec^(-1)`
    B
    `0.0231 sec^(-1)`
    C
    `75 sec^(-1)`
    D
    `12 sec^(-1)`
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    The decomposition of N_(2)O_(5)(g) is a first order reaction with a rate constant of 5xx10^(-4)s^(-1) at 45^(@)C . i.e., 2N_(2)O_(5)(g) to 4NO_(2)(g)+O_(2)(g) . If initial concentration of N_(2)O_(5) is 0.25M, calculate its concentration after 2 minutes. Also calculate the half-life for decomposition of N_(2)O_(5)(g) .

    Using the concentration time equation for a first order reaction : The decomposition of N_(2)O_(5) to NO_(2) and O_(2) is first order, with a rate constant of 4.80xx10^(-4)//s att 45^(@)C N_(2)O_(5)(g)rarr2NO_(2)(g)+(1)/(2)O_(2)(g) (a) If the initial concentration of N_(2)O_(5) is 1.65xx10^(-2)mol//L , what is its concentration after 825 s ? (b) How long would it take for the concentration of N_(2)O_(5) to decrease to 100xx10^(-2) mol L^(-1) from its initiqal value, given in (a) ? Strategy : Since this reaction has a first order rate law, d[N_(2)O_(5)]//dt = k[N_(2)O_(5)] , we can use the corresaponding concentration time equation for a first order reaction : k = (2.303)/(t) log ([N_(2)O_(5)]_(0))/([N_(2)O_(5)]_(t)) In each part, we substitute the know quantities into this equation and solve for the unkbnown.

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