At 250° C, the time for half-change for the thermal dissociation of N2O5 is 5.7 hours. The specific rate constant is given by:

Question

At 250° C, the time for half-change for the thermal dissociation of N2O5 is 5.7 hours. The specific rate constant is given by: (A)  (1/5.7) texthou textr-1  (B)  ( ln 2/5.7) texthou textr-1  (C)  ln (2/5.7) texthou textr-1  (D)  0.5 texthou textr-1

Tardigrade Admin · Accepted Answer

The thermal dissociation of N2O5 is a first order reaction. Hence, for it k=(1/t) ln (a/(a-x)) For half-change, k=(1/t1/2) ln (a/( a-(a)2 )) =(1/t1/2)⋅ ln 2 ∴ k=( ln 2/5.7) texthou textr-1

Q. At $250^{\circ} C,$ the time for half-change for the thermal dissociation of $N_{2} O_{5}$ is 5.7 hours. The specific rate constant is given by:

$\frac{1}{5.7} hou r^{- 1}$

$\frac{l n 2}{5.7} hou r^{- 1}$

$ln \frac{2}{5.7} hou r^{- 1}$

$0.5 hou r^{- 1}$

The thermal dissociation of $N_{2} O_{5}$ is a first order reaction. Hence, for it $k = \frac{1}{t} ln \frac{a}{( a - x )}$ For half-change, $k = \frac{1}{t _{1/2}} ln \frac{a}{( a - \frac{a}{2} )}$ $= \frac{1}{t _{1/2}} \cdot ln 2$ $∴$ $k = \frac{l n 2}{5.7} hou r^{- 1}$

Q. At 250∘C, the time for half-change for the thermal dissociation of N2​O5​ is 5.7 hours. The specific rate constant is given by:

5.71​hour−1

5.7ln2​hour−1

ln5.72​hour−1

0.5hour−1